Exercise Chapter 9 - Question 1

Question: Draw a quadrilateral in the Cartesian plane, whose vertices are $(-4, 5)$, $(0, 7)$, $(5, -5)$ and $(-4, -2)$. Also, find its area.

Solution:

Let the given vertices of the quadrilateral be $A(-4, 5)$, $B(0, 7)$, $C(5, -5)$, and $D(-4, -2)$.

First, let us visualize the quadrilateral in the Cartesian plane:

To find the area of the quadrilateral $ABCD$, we can split it into two triangles by drawing the diagonal $AC$. Thus,

$$\text{Area of quadrilateral } ABCD = \text{Area}(\triangle ABC) + \text{Area}(\triangle ADC)$$

Step 1: Find the area of $\triangle ABC$ with vertices $A(-4, 5)$, $B(0, 7)$, and $C(5, -5)$

The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:

$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$

Substituting the coordinates of $A(-4, 5)$, $B(0, 7)$, $C(5, -5)$:

$$\text{Area}(\triangle ABC) = \frac{1}{2} |(-4)(7 - (-5)) + 0((-5) - 5) + 5(5 - 7)|$$ $$\text{Area}(\triangle ABC) = \frac{1}{2} |(-4)(12) + 0 + 5(-2)|$$ $$\text{Area}(\triangle ABC) = \frac{1}{2} |-48 - 10| = \frac{1}{2} |-58| = 29 \text{ square units}$$

Step 2: Find the area of $\triangle ADC$ with vertices $A(-4, 5)$, $D(-4, -2)$, and $C(5, -5)$

Substituting the coordinates into the area formula:

$$\text{Area}(\triangle ADC) = \frac{1}{2} |(-4)((-2) - (-5)) + (-4)((-5) - 5) + 5(5 - (-2))|$$ $$\text{Area}(\triangle ADC) = \frac{1}{2} |(-4)(3) + (-4)(-10) + 5(7)|$$ $$\text{Area}(\triangle ADC) = \frac{1}{2} |-12 + 40 + 35|$$ $$\text{Area}(\triangle ADC) = \frac{1}{2} |63| = 31.5 \text{ square units}$$

Step 3: Total Area of the Quadrilateral

$$\text{Area of quadrilateral } ABCD = 29 + 31.5 = 60.5 \text{ square units}$$

Exercise Chapter 9 - Question 2

Question: The base of an equilateral triangle with side $2a$ lies along the $y$-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Solution:

Let $\triangle ABC$ be the equilateral triangle with side length $2a$.

According to the problem, the base lies along the $y$-axis and its mid-point is at the origin $(0, 0)$.

Since the length of the side is $2a$ and the mid-point of the base is at the origin, the endpoints of the base (let's call them $B$ and $C$) are at a distance of $a$ from the origin along the $y$-axis.

Therefore, the coordinates of vertices $B$ and $C$ are:

$$B = (0, a) \quad \text{and} \quad C = (0, -a)$$

(or vice versa).

Now, let the third vertex be $A(x, y)$. Since the triangle is equilateral and the base lies on the $y$-axis, the vertex $A$ must lie on the $x$-axis (perpendicular bisector of the base). Hence, the $y$-coordinate of $A$ is $0$, so $A = (x, 0)$.

The distance from vertex $A(x, 0)$ to either $B(0, a)$ or $C(0, -a)$ must equal the side length $2a$:

$$(AB)^2 = (2a)^2$$ $$(x - 0)^2 + (0 - a)^2 = 4a^2$$ $$x^2 + a^2 = 4a^2$$ $$x^2 = 3a^2 \implies x = \pm a\sqrt{3}$$

Thus, the third vertex $A$ can be $(\pm a\sqrt{3}, 0)$.

Depending on whether $A$ lies on the positive or negative $x$-axis, we have two possible sets of vertices for the equilateral triangle:

1. $(a\sqrt{3}, 0), (0, a), (0, -a)$
2. $(-a\sqrt{3}, 0), (0, a), (0, -a)$

Exercise Chapter 9 - Question 3

Question: Find the distance between $P (x_1, y_1)$ and $Q (x_2, y_2)$ when : (i) $PQ$ is parallel to the $y$-axis, (ii) $PQ$ is parallel to the $x$-axis.

Solution:

The general distance formula between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by:

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

(i) When $PQ$ is parallel to the $y$-axis:

If the line segment $PQ$ is parallel to the $y$-axis, the $x$-coordinates of both points must be equal. That is,

$$x_1 = x_2$$

Substituting $x_1 = x_2$ into the distance formula:

$$d = \sqrt{(x_1 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{0 + (y_2 - y_1)^2} = |y_2 - y_1|$$

(ii) When $PQ$ is parallel to the $x$-axis:

If the line segment $PQ$ is parallel to the $x$-axis, the $y$-coordinates of both points must be equal. That is,

$$y_1 = y_2$$

Substituting $y_1 = y_2$ into the distance formula:

$$d = \sqrt{(x_2 - x_1)^2 + (y_1 - y_1)^2} = \sqrt{(x_2 - x_1)^2 + 0} = |x_2 - x_1|$$

Exercise Chapter 9 - Question 4

Question: Find a point on the $x$-axis, which is equidistant from the points $(7, 6)$ and $(3, 4)$.

Solution:

Let the required point on the $x$-axis be $P(x, 0)$. Since the point lies on the $x$-axis, its $y$-coordinate is $0$.

Let the given points be $A(7, 6)$ and $B(3, 4)$. According to the problem, point $P$ is equidistant from $A$ and $B$, which means:

$$PA = PB \implies PA^2 = PB^2$$

Using the distance formula, we write:

$$(x - 7)^2 + (0 - 6)^2 = (x - 3)^2 + (0 - 4)^2$$

Expanding both sides:

$$x^2 - 14x + 49 + 36 = x^2 - 6x + 9 + 16$$

Subtracting $x^2$ from both sides and simplifying:

$$-14x + 85 = -6x + 25$$

Rearranging the terms:

$$85 - 25 = 14x - 6x$$ $$60 = 8x$$ $$x = \frac{60}{8} = \frac{15}{2} = 7.5$$

Thus, the required point on the $x$-axis is $\left(\frac{15}{2}, 0\right)$.

Exercise - Question 5

Question: Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points $P(0, -4)$ and $B(8, 0)$.

Solution:

First, find the mid-point of the line segment joining $P(0, -4)$ and $B(8, 0)$. Let the mid-point be $M(x_m, y_m)$:

$$x_m = \frac{0 + 8}{2} = \frac{8}{2} = 4$$ $$y_m = \frac{-4 + 0}{2} = \frac{-4}{2} = -2$$

So, the mid-point $M$ is $(4, -2)$.

The line passes through the origin $O(0, 0)$ and the point $M(4, -2)$. The slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

Substituting $O(0,0)$ and $M(4,-2)$:

$$m = \frac{-2 - 0}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}$$

Thus, the slope of the line is $-\frac{1}{2}$.

Exercise - Question 6

Question: Without using the Pythagoras theorem, show that the points $(4, 4)$, $(3, 5)$ and $(-1, -1)$ are the vertices of a right angled triangle.

Solution:

Let the given points be $A(4, 4)$, $B(3, 5)$, and $C(-1, -1)$. We will find the slopes of the sides $AB$, $BC$, and $AC$.

1. Slope of side $AB$ ($m_1$):

$$m_1 = \frac{5 - 4}{3 - 4} = \frac{1}{-1} = -1$$

2. Slope of side $BC$ ($m_2$):

$$m_2 = \frac{-1 - 5}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}$$

3. Slope of side $AC$ ($m_3$):

$$m_3 = \frac{-1 - 4}{-1 - 4} = \frac{-5}{-5} = 1$$

Notice the product of the slopes of $AB$ and $AC$:

$$m_1 \times m_3 = (-1) \times 1 = -1$$

Since the product of their slopes is $-1$, the line segments $AB$ and $AC$ are perpendicular to each other ($AB \perp AC$). Therefore, $\angle A = 90^\circ$, which shows that $\triangle ABC$ is a right-angled triangle without using the Pythagoras theorem.

Exercise - Question 7

Question: Find the slope of the line, which makes an angle of $30^\circ$ with the positive direction of $y$-axis measured anticlockwise.

Solution:

The slope of a line is defined as $\tan \theta$, where $\theta$ is the inclination of the line measured in the anticlockwise direction from the positive direction of the $x$-axis.

The problem states that the line makes an angle of $30^\circ$ with the positive direction of the $y$-axis measured anticlockwise.

Since the positive $y$-axis is at an angle of $90^\circ$ with the positive $x$-axis, the total angle $\theta$ with the positive $x$-axis in the anticlockwise direction is:

$$\theta = 90^\circ + 30^\circ = 120^\circ$$

Thus, the slope $m$ of the line is:

$$m = \tan(120^\circ) = \tan(180^\circ - 60^\circ) = -\tan(60^\circ) = -\sqrt{3}$$

Therefore, the slope of the line is $-\sqrt{3}$.

Exercise - Question 8

Question: Without using distance formula, show that points $(-2, -1)$, $(4, 0)$, $(3, 3)$ and $(-3, 2)$ are the vertices of a parallelogram.

Solution:
Let the given points be $A(-2, -1)$, $B(4, 0)$, $C(3, 3)$, and $D(-3, 2)$.
To show that $ABCD$ is a parallelogram without using the distance formula, we need to prove that the opposite sides are parallel, which is equivalent to showing that their slopes are equal.

1. Slope of side $AB$:
Points are $A(-2, -1)$ and $B(4, 0)$.
$$m_{AB} = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{4 + 2} = \frac{1}{6}$$

2. Slope of side $CD$:
Points are $C(3, 3)$ and $D(-3, 2)$.
$$m_{CD} = \frac{2 - 3}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6}$$
Since $m_{AB} = m_{CD}$, the line segment $AB$ is parallel to $CD$ ($AB \parallel CD$).

3. Slope of side $BC$:
Points are $B(4, 0)$ and $C(3, 3)$.
$$m_{BC} = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$$

4. Slope of side $DA$:
Points are $D(-3, 2)$ and $A(-2, -1)$.
$$m_{DA} = \sfrac{-1 - 2}{-2 - (-3)} = \frac{-3}{-2 + 3} = \frac{-3}{1} = -3$$
Since $m_{BC} = m_{DA}$, the line segment $BC$ is parallel to $DA$ ($BC \parallel DA$).

Since both pairs of opposite sides are parallel ($AB \parallel CD$ and $BC \parallel DA$), the quadrilateral $ABCD$ is a parallelogram.

Exercise Chapter 9 - Question 9

Question: Find the angle between the $x$-axis and the line joining the points $(3, -1)$ and $(4, -2)$.

Solution:
Let the given points be $P(3, -1)$ and $Q(4, -2)$.
The slope ($m$) of the line passing through $P(x_1, y_1)$ and $Q(x_2, y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substitute the given coordinates: $$m = \frac{-2 - (-1)}{4 - 3} = \frac{-2 + 1}{1} = -1$$ We know that the slope of a line is equal to the tangent of the angle $\theta$ it makes with the positive direction of the $x$-axis ($m = \tan \theta$).
$$\tan \theta = -1$$ Since $\tan \theta = -1$, and $\theta$ lies in $[0, \pi]$, we have: $$\theta = 180^\circ - 45^\circ = 135^\circ \quad \text{(or } \frac{3\pi}{4} \text{ radians)}$$ Thus, the angle between the $x$-axis and the line is $135^\circ$.

Exercise Chapter 9 - Question 10

Question: The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{1}{3}$, find the slopes of the lines.

Solution:
Let the slope of the first line be $m_1 = m$.
According to the question, the slope of the second line is $m_2 = 2m$.
The acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula: $$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$$ Given that $\tan \theta = \frac{1}{3}$, we substitute $m_1 = m$ and $m_2 = 2m$: $$\frac{1}{3} = \left| \frac{2m - m}{1 + m(2m)} \right|$$ $$\frac{1}{3} = \left| \frac{m}{1 + 2m^2} \right|$$ Removing the absolute value gives two cases:

Case 1: $$\frac{m}{1 + 2m^2} = \frac{1}{3}$$ $$3m = 1 + 2m^2$$ $$2m^2 - 3m + 1 = 0$$ Factoring the quadratic equation: $$(2m - 1)(m - 1) = 0$$ This gives: $$m = \frac{1}{2} \quad \text{or} \quad m = 1$$ If $m = \frac{1}{2}$, the slopes are $m_1 = \frac{1}{2}$ and $m_2 = 2\left(\frac{1}{2}\right) = 1$.
If $m = 1$, the slopes are $m_1 = 1$ and $m_2 = 2(1) = 2$.

Case 2: $$\frac{m}{1 + 2m^2} = -\frac{1}{3}$$ $$3m = -1 - 2m^2$$ $$2m^2 + 3m + 1 = 0$$ Factoring the quadratic equation: $$(2m + 1)(m + 1) = 0$$ This gives: $$m = -\frac{1}{2} \quad \text{or} \quad m = -1$$ If $m = -\frac{1}{2}$, the slopes are $m_1 = -\frac{1}{2}$ and $m_2 = -1$.
If $m = -1$, the slopes are $m_1 = -1$ and $m_2 = -2$.

Thus, the possible pairs of slopes of the lines are: 1. $\left(\frac{1}{2}, 1\right)$ 2. $(1, 2)$ 3. $\left(-\frac{1}{2}, -1\right)$ 4. $(-1, -2)$

Exercise Chapter 9 - Question 11

Question: A line passes through $(x_1, y_1)$ and $(h, k)$. If slope of the line is $m$, show that $k - y_1 = m (h - x_1)$.

Solution:
We are given: - A point on the line: $(x_1, y_1)$ - Another point on the line: $(h, k)$ - The slope of the line: $m$

The formula for the slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substituting our specific points $(x_1, y_1)$ and $(h, k)$ into the slope formula: $$m = \frac{k - y_1}{h - x_1}$$ Now, multiply both sides of the equation by $(h - x_1)$ (assuming $h \neq x_1$): $$m(h - x_1) = k - y_1$$ Rearranging the terms: $$k - y_1 = m(h - x_1)$$ Hence proved.

Exercise Chapter 9 - Question 1

Question: Write the equations for the $x$- and $y$-axes.

Solution:

(i) For the $x$-axis, the $y$-coordinate of every point lying on it is $0$. Therefore, the equation of the $x$-axis is given by: $$y = 0$$

(ii) For the $y$-axis, the $x$-coordinate of every point lying on it is $0$. Therefore, the equation of the $y$-axis is given by: $$x = 0$$

Exercise Chapter 9 - Question 2

Question: Find the equation of the line passing through the point $(-4, 3)$ with slope $\frac{1}{2}$.

Solution:

We use the point-slope form of the equation of a line, which is given by: $$y - y_1 = m(x - x_1)$$

Here, the given point is $(x_1, y_1) = (-4, 3)$ and the slope is $m = \frac{1}{2}$. Substituting these values into the formula: $$y - 3 = \frac{1}{2}(x - (-4))$$ $$y - 3 = \frac{1}{2}(x + 4)$$

Multiplying both sides by $2$ to clear the fraction: $$2(y - 3) = x + 4$$ $$2y - 6 = x + 4$$ Rearranging the terms into standard form ($Ax + By + C = 0$): $$x - 2y + 4 + 6 = 0$$ $$x - 2y + 10 = 0$$

Answer: The equation of the line is $x - 2y + 10 = 0$.

Exercise Chapter 9 - Question 3

Question: Find the equation of the line passing through $(0, 0)$ with slope $m$.

Solution:

We use the point-slope form of the equation of a line passing through a point $(x_1, y_1)$ with slope $m$: $$y - y_1 = m(x - x_1)$$

Substituting the given point $(x_1, y_1) = (0, 0)$ and slope $m$: $$y - 0 = m(x - 0)$$ $$y = mx$$

Answer: The equation of the line is $y = mx$.

Exercise Chapter 9 - Question 4

Question: Find the equation of the line passing through $(2, 2\sqrt{3})$ and inclined with the $x$-axis at an angle of $75^\circ$.

Solution:

The slope $m$ of a line making an angle $\theta$ with the positive direction of the $x$-axis is given by $m = \tan\theta$.
Here, $\theta = 75^\circ$, so: $$m = \tan(75^\circ) = \tan(45^\circ + 30^\circ)$$

Using the trigonometric identity $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$: $$m = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - (1)\left(\frac{1}{\sqrt{3}}\right)} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

Rationalizing the denominator: $$m = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

Now, using the point-slope form for the point $(x_1, y_1) = (2, 2\sqrt{3})$ and slope $m = 2 + \sqrt{3}$: $$y - y_1 = m(x - x_1)$$ $$y - 2\sqrt{3} = (2 + \sqrt{3})(x - 2)$$ $$y - 2\sqrt{3} = (2 + \sqrt{3})x - 2(2 + \sqrt{3})$$ $$y - 2\sqrt{3} = (2 + \sqrt{3})x - 4 - 2\sqrt{3}$$

Simplifying and rearranging terms: $$(2 + \sqrt{3})x - y - 4 = 0$$

Answer: The equation of the line is $(2 + \sqrt{3})x - y - 4 = 0$.

Exercise Chapter 9 - Question 5

Question: Find the equation of the line intersecting the $x$-axis at a distance of $3$ units to the left of origin with slope $-2$.

Solution:

1. Since the line intersects the $x$-axis at a distance of $3$ units to the left of the origin, its $x$-intercept is $-3$. Thus, the line passes through the point $(-3, 0)$.
2. We are given the slope of the line, $m = -2$.
3. Using the point-slope form of the equation of a line, given by $y - y_1 = m(x - x_1)$, we substitute the point $(-3, 0)$ and $m = -2$: $$y - 0 = -2(x - (-3))$$ $$y = -2(x + 3)$$ $$y = -2x - 6$$ $$2x + y + 6 = 0$$

Answer: The equation of the line is $2x + y + 6 = 0$.

Exercise - Question 6

Question: Find the equation of the line intersecting the $y$-axis at a distance of $2$ units above the origin and making an angle of $30^\circ$ with positive direction of the $x$-axis.

Solution:

1. Since the line intersects the $y$-axis at $2$ units above the origin, the $y$-intercept is $c = 2$. Thus, the line passes through the point $(0, 2)$.
2. The angle made with the positive direction of the $x$-axis is $\theta = 30^\circ$. Therefore, the slope of the line is: $$m = \tan 30^\circ = \frac{1}{\sqrt{3}}$$
3. Using the slope-intercept form $y = mx + c$: $$y = \left(\frac{1}{\sqrt{3}}\right)x + 2$$
4. Multiplying the entire equation by $\sqrt{3}$ to clear the denominator: $$\sqrt{3}y = x + 2\sqrt{3} \implies x - \sqrt{3}y + 2\sqrt{3} = 0$$

Answer: The equation of the line is $x - \sqrt{3}y + 2\sqrt{3} = 0$.

Exercise - Question 7

Question: Find the equation of the line passing through the points $(-1, 1)$ and $(2, -4)$.

Solution:

1. Let the given points be $(x_1, y_1) = (-1, 1)$ and $(x_2, y_2) = (2, -4)$.
2. The slope $m$ of the line passing through two points is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 1}{2 - (-1)} = \frac{-5}{2 + 1} = \frac{-5}{3}$$
3. Using the point-slope form with point $(-1, 1)$ and slope $m = -\frac{5}{3}$: $$y - 1 = -\frac{5}{3}(x - (-1))$$ $$3(y - 1) = -5(x + 1)$$ $$3y - 3 = -5x - 5$$ $$5x + 3y - 3 + 5 = 0$$ $$5x + 3y + 2 = 0$$

Answer: The equation of the line is $5x + 3y + 2 = 0$.

Exercise - Question 8

Question: The vertices of $\Delta PQR$ are $P(2, 1)$, $Q(-2, 3)$ and $R(4, 5)$. Find equation of the median through the vertex $R$.

Solution:

1. Let the median through vertex $R(4, 5)$ meet the side $PQ$ at point $S$.
2. Since $S$ is the midpoint of $PQ$, we use the midpoint formula to find the coordinates of $S$: $$S = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) = \left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) = \left(\frac{0}{2}, \frac{4}{2}\right) = (0, 2)$$
3. Now, we need to find the equation of the median $RS$ passing through points $R(4, 5)$ and $S(0, 2)$.
4. The slope of $RS$ is: $$m = \frac{2 - 5}{0 - 4} = \frac{-3}{-4} = \frac{3}{4}$$
5. Using the slope-intercept form with $y$-intercept $c = 2$ (since $S(0, 2)$ is on the $y$-axis): $$y = \frac{3}{4}x + 2$$ $$4y = 3x + 8$$ $$3x - 4y + 8 = 0$$

Answer: The equation of the median through $R$ is $3x - 4y + 8 = 0$.

Exercise - Question 9

Question: Find the equation of the line passing through $(-3, 5)$ and perpendicular to the line through the points $(2, 5)$ and $(-3, 6)$.

Solution:

1. First, let us find the slope ($m_1$) of the line passing through the points $(x_1, y_1) = (2, 5)$ and $(x_2, y_2) = (-3, 6)$.
2. The slope formula is given by: $$m_1 = \frac{y_2 - y_1}{x_2 - x_1}$$
3. Substituting the given coordinates: $$m_1 = \frac{6 - 5}{-3 - 2} = \frac{1}{-5} = -\frac{1}{5}$$
4. Since the required line is perpendicular to this line, its slope ($m_2$) must satisfy the condition for perpendicular lines, which is $m_1 \cdot m_2 = -1$. Therefore: $$m_2 = -\frac{1}{m_1} = -\frac{1}{-1/5} = 5$$
5. Now, using the point-slope form with point $(-3, 5)$ and slope $m_2 = 5$: $$y - 5 = 5(x - (-3))$$ $$y - 5 = 5(x + 3)$$ $$y - 5 = 5x + 15$$ $$5x - y + 20 = 0$$

Answer: The equation of the line is $5x - y + 20 = 0$.

Exercise Chapter 9 - Question 10

Question: A line perpendicular to the line segment joining the points $(1, 0)$ and $(2, 3)$ divides it in the ratio $1 : n$. Find the equation of the line.

Solution:

Let the given points be $A(1, 0)$ and $B(2, 3)$.

First, let's find the coordinates of the point $P(x, y)$ that divides the line segment $AB$ in the ratio $1 : n$ internally using the section formula:

$$x = \frac{m_2x_1 + m_1x_2}{m_1 + m_2}, \quad y = \frac{m_2y_1 + m_1y_2}{m_1 + m_2}$$

Here, $m_1 : m_2 = 1 : n$, $(x_1, y_1) = (1, 0)$, and $(x_2, y_2) = (2, 3)$. Thus:

$$x = \frac{n(1) + 1(2)}{1 + n} = \frac{n + 2}{n + 1}$$

$$y = \frac{n(0) + 1(3)}{1 + n} = \frac{3}{n + 1}$$

So, the point of division is $P\left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)$.

Next, let's find the slope of the line segment $AB$ ($m_1$):

$$m_1 = \frac{3 - 0}{2 - 1} = \frac{3}{1} = 3$$

Since the required line is perpendicular to $AB$, the slope of the required line ($m$) is:

$$m = -\frac{1}{3}$$

Now, using the point-slope form for the line passing through $P\left(\frac{n+2}{n+1}, \frac{3}{n+1}\right)$ with slope $m = -\frac{1}{3}$:

$$y - \frac{3}{n + 1} = -\frac{1}{3}\left(x - \frac{n + 2}{n + 1}\right)$$

Multiplying the entire equation by $3(n + 1)$ to clear the denominators:

$$3(n + 1)y - 9 = - (n + 1) \left[ \frac{3(n + 1)x - 3(n + 2)}{3(n + 1)} \right]$$

$$3(n + 1)y - 9 = -[x(n + 1) - (n + 2)]$$

$$3(n + 1)y - 9 = -(n + 1)x + n + 2$$

Rearranging into standard form: $$(n + 1)x + 3(n + 1)y - n - 11 = 0$$

Answer: The equation of the line is $(n + 1)x + 3(n + 1)y - (n + 11) = 0$.

Exercise Chapter 9 - Question 11

Question: Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point $(2, 3)$.

Solution:

The intercept form of the equation of a line is given by: $$\frac{x}{a} + \frac{y}{b} = 1$$ where $a$ and $b$ are the intercepts made by the line on the x-axis and y-axis respectively.

Since the line cuts off equal intercepts on the coordinate axes, we have $a = b$.

Thus, the equation becomes: $$\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a$$

Given that the line passes through the point $(2, 3)$, we substitute $x = 2$ and $y = 3$ into the equation to find $a$:

$$2 + 3 = a \implies a = 5$$

Substitute $a = 5$ back into the line equation: $$x + y = 5$$

Answer: The equation of the line is $x + y = 5$.

Exercise Chapter 9 - Question 12

Question: Find equation of the line passing through the point $(2, 2)$ and cutting off intercepts on the axes whose sum is $9$.

Solution:

Let the x-intercept be $a$ and the y-intercept be $b$. According to the problem, the sum of the intercepts is $9$: $$a + b = 9 \implies b = 9 - a$$

The intercept form of the equation of the line is: $$\frac{x}{a} + \frac{y}{b} = 1$$

Substituting $b = 9 - a$: $$\frac{x}{a} + \frac{y}{9 - a} = 1$$

Since the line passes through the point $(2, 2)$, we substitute $x = 2$ and $y = 2$ into the equation: $$\frac{2}{a} + \frac{2}{9 - a} = 1$$

Taking the common denominator and simplifying: $$\frac{2(9 - a) + 2a}{a(9 - a)} = 1$$ $$\frac{18 - 2a + 2a}{9a - a^2} = 1$$ $$\frac{18}{9a - a^2} = 1$$

Cross-multiplying gives: $$18 = 9a - a^2$$ $$a^2 - 9a + 18 = 0$$

Factoring the quadratic equation: $$(a - 3)(a - 6) = 0$$ This gives two possible values for $a$: $$a = 3 \quad \text{or} \quad a = 6$$

Case 1: When $a = 3$, $b = 9 - 3 = 6$.
The equation of the line is: $$\frac{x}{3} + \frac{y}{6} = 1 \implies 2x + y = 6$$

Case 2: When $a = 6$, $b = 9 - 6 = 3$.
The equation of the line is: $$\frac{x}{6} + \frac{y}{3} = 1 \implies x + 2y = 6$$

Answer: The equations of the lines are $2x + y = 6$ and $x + 2y = 6$.

Exercise Chapter 9 - Question 13

Question: Find equation of the line through the point $(0, 2)$ making an angle $\frac{2\pi}{3}$ with the positive $x$-axis. Also, find the equation of line parallel to it and crossing the $y$-axis at a distance of $2$ units below the origin.

Solution:

Step 1: Find the slope of the given line.
The angle made by the line with the positive direction of the $x$-axis is given as $\theta = \frac{2\pi}{3}$.
The slope $m$ of the line is given by: $$m = \tan(\theta) = \tan\left(\frac{2\pi}{3}\right) = \tan\left(\pi - \frac{\pi}{3}\right) = -\tan\left(\frac{\pi}{3}\right) = -\sqrt{3}$$

Step 2: Find the equation of the first line.
The line passes through the point $(x_1, y_1) = (0, 2)$ and has slope $m = -\sqrt{3}$. Using the point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 2 = -\sqrt{3}(x - 0)$$ $$y - 2 = -\sqrt{3}x \implies \sqrt{3}x + y - 2 = 0$$

Step 3: Find the equation of the parallel line.
Since the required line is parallel to the first line, its slope will be the same, i.e., $m = -\sqrt{3}$.
It crosses the $y$-axis at a distance of $2$ units below the origin, which means its $y$-intercept $c$ is $-2$, so it passes through $(0, -2)$.
Using slope-intercept form $y = mx + c$: $$y = -\sqrt{3}x - 2 \implies \sqrt{3}x + y + 2 = 0$$

Answer: The equations of the lines are $\sqrt{3}x + y - 2 = 0$ and $\sqrt{3}x + y + 2 = 0$.

Exercise - Chapter 9 - Question 14

Question: The perpendicular from the origin to a line meets it at the point $(-2, 9)$, find the equation of the line.

Solution:

Step 1: Find the slope of the perpendicular segment.
Let the origin be $O(0, 0)$ and the given point be $P(-2, 9)$.
The slope of the line segment $OP$ joining $(0, 0)$ and $(-2, 9)$ is: $$m_1 = \frac{9 - 0}{-2 - 0} = -\frac{9}{2}$$

Step 2: Find the slope of the required line.
Since the line is perpendicular to the segment $OP$, its slope $m$ is the negative reciprocal of $m_1$: $$m = -\frac{1}{m_1} = -\frac{1}{-\frac{9}{2}} = \frac{2}{9}$$

Step 3: Find the equation of the line.
The line passes through the point $P(-2, 9)$ and has slope $m = \frac{2}{9}$. Using the point-slope form: $$y - 9 = \frac{2}{9}(x - (-2))$$ $$9(y - 9) = 2(x + 2)$$ $$9y - 81 = 2x + 4$$ $$2x - 9y + 85 = 0$$

Exercise - Chapter 9 - Question 15

Question: The length $L$ (in centimetre) of a copper rod is a linear function of its Celsius temperature $C$. In an experiment, if $L = 124.942$ when $C = 20$ and $L = 125.134$ when $C = 110$, express $L$ in terms of $C$.

Solution:

Step 1: Formulate the points representing the linear relationship.
Since $L$ is a linear function of $C$, the relationship can be represented in the coordinate plane as a straight line passing through the points $(C_1, L_1) = (20, 124.942)$ and $(C_2, L_2) = (110, 125.134)$, where $C$ acts as the independent variable ($x$-axis) and $L$ as the dependent variable ($y$-axis).

Step 2: Find the slope of the line.
$$m = \frac{L_2 - L_1}{C_2 - C_1} = \frac{125.134 - 124.942}{110 - 20} = \frac{0.192}{90} = \frac{192}{90000} = \frac{8}{3750} = 0.0021333$$

Step 3: Write the equation using the point-slope form.
Using point $(20, 124.942)$: $$L - 124.942 = \frac{8}{3750}(C - 20)$$ $$L = \frac{8}{3750}C - \frac{160}{3750} + 124.942$$ $$L = 0.002133 C - 0.04267 + 124.942$$ $$L = 0.002133 C + 124.89933$$ Or expressed with fractions: $$L = \frac{8}{3750}C + 124.89933$$

Exercise - Chapter 9 - Question 16

Question: The owner of a milk store finds that, he can sell $980$ litres of milk each week at Rs $14$/litre and $1220$ litres of milk each week at Rs $16$/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs $17$/litre?

Solution:

Step 1: Set up the coordinate points.
Let $p$ be the selling price per litre (in Rs) and $y$ be the demand (in litres). We are given a linear relationship between $p$ and $y$.
The two points given are $(p_1, y_1) = (14, 980)$ and $(p_2, y_2) = (16, 1220)$.

Step 2: Find the equation of the line.
First, calculate the slope $m$ of the line passing through $(14, 980)$ and $(16, 1220)$: $$m = \frac{y_2 - y_1}{p_2 - p_1} = \frac{1220 - 980}{16 - 14} = \frac{240}{2} = 120$$ Now, using the point-slope form with point $(14, 980)$: $$y - y_1 = m(p - p_1)$$ $$y - 980 = 120(p - 14)$$ $$y - 980 = 120p - 1680$$ $$y = 120p - 1680 + 980$$ $$y = 120p - 700$$

Step 3: Calculate the demand at $p = 17$.
Substitute $p = 17$ into the linear equation: $$y = 120(17) - 700$$ $$y = 2040 - 700 = 1340$$

Answer: He could sell 1340 litres of milk weekly at Rs $17$/litre.

Miscellaneous Exercise - Chapter 9 - Question 17

Question: $P (a, b)$ is the mid-point of a line segment between axes. Show that equation of the line is $\frac{x}{a} + \frac{y}{b} = 2$.

Solution:

Let the line meet the $x$-axis at point $A(A_x, 0)$ and the $y$-axis at point $B(0, B_y)$.

Since $P(a, b)$ is given as the mid-point of the line segment $AB$, using the mid-point formula:

$$a = \frac{A_x + 0}{2} \implies A_x = 2a$$ $$b = \frac{0 + B_y}{2} \implies B_y = 2b$$

Thus, the $x$-intercept is $2a$ and the $y$-intercept is $2b$.

Using the intercept form of the equation of a line, $\frac{x}{A_x} + \frac{y}{B_y} = 1$:

$$\frac{x}{2a} + \frac{y}{2b} = 1$$ $$\implies \frac{x}{a} + \frac{y}{b} = 2$$

Hence proved.

Miscellaneous Exercise - Question 18

Question: Point $R (h, k)$ divides a line segment between the axes in the ratio $1 : 2$. Find equation of the line.

Solution:

Let the line meet the $x$-axis at $A(a, 0)$ and the $y$-axis at $B(0, b)$.

The point $R(h, k)$ divides the line segment $AB$ joining $A(a, 0)$ and $B(0, b)$ in the ratio $1:2$. Using the section formula:

$$h = \frac{1(0) + 2(a)}{1 + 2} = \frac{2a}{3} \implies a = \frac{3h}{2}$$ $$k = \frac{1(b) + 2(0)}{1 + 2} = \frac{b}{3} \implies b = 3k$$

The intercept form of the line is:

$$\frac{x}{a} + \frac{y}{b} = 1$$

Substituting the values of $a$ and $b$:

$$\frac{x}{\frac{3h}{2}} + \frac{y}{3k} = 1$$ $$\implies \frac{2x}{3h} + \frac{y}{3k} = 1$$

Multiplying the entire equation by $3$, we get:

$$\frac{2x}{h} + \frac{y}{k} = 3$$

Miscellaneous Exercise - Question 19

Question: By using the concept of equation of a line, prove that the three points $(3, 0)$, $(-2, -2)$ and $(8, 2)$ are collinear.

Solution:

Let the points be $A(3, 0)$, $B(-2, -2)$, and $C(8, 2)$.

First, we find the equation of the line passing through points $A(3, 0)$ and $B(-2, -2)$ using the two-point form:

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$ $$y - 0 = \frac{-2 - 0}{-2 - 3}(x - 3)$$ $$y = \frac{-2}{-5}(x - 3)$$ $$y = \frac{2}{5}(x - 3)$$ $$5y = 2x - 6 \implies 2x - 5y - 6 = 0 \quad \text{--- (1)}$$

Now, we substitute the coordinates of the third point $C(8, 2)$ into the left-hand side of equation (1):

$$\text{LHS} = 2(8) - 5(2) - 6 = 16 - 10 - 6 = 0 = \text{RHS}$$

Since the coordinates of point $C$ satisfy the equation of the line passing through $A$ and $B$, all three points lie on the same straight line.

Hence, the points $(3, 0)$, $(-2, -2)$, and $(8, 2)$ are collinear.

Exercise Chapter 9 - Question 1

Question: Reduce the following equations into slope-intercept form and find their slopes and the $y$-intercepts.
(i) $x + 7y = 0$, (ii) $6x + 3y - 5 = 0$, (iii) $y = 0$.

Solution:

The slope-intercept form of a line is given by $y = mx + c$, where $m$ is the slope and $c$ is the $y$-intercept.

---

(i) $x + 7y = 0$

Rearranging into $y = mx + c$:

$$7y = -x \implies y = -\frac{1}{7}x + 0$$

Comparing with $y = mx + c$:

• Slope ($m$) = $-\frac{1}{7}$
• $y$-intercept ($c$) = $0$

---

(ii) $6x + 3y - 5 = 0$

Rearranging into $y = mx + c$:

$$3y = -6x + 5$$ $$y = \frac{-6x + 5}{3} \implies y = -2x + \frac{5}{3}$$

Comparing with $y = mx + c$:

• Slope ($m$) = $-2$
• $y$-intercept ($c$) = $\frac{5}{3}$

---

(iii) $y = 0$

Writing in the slope-intercept form:

$$y = 0 \cdot x + 0$$

Comparing with $y = mx + c$:

• Slope ($m$) = $0$
• $y$-intercept ($c$) = $0$

Exercise - Question 2

Question: Reduce the following equations into intercept form and find their intercepts on the axes:
(i) $3x + 2y - 12 = 0$
(ii) $4x - 3y = 6$
(iii) $3y + 2 = 0$

Solution:
The intercept form of a line is given by $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the $x$-intercept and $b$ is the $y$-intercept.

(i) $3x + 2y - 12 = 0$

$$3x + 2y = 12$$ Divide both sides by $12$: $$\frac{3x}{12} + \frac{2y}{12} = 1 \implies \frac{x}{4} + \frac{y}{6} = 1$$ Thus, the $x$-intercept $a = 4$ and the $y$-intercept $b = 6$.

(ii) $4x - 3y = 6$

Divide both sides by $6$: $$\frac{4x}{6} - \frac{3y}{6} = 1 \implies \frac{x}{\frac{6}{4}} + \frac{y}{-2} = 1 \implies \frac{x}{\frac{3}{2}} + \frac{y}{-2} = 1$$ Thus, the $x$-intercept $a = \frac{3}{2}$ and the $y$-intercept $b = -2$.

(iii) $3y + 2 = 0$

$$3y = -2 \implies y = -\frac{2}{3}$$ This represents a horizontal line parallel to the $x$-axis. It does not intersect the $x$-axis (no $x$-intercept), and the $y$-intercept is $-\frac{2}{3}$. In intercept form, this cannot be uniquely expressed in the standard two-intercept form since $a$ is undefined.

Exercise - Question 3

Question: Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$.

Solution:
First, simplify the given equation of the line into the general form $Ax + By + C = 0$: $$12x + 72 = 5y - 10$$ $$12x - 5y + 82 = 0$$ Here, $A = 12$, $B = -5$, and $C = 82$.
The given point is $(x_1, y_1) = (-1, 1)$.
The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by the formula: $$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$ Substitute the values: $$d = \frac{|12(-1) - 5(1) + 82|}{\sqrt{12^2 + (-5)^2}}$$ $$d = \frac{|-12 - 5 + 82|}{\sqrt{144 + 25}} = \frac{|65|}{\sqrt{169}} = \frac{65}{13} = 5$$ Thus, the distance of the point from the line is $5$ units.

Exercise - Question 4

Question: Find the points on the $x$-axis, whose distances from the line $\frac{x}{3} + \frac{y}{4} = 1$ are $4$ units.

Solution:
Let any point on the $x$-axis be $(a, 0)$.
First, convert the equation of the line $\frac{x}{3} + \frac{y}{4} = 1$ into general form by multiplying by $12$: $$4x + 3y - 12 = 0$$ Here, $A = 4$, $B = 3$, and $C = -12$.
The distance of the point $(a, 0)$ from this line is given as $4$ units: $$\frac{|4(a) + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = 4$$ $$\frac{|4a - 12|}{5} = 4$$ $$|4a - 12| = 20$$ This gives two cases: 1. $4a - 12 = 20 \implies 4a = 32 \implies a = 8$ 2. $4a - 12 = -20 \implies 4a = -8 \implies a = -2$ Thus, the required points on the $x$-axis are $(8, 0)$ and $(-2, 0)$.

Exercise - Question 5

Question: Find the distance between parallel lines
(i) $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$
(ii) $l(x + y) + p = 0$ and $l(x + y) - r = 0$.

Solution:
The distance $d$ between two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ is given by: $$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$

(i) $15x + 8y - 34 = 0$ and $15x + 8y + 31 = 0$

Here, $A = 15$, $B = 8$, $C_1 = -34$, and $C_2 = 31$. $$d = \frac{|(-34) - 31|}{\sqrt{15^2 + 8^2}} = \frac{|-65|}{\sqrt{225 + 64}} = \frac{65}{\sqrt{289}} = \frac{65}{17} \text{ units}$$

(ii) $l(x + y) + p = 0$ and $l(x + y) - r = 0$

Rewrite the equations in standard form: $$lx + ly + p = 0 \quad \text{and} \quad lx + ly - r = 0$$ Here, $A = l$, $B = l$, $C_1 = p$, and $C_2 = -r$. $$d = \frac{|p - (-r)|}{\sqrt{l^2 + l^2}} = \frac{|p + r|}{\sqrt{2l^2}} = \frac{|p + r|}{|l|\sqrt{2}} \text{ units}$$

Exercise - Question 6

Question: Find equation of the line parallel to the line $3x - 4y + 2 = 0$ and passing through the point $(-2, 3)$.

Solution:
1. Given line is $3x - 4y + 2 = 0$.
The slope ($m$) of this line is given by $-\frac{\text{coefficient of } x}{\text{coefficient of } y} = -\frac{3}{-4} = \frac{3}{4}$.
2. Since the required line is parallel to the given line, its slope will also be equal to $\frac{3}{4}$.
3. We are given that the line passes through the point $(x_1, y_1) = (-2, 3)$.
4. Using the point-slope form, the equation of the line is: $$y - y_1 = m(x - x_1)$$ $$y - 3 = \frac{3}{4}(x - (-2))$$ $$y - 3 = \frac{3}{4}(x + 2)$$ 5. Simplifying the equation: $$4(y - 3) = 3(x + 2)$$ $$4y - 12 = 3x + 6$$ $$3x - 4y + 18 = 0$$

Exercise Chapter 9 - Question 7

Question: Find equation of the line perpendicular to the line $x - 7y + 5 = 0$ and having $x$-intercept $3$.

Solution:
1. Given line: $x - 7y + 5 = 0$.
Slope of this line $m_1 = -\frac{1}{-7} = \frac{1}{7}$.
2. Let the slope of the perpendicular line be $m_2$. Since the lines are perpendicular, $m_1 \cdot m_2 = -1$: $$\frac{1}{7} \cdot m_2 = -1 \implies m_2 = -7$$ 3. The line has an $x$-intercept of $3$, which means it passes through the point $(3, 0)$.
4. Using the point-slope form with slope $-7$ and point $(3, 0)$: $$y - 0 = -7(x - 3)$$ $$y = -7x + 21$$ $$7x + y - 21 = 0$$

Exercise Chapter 9 - Question 8

Question: Find angles between the lines $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$.

Solution:
1. Let the given lines be $L_1: \sqrt{3}x + y - 1 = 0$ and $L_2: x + \sqrt{3}y - 1 = 0$.
2. Slope of $L_1$ ($m_1$) = $-\frac{\sqrt{3}}{1} = -\sqrt{3}$.
3. Slope of $L_2$ ($m_2$) = $-\frac{1}{\sqrt{3}}$.
4. Let $\theta$ be the acute angle between the two lines. The formula for the angle between two lines with slopes $m_1$ and $m_2$ is: $$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$$ 5. Substituting the values: $$\tan \theta = \left| \frac{-\frac{1}{\sqrt{3}} - (-\sqrt{3})}{1 + (-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)} \right| = \left| \frac{-\frac{1}{\sqrt{3}} + \sqrt{3}}{1 + 1} \right| = \left| \frac{\frac{2}{\sqrt{3}}}{2} \right| = \frac{1}{\sqrt{3}}$$ 6. Therefore, $\theta = \frac{\pi}{6}$ (or $30^\circ$).
The angles between the lines are $\frac{\pi}{6}$ and $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Exercise Chapter 9 - Question 9

Question: The line through the points $(h, 3)$ and $(4, 1)$ intersects the line $7x - 9y - 19 = 0$ at right angle. Find the value of $h$.

Solution:
1. Let $L_1$ be the line passing through $(h, 3)$ and $(4, 1)$.
The slope of $L_1$ ($m_1$) is: $$m_1 = \frac{1 - 3}{4 - h} = \frac{-2}{4 - h}$$ 2. Let $L_2$ be the given line $7x - 9y - 19 = 0$.
The slope of $L_2$ ($m_2$) is: $$m_2 = -\frac{7}{-9} = \frac{7}{9}$$ 3. Since $L_1$ and $L_2$ intersect at right angles, the product of their slopes must be $-1$: $$m_1 \cdot m_2 = -1$$ $$\left(\frac{-2}{4 - h}\right) \left(\frac{7}{9}\right) = -1$$ $$\frac{-14}{9(4 - h)} = -1$$ $$-14 = -9(4 - h)$$ $$14 = 36 - 9h$$ $$9h = 36 - 14$$ $$9h = 22 \implies h = \frac{22}{9}$$

Exercise Chapter 9 - Question 10

Question: Prove that the line through the point $(x_1, y_1)$ and parallel to the line $Ax + By + C = 0$ is $A(x - x_1) + B(y - y_1) = 0$.

Solution:

Given the equation of the line: $$Ax + By + C = 0$$ We can rewrite this equation in slope-intercept form ($y = mx + c$) to find its slope: $$By = -Ax - C \implies y = -\frac{A}{B}x - \frac{C}{B}$$ Thus, the slope of the given line is: $$m = -\frac{A}{B}$$

Since parallel lines have equal slopes, the slope of the required line is also $m = -\frac{A}{B}$.

Now, using the point-slope form of a line passing through the point $(x_1, y_1)$ with slope $m$, the equation of the line is: $$y - y_1 = m(x - x_1)$$ Substitute $m = -\frac{A}{B}$: $$y - y_1 = -\frac{A}{B}(x - x_1)$$ Cross-multiplying by $B$: $$B(y - y_1) = -A(x - x_1)$$ Rearranging the terms, we get: $$A(x - x_1) + B(y - y_1) = 0$$ Hence proved.

Exercise Chapter 9 - Question 11

Question: Two lines passing through the point $(2, 3)$ intersect each other at an angle of $60^\circ$. If slope of one line is $2$, find equation of the other line.

Solution:

Let the slope of the first line be $m_1 = 2$, and let the slope of the second line be $m_2 = m$.

The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula: $$\tan \theta = \left|\frac{m_2 - m_1}{1 + m_1 m_2}\right|$$

Given that $\theta = 60^\circ$ and $m_1 = 2$: $$\tan 60^\circ = \left|\frac{m - 2}{1 + 2m}\right|$$ $$\sqrt{3} = \left|\frac{m - 2}{1 + 2m}\right|$$

Removing the absolute value gives two possible cases:

Case 1: $$\frac{m - 2}{1 + 2m} = \sqrt{3}$$ $$m - 2 = \sqrt{3}(1 + 2m)$$ $$m - 2 = \sqrt{3} + 2\sqrt{3}m$$ $$m - 2\sqrt{3}m = \sqrt{3} + 2$$ $$m(1 - 2\sqrt{3}) = 2 + \sqrt{3}$$ $$m = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}$$ Rationalizing the denominator: $$m = \frac{(2 + \sqrt{3})(1 + 2\sqrt{3})}{(1 - 2\sqrt{3})(1 + 2\sqrt{3})} = \frac{2 + 4\sqrt{3} + \sqrt{3} + 6}{1 - 12} = \frac{8 + 5\sqrt{3}}{-11} = -\frac{8 + 5\sqrt{3}}{11}$$

Case 2: $$\frac{m - 2}{1 + 2m} = -\sqrt{3}$$ $$m - 2 = -\sqrt{3}(1 + 2m)$$ $$m - 2 = -\sqrt{3} - 2\sqrt{3}m$$ $$m + 2\sqrt{3}m = 2 - \sqrt{3}$$ $$m(1 + 2\sqrt{3}) = 2 - \sqrt{3}$$ $$m = \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}}$$ Rationalizing the denominator: $$m = \frac{(2 - \sqrt{3})(1 - 2\sqrt{3})}{(1 + 2\sqrt{3})(1 - 2\sqrt{3})} = \frac{2 - 4\sqrt{3} - \sqrt{3} + 6}{1 - 12} = \frac{8 - 5\sqrt{3}}{-11} = \frac{5\sqrt{3} - 8}{11}$$

Now, using the point-slope form for a line passing through $(2, 3)$ with slope $m$:

Equation corresponding to Case 1 ($m = -\frac{8 + 5\sqrt{3}}{11}$): $$y - 3 = -\frac{8 + 5\sqrt{3}}{11}(x - 2)$$ $$11(y - 3) + (8 + 5\sqrt{3})(x - 2) = 0$$ $$(8 + 5\sqrt{3})x + 11y - (16 + 10\sqrt{3} + 33) = 0$$ $$(8 + 5\sqrt{3})x + 11y - (49 + 10\sqrt{3}) = 0$$

Equation corresponding to Case 2 ($m = \frac{5\sqrt{3} - 8}{11}$): $$y - 3 = \frac{5\sqrt{3} - 8}{11}(x - 2)$$ $$11(y - 3) = (5\sqrt{3} - 8)(x - 2)$$ $$(8 - 5\sqrt{3})x + 11y + (10\sqrt{3} - 16 - 33) = 0$$ $$(8 - 5\sqrt{3})x + 11y + (10\sqrt{3} - 49) = 0$$

Exercise Chapter 9 - Question 12

Question: Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and $(-1, 2)$.

Solution:

Let the given points be $A(3, 4)$ and $B(-1, 2)$.

Step 1: Find the midpoint of segment $AB$.
Using the midpoint formula $(x_m, y_m) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$: $$M = \left(\frac{3 + (-1)}{2}, \frac{4 + 2}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3)$$

Step 2: Find the slope of the line segment $AB$.
$$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}$$

Step 3: Find the slope of the right bisector.
Since the right bisector is perpendicular to $AB$, its slope $m$ is the negative reciprocal of $m_{AB}$: $$m = -\frac{1}{m_{AB}} = -\frac{1}{1/2} = -2$$

Step 4: Find the equation of the right bisector.
The right bisector passes through the midpoint $M(1, 3)$ and has slope $m = -2$. Using the point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 3 = -2(x - 1)$$ $$y - 3 = -2x + 2$$ $$2x + y - 5 = 0$$

Answer: The equation of the right bisector is $2x + y - 5 = 0$.

Exercise Chapter 9 - Question 13

Question: Find the coordinates of the foot of perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$.

Solution:

Let $P(-1, 3)$ be the given point and let $N(x_1, y_1)$ be the foot of the perpendicular from $P$ to the line: $$3x - 4y - 16 = 0 \quad \text{--- (1)}$$

Step 1: Find the slope of the given line and perpendicular line.
Rewriting equation (1) in slope-intercept form: $$4y = 3x - 16 \implies y = \frac{3}{4}x - 4$$ The slope of the line is $m_1 = \frac{3}{4}$.

Since $PN$ is perpendicular to the given line, the slope of $PN$ ($m_2$) is: $$m_2 = -\frac{1}{m_1} = -\frac{4}{3}$$

Step 2: Write the equation of the perpendicular line $PN$.
Using the point-slope form passing through $P(-1, 3)$ with slope $-\frac{4}{3}$: $$y - 3 = -\frac{4}{3}(x - (-1))$$ $$3(y - 3) = -4(x + 1)$$ $$3y - 9 = -4x - 4$$ $$4x + 3y - 5 = 0 \quad \text{--- (2)}$$

Step 3: Solve the simultaneous linear equations (1) and (2) to find point $N(x_1, y_1)$.
Multiply equation (1) by $4$ and equation (2) by $3$: $$(3x - 4y - 16 = 0) \times 4 \implies 12x - 16y - 64 = 0$$ $$(4x + 3y - 5 = 0) \times 3 \implies 12x + 9y - 15 = 0$$

Subtracting the two equations: $$(12x - 16y - 64) - (12x + 9y - 15) = 0$$ $$-25y - 49 = 0 \implies 25y = -49 \implies y = -\frac{49}{25}$$

Substitute $y = -\frac{49}{25}$ into equation (2): $$4x + 3\left(-\frac{49}{25}\right) - 5 = 0$$ $$4x - \frac{147}{25} - 5 = 0$$ $$4x = \frac{147}{25} + 5 = \frac{147 + 125}{25} = \frac{272}{25}$$ $$x = \frac{272}{100} = \frac{68}{25}$$

Answer: The coordinates of the foot of the perpendicular are $\left(\frac{68}{25}, -\frac{49}{25}\right)$.

Exercise Chapter 9 - Question 14

Question: The perpendicular from the origin to the line $y = mx + c$ meets it at the point $(-1, 2)$. Find the values of $m$ and $c$.

Solution:

1. Since the line passes through the point $(-1, 2)$, the coordinates of this point must satisfy the equation of the line $y = mx + c$: $$2 = m(-1) + c \implies -m + c = 2 \quad \text{--- (1)}$$
2. The given line is $y = mx + c$, which can be written in slope-intercept form as $mx - y + c = 0$. The slope of this line is $m$.
3. The perpendicular to this line drawn from the origin $(0, 0)$ passes through $(-1, 2)$. Therefore, the slope of this perpendicular line is given by the slope of the segment joining $(0, 0)$ and $(-1, 2)$: $$\text{Slope} = \frac{2 - 0}{-1 - 0} = -2$$
4. Since two perpendicular lines have the product of their slopes equal to $-1$: $$m \times (-2) = -1 \implies m = \frac{1}{2}$$
5. Substitute $m = \frac{1}{2}$ into equation (1): $$-\frac{1}{2} + c = 2 \implies c = 2 + \frac{1}{2} = \frac{5}{2}$$
6. Thus, the values are $m = \frac{1}{2}$ and $c = \frac{5}{2}$.

Exercise Chapter 9 - Question 15

Question: If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x \cos\theta - y \sin\theta = k \cos 2\theta$ and $x \sec\theta + y \csc\theta = k$, respectively, prove that $p^2 + 4q^2 = k^2$.

Solution:

1. The length of the perpendicular from the origin $(0,0)$ to the line $Ax + By + C = 0$ is given by $p = \left|\frac{C}{\sqrt{A^2 + B^2}}\right|$.
2. For the first line $x \cos\theta - y \sin\theta - k \cos 2\theta = 0$: $$p = \left|\frac{-k \cos 2\theta}{\sqrt{\cos^2\theta + (-\sin\theta)^2}}\right| = |k \cos 2\theta|$$ Squaring both sides: $$p^2 = k^2 \cos^2 2\theta \quad \text{--- (1)}$$
3. For the second line $x \sec\theta + y \csc\theta - k = 0$, rewriting in terms of sine and cosine: $$\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = k \implies x \sin\theta + y \cos\theta - k \sin\theta \cos\theta = 0$$ The length of the perpendicular $q$ from $(0,0)$ is: $$q = \left|\frac{-k \sin\theta \cos\theta}{\sqrt{\sin^2\theta + \cos^2\theta}}\right| = |k \sin\theta \cos\theta|$$ Squaring both sides: $$q^2 = k^2 \sin^2\theta \cos^2\theta = \frac{k^2}{4} (2\sin\theta \cos\theta)^2 = \frac{k^2}{4} \sin^2 2\theta$$ Therefore: $$4q^2 = k^2 \sin^2 2\theta \quad \text{--- (2)}$$
4. Adding equations (1) and (2): $$p^2 + 4q^2 = k^2 \cos^2 2\theta + k^2 \sin^2 2\theta = k^2(\cos^2 2\theta + \sin^2 2\theta) = k^2$$ Hence proved.

Exercise Chapter 9 - Question 16

Question: In the triangle $ABC$ with vertices $A(2, 3)$, $B(4, -1)$ and $C(1, 2)$, find the equation and length of altitude from the vertex $A$.

Solution:

1. First, find the slope of the line segment $BC$ where $B(4, -1)$ and $C(1, 2)$: $$m_{BC} = \frac{2 - (-1)}{1 - 4} = \frac{3}{-3} = -1$$
2. Since the altitude from $A$ is perpendicular to $BC$, its slope $m$ is the negative reciprocal of $m_{BC}$: $$m = -\left(\frac{1}{-1}\right) = 1$$
3. The equation of the altitude passing through $A(2, 3)$ with slope $1$ is given by: $$y - y_1 = m(x - x_1) \implies y - 3 = 1(x - 2) \implies x - y + 1 = 0$$
4. The length of the altitude is the perpendicular distance from point $A(2, 3)$ to the line $BC$. First, find the equation of line $BC$: $$y - 2 = -1(x - 1) \implies x + y - 3 = 0$$
5. Calculate the perpendicular distance from $A(2, 3)$ to $x + y - 3 = 0$: $$\text{Length} = \left|\frac{1(2) + 1(3) - 3}{\sqrt{1^2 + 1^2}}\right| = \frac{|2 + 3 - 3|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$
6. Thus, the equation of the altitude is $x - y + 1 = 0$ and its length is $\sqrt{2}$.

Exercise Chapter 9 - Question 17

Question: If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$, then show that $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.

Solution:

1. The equation of a line with $x$-intercept $a$ and $y$-intercept $b$ in intercept form is: $$\frac{x}{a} + \frac{y}{b} = 1$$ Multiplying by $ab$, we get: $$bx + ay - ab = 0$$
2. The length of the perpendicular $p$ from the origin $(0, 0)$ to this line is: $$p = \left|\frac{-ab}{\sqrt{b^2 + a^2}}\right| = \frac{|ab|}{\sqrt{a^2 + b^2}}$$
3. Squaring both sides: $$p^2 = \frac{a^2 b^2}{a^2 + b^2}$$
4. Taking the reciprocal of both sides: $$\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2}$$ $$\frac{1}{p^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}$$ $$\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$$ Hence proved.

Exercise Chapter 9 - Question 1

Question: Find the values of $k$ for which the line $(k - 3)x - (4 - k^2)y + k^2 - 7k + 6 = 0$ is (a) Parallel to the $x$-axis, (b) Parallel to the $y$-axis, (c) Passing through the origin.

Solution: Given equation of the line is: $$(k - 3)x - (4 - k^2)y + k^2 - 7k + 6 = 0$$ **(a) Parallel to the $x$-axis:** A line parallel to the $x$-axis has a slope of $0$, which means its equation contains only the $y$-term (coefficient of $x$ is zero). Setting the coefficient of $x$ to $0$: $$k - 3 = 0 \implies k = 3$$ We must also check that the coefficient of $y$ does not vanish for this value of $k$: $$-\left(4 - k^2\right) = -(4 - 3^2) = -(4 - 9) = 5 \neq 0$$ Thus, $k = 3$ is valid. **(b) Parallel to the $y$-axis:** A line parallel to the $y$-axis has an undefined slope, meaning its equation contains only the $x$-term (coefficient of $y$ is zero). Setting the coefficient of $y$ to $0$: $$4 - k^2 = 0 \implies k^2 = 4 \implies k = \pm 2$$ We check the coefficient of $x$ for these values: - For $k = 2$: $k - 3 = 2 - 3 = -1 \neq 0$ (Valid) - For $k = -2$: $k - 3 = -2 - 3 = -5 \neq 0$ (Valid) Thus, $k = \pm 2$. **(c) Passing through the origin:** If a line passes through the origin $(0, 0)$, its constant term must be equal to $0$. Setting the constant term to $0$: $$k^2 - 7k + 6 = 0$$ Factorizing the quadratic equation: $$(k - 1)(k - 6) = 0 \implies k = 1 \text{ or } k = 6$$ Thus, $k = 1, 6$.

Exercise Chapter 9 - Question 2

Question: Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $1$ and $-6$, respectively.

Solution: Let the $x$-intercept and $y$-intercept of the line be $a$ and $b$ respectively. According to the given conditions: $$a + b = 1 \quad \text{--(1)}$$ $$ab = -6 \quad \text{--(2)}$$ From equation (1), we have $b = 1 - a$. Substituting this into equation (2): $$a(1 - a) = -6 \implies a - a^2 = -6 \implies a^2 - a - 6 = 0$$ Factorizing the quadratic equation: $$(a - 3)(a + 2) = 0 \implies a = 3 \text{ or } a = -2$$ - **Case 1:** If $a = 3$, then $b = 1 - 3 = -2$. The intercept form of the line equation is $\frac{x}{a} + \frac{y}{b} = 1$: $$\frac{x}{3} + \frac{y}{-2} = 1 \implies 2x - 3y = 6$$ - **Case 2:** If $a = -2$, then $b = 1 - (-2) = 3$. $$\frac{x}{-2} + \frac{y}{3} = 1 \implies -3x + 2y = 6 \implies 3x - 2y + 6 = 0$$ Thus, the required equations of the lines are $2x - 3y - 6 = 0$ and $3x - 2y + 6 = 0$.

Exercise Chapter 9 - Question 3

Question: What are the points on the $y$-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is $4$ units.

Solution: First, simplify the given line equation $\frac{x}{3} + \frac{y}{4} = 1$ by multiplying by $12$: $$4x + 3y - 12 = 0$$ Any point on the $y$-axis is of the form $(0, y_1)$. The perpendicular distance $d$ from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by: $$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$ Here, $A = 4$, $B = 3$, $C = -12$, $(x_1, y_1) = (0, y_1)$, and $d = 4$: $$4 = \frac{|4(0) + 3(y_1) - 12|}{\sqrt{4^2 + 3^2}}$$ $$4 = \frac{|3y_1 - 12|}{5}$$ $$|3y_1 - 12| = 20$$ This gives two cases: 1. $3y_1 - 12 = 20 \implies 3y_1 = 32 \implies y_1 = \frac{32}{3}$ 2. $3y_1 - 12 = -20 \implies 3y_1 = -8 \implies y_1 = -\frac{8}{3}$ Thus, the required points on the $y$-axis are $\left(0, \frac{32}{3}\right)$ and $\left(0, -\frac{8}{3}\right)$.

Exercise Chapter 9 - Question 4

Question: Find perpendicular distance from the origin to the line joining the points $(\cos\theta, \sin\theta)$ and $(\cos\phi, \sin\phi)$.

Solution: The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$ Substituting the given points $(\cos\theta, \sin\theta)$ and $(\cos\phi, \sin\phi)$: $$y - \sin\theta = \frac{\sin\phi - \sin\theta}{\cos\phi - \cos\theta}(x - \cos\theta)$$ Using trigonometric transformations for sum-to-product: $$\sin\phi - \sin\theta = 2 \cos\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right)$$ $$\cos\phi - \cos\theta = -2 \sin\left(\frac{\phi+\theta}{2}\right)\sin\left(\frac{\phi-\theta}{2}\right)$$ Thus, the slope is: $$m = -\frac{\cos\left(\frac{\phi+\theta}{2}\right)}{\sin\left(\frac{\phi+\theta}{2}\right)}$$ Substituting this back and simplifying yields the general equation of the line: $$x\sin\left(\frac{\theta+\phi}{2}\right) + y\cos\left(\frac{\theta+\phi}{2}\right) = \sin\left(\frac{\theta+\phi}{2}\right)\cos\theta - \cos\left(\frac{\theta+\phi}{2}\right)\sin\theta$$ $$x\sin\left(\frac{\theta+\phi}{2}\right) + y\cos\left(\frac{\theta+\phi}{2}\right) = \sin\left(\frac{\theta+\phi}{2} - \theta\right) = \sin\left(\frac{\phi-\theta}{2}\right)$$ Now, the perpendicular distance $p$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is: $$p = \frac{|C|}{\sqrt{A^2 + B^2}}$$ Here, $A = \sin\left(\frac{\theta+\phi}{2}\right)$, $B = \cos\left(\frac{\theta+\phi}{2}\right)$, and $C = -\sin\left(\frac{\phi-\theta}{2}\right)$: $$p = \frac{\left|-\sin\left(\frac{\phi-\theta}{2}\right)\right|}{\sqrt{\sin^2\left(\frac{\theta+\phi}{2}\right) + \cos^2\left(\frac{\theta+\phi}{2}\right)}} = \left|\sin\left(\frac{\phi-\theta}{2}\right)\right|$$ Thus, the perpendicular distance from the origin is $\left|\sin\left(\frac{\phi-\theta}{2}\right)\right|$.

Exercise Chapter 9 - Question 5

Question: Find the equation of the line parallel to $y$-axis and drawn through the point of intersection of the lines $x - 7y + 5 = 0$ and $3x + y = 0$.

Solution:

First, we find the point of intersection of the given two lines by solving them simultaneously:

1. $x - 7y + 5 = 0$ $\implies x = 7y - 5$
2. $3x + y = 0$

Substitute $x = 7y - 5$ into the second equation:

$$3(7y - 5) + y = 0$$ $$21y - 15 + y = 0$$ $$22y = 15 \implies y = \frac{15}{22}$$

Now, substitute $y = \frac{15}{22}$ back to find $x$:

$$x = 7\left(\frac{15}{22}\right) - 5 = \frac{105}{22} - \frac{110}{22} = -\frac{5}{22}$$

Thus, the point of intersection is $\left(-\frac{5}{22}, \frac{15}{22}\right)$.

We need the equation of a line parallel to the $y$-axis passing through this point. Any line parallel to the $y$-axis has the general equation:

$$x = c$$

Since it passes through $\left(-\frac{5}{22}, \frac{15}{22}\right)$, the $x$-coordinate is constant. Therefore, the required equation is:

$$x = -\frac{5}{22} \implies 22x + 5 = 0$$

Exercise - Chapter 9 - Question 6

Question: Find the equation of a line drawn perpendicular to the line $\frac{x}{4} + \frac{y}{6} = 1$ through the point, where it meets the $y$-axis.

Solution:

Given line:

$$\frac{x}{4} + \frac{y}{6} = 1$$

To find the point where it meets the $y$-axis, set $x = 0$:

$$\frac{0}{4} + \frac{y}{6} = 1 \implies y = 6$$

So, the line meets the $y$-axis at the point $P(0, 6)$.

Next, we find the slope ($m_1$) of the given line by rewriting it in slope-intercept form ($y = mx + c$):

$$\frac{y}{6} = 1 - \frac{x}{4} \implies y = 6 - \frac{6}{4}x = -\frac{3}{2}x + 6$$

Thus, the slope of the given line is $m_1 = -\frac{3}{2}$.

The slope ($m_2$) of the line perpendicular to it is:

$$m_2 = -\frac{1}{m_1} = -\frac{1}{-3/2} = \frac{2}{3}$$

Using the point-slope form with point $(0, 6)$ and slope $\frac{2}{3}$:

$$y - y_1 = m_2(x - x_1)$$ $$y - 6 = \frac{2}{3}(x - 0)$$ $$3y - 18 = 2x \implies 2x - 3y + 18 = 0$$

Exercise - Chapter 9 - Question 7

Question: Find the area of the triangle formed by the lines $y - x = 0$, $x + y = 0$ and $x - k = 0$.

Solution:

Let us find the vertices of the triangle by finding the intersection points of the three lines taken in pairs:

1. Intersection of $y - x = 0$ ($y = x$) and $x + y = 0$ ($y = -x$):
   $x + x = 0 \implies x = 0$, so $y = 0$. Vertex $A(0, 0)$.
2. Intersection of $y - x = 0$ and $x - k = 0$ ($x = k$):
   Since $x = k$, $y = k$. Vertex $B(k, k)$.
3. Intersection of $x + y = 0$ and $x - k = 0$ ($x = k$):
   $k + y = 0 \implies y = -k$. Vertex $C(k, -k)$.

The vertices of the triangle are $A(0, 0)$, $B(k, k)$, and $C(k, -k)$.

Notice that side $BC$ lies on the vertical line $x = k$, so the length of the base $BC$ is the distance between $(k, k)$ and $(k, -k)$:

$$\text{Base } BC = |k - (-k)| = |2k| = 2|k|$$

The height of the triangle corresponding to base $BC$ is the perpendicular distance from vertex $A(0,0)$ to the line $x = k$, which is:

$$\text{Height} = |k-0| = |k|$$

Therefore, the area of the triangle is:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times |2k| \times |k| = k^2 \text{ square units}.$$

Exercise - Chapter 9 - Question 8

Question: Find the value of $p$ so that the three lines $3x + y - 2 = 0$, $px + 2y - 3 = 0$ and $2x - y - 3 = 0$ may intersect at one point.

Solution:

If the three lines intersect at one point (concurrent lines), then the point of intersection of any two lines must also lie on the third line.

Let us solve the first and third equations simultaneously to find their point of intersection:

1. $3x + y - 2 = 0$
2. $2x - y - 3 = 0$

Adding equation (1) and equation (2):

$$(3x + y - 2) + (2x - y - 3) = 0$$ $$5x - 5 = 0 \implies x = 1$$

Substitute $x = 1$ into equation (1):

$$3(1) + y - 2 = 0 \implies 1 + y = 0 \implies y = -1$$

So, the point of intersection of the first and third lines is $(1, -1)$.

Since all three lines intersect at this single point, $(1, -1)$ must satisfy the second equation, $px + 2y - 3 = 0$:

$$p(1) + 2(-1) - 3 = 0$$ $$p - 2 - 3 = 0$$ $$p - 5 = 0 \implies p = 5$$

Exercise - Chapter 9 - Question 9

Question: If three lines whose equations are $y = m_1x + c_1$, $y = m_2x + c_2$ and $y = m_3x + c_3$ are concurrent, then show that $m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$.

Solution:

Three lines are concurrent if they pass through a common point, say $(x_0, y_0)$. Since $(x_0, y_0)$ lies on all three lines, it satisfies their equations:

$$y_0 = m_1x_0 + c_1 \quad \text{--- (1)}$$ $$y_0 = m_2x_0 + c_2 \quad \text{--- (2)}$$ $$y_0 = m_3x_0 + c_3 \quad \text{--- (3)}$$

Subtracting equation (2) from equation (1):

$$0 = (m_1 - m_2)x_0 + (c_1 - c_2) \implies (m_1 - m_2)x_0 = c_2 - c_1 \implies x_0 = \frac{c_2 - c_1}{m_1 - m_2}$$

Similarly, subtracting equation (3) from equation (2):

$$0 = (m_2 - m_3)x_0 + (c_2 - c_3) \implies x_0 = \frac{c_3 - c_2}{m_2 - m_3}$$

Equating the two expressions for $x_0$:

$$\frac{c_2 - c_1}{m_1 - m_2} = \frac{c_3 - c_2}{m_2 - m_3}$$

Cross-multiplying gives:

$$(c_2 - c_1)(m_2 - m_3) = (c_3 - c_2)(m_1 - m_2)$$ $$c_2m_2 - c_2m_3 - c_1m_2 + c_1m_3 = c_3m_1 - c_3m_2 - c_2m_1 + c_2m_2$$

Canceling $c_2m_2$ from both sides and rearranging terms to one side:

$$m_1(c_2 - c_3) + m_2(c_3 - c_1) + m_3(c_1 - c_2) = 0$$

Hence proved.

Exercise Chapter 9 - Question 10

Question: Find the equation of the lines through the point $(3, 2)$ which make an angle of $45^\circ$ with the line $x - 2y = 3$.

Solution:

Let the slope of the required line be $m$. The given line is $x - 2y = 3$, which can be rewritten as $2y = x - 3 \implies y = \frac{1}{2}x - \frac{3}{2}$.

The slope of the given line is $m_1 = \frac{1}{2}$.

The angle $\theta$ between the two lines is $45^\circ$. Using the formula for the angle between two lines:

$$\tan\theta = \left|\frac{m - m_1}{1 + mm_1}\right|$$ $$\tan 45^\circ = \left|\frac{m - \frac{1}{2}}{1 + \frac{1}{2}m}\right|$$ $$1 = \left|\frac{2m - 1}{2 + m}\right|$$

Removing the absolute value gives two cases:

Case 1: $\frac{2m - 1}{2 + m} = 1$

$$2m - 1 = 2 + m \implies m = 3$$

Equation of the line passing through $(3, 2)$ with slope $3$:

$$y - 2 = 3(x - 3) \implies y - 2 = 3x - 9 \implies 3x - y - 7 = 0$$

Case 2: $\frac{2m - 1}{2 + m} = -1$

$$2m - 1 = -(2 + m) \implies 2m - 1 = -2 - m \implies 3m = -1 \implies m = -\frac{1}{3}$$

Equation of the line passing through $(3, 2)$ with slope $-\frac{1}{3}$:

$$y - 2 = -\frac{1}{3}(x - 3) \implies 3(y - 2) = -(x - 3) \implies 3y - 6 = -x + 3 \implies x + 3y - 9 = 0$$

Thus, the required equations of the lines are $3x - y - 7 = 0$ and $x + 3y - 9 = 0$.

Exercise Chapter 9 - Question 11

Question: Find the equation of the line passing through the point of intersection of the lines $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$ that has equal intercepts on the axes.

Solution:

First, find the point of intersection of the given lines by solving them simultaneously:

$$4x + 7y - 3 = 0 \quad \text{--- (1)}$$ $$2x - 3y + 1 = 0 \implies 4x - 6y + 2 = 0 \quad \text{--- (2)}$$

Subtracting (2) from (1):

$$(7y - (-6y)) + (-3 - 2) = 0 \implies 13y - 5 = 0 \implies y = \frac{5}{13}$$

Substituting $y = \frac{5}{13}$ into equation (2):

$$2x - 3\left(\frac{5}{13}\right) + 1 = 0 \implies 2x = \frac{15}{13} - 1 = \frac{2}{13} \implies x = \frac{1}{13}$$

So, the point of intersection is $\left(\frac{1}{13}, \frac{5}{13}\right)$.

Let the equation of the line with equal intercepts $a$ (where $a \neq 0$) be:

$$\frac{x}{a} + \frac{y}{a} = 1 \implies x + y = a$$

Since this line passes through $\left(\frac{1}{13}, \frac{5}{13}\right)$, we substitute the coordinates to find $a$:

$$\frac{1}{13} + \frac{5}{13} = a \implies a = \frac{6}{13}$$

Substituting $a$ back into the intercept form equation:

$$x + y = \frac{6}{13} \implies 13x + 13y = 6 \implies 13x + 13y - 6 = 0$$

Exercise Chapter 9 - Question 12

Question: Show that the equation of the line passing through the origin and making an angle $\theta$ with the line $y = mx + c$ is $\frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m \tan\theta}$.

Solution:

Let the equation of the line passing through the origin be $y = m'x$, where $m'$ is the slope of this line. Thus, $\frac{y}{x} = m'$.

The given line is $y = mx + c$, which has slope $m_1 = m$.

The angle between the line $y = m'x$ and $y = mx + c$ is given as $\theta$. Using the angle formula between two lines:

$$\tan\theta = \left|\frac{m' - m}{1 + m'm}\right|$$

Removing the absolute value gives two possibilities:

$$\tan\theta = \frac{m' - m}{1 + mm'} \quad \text{or} \quad \tan\theta = -\frac{m' - m}{1 + mm'}$$

Case 1:

$$\tan\theta (1 + mm') = m' - m$$ $$\tan\theta + mm'\tan\theta = m' - m$$ $$m + \tan\theta = m'(1 - m\tan\theta)$$ $$m' = \frac{m + \tan\theta}{1 - m\tan\theta}$$

Case 2:

$$\tan\theta (1 + mm') = -(m' - m) = m - m'$$ $$\tan\theta + mm'\tan\theta = m - m'$$ $$m' + mm'\tan\theta = m - \tan\theta$$ $$m'(1 + m\tan\theta) = m - \tan\theta$$ $$m' = \frac{m - \tan\theta}{1 + m\tan\theta}$$

Combining both cases, we get:

$$m' = \frac{m \pm \tan\theta}{1 \mp m \tan\theta}$$

Since $m' = \frac{y}{x}$, it follows that:

$$\frac{y}{x} = \frac{m \pm \tan\theta}{1 \mp m \tan\theta}$$

Hence proved.

Exercise Chapter 9 - Question 13

Question: In what ratio, the line joining $(-1, 1)$ and $(5, 7)$ is divided by the line $x + y = 4$?

Solution:

Let the line $x + y = 4$ divide the line segment joining the points $A(-1, 1)$ and $B(5, 7)$ in the ratio $k:1$ at point $P$.

Using the section formula, the coordinates of the point of intersection $P$ are given by:

$$P\left(\frac{5k - 1}{k + 1}, \frac{7k + 1}{k + 1}\right)$$

Since the point $P$ lies on the line $x + y = 4$, its coordinates must satisfy the equation of the line:

$$\frac{5k - 1}{k + 1} + \frac{7k + 1}{k + 1} = 4$$

Multiplying both sides by $(k + 1)$ (assuming $k \neq -1$):

$$(5k - 1) + (7k + 1) = 4(k + 1)$$

$$12k = 4k + 4$$

$$8k = 4 \implies k = \frac{4}{8} = \frac{1}{2}$$

Thus, the required ratio is $k : 1 = \frac{1}{2} : 1 = 1 : 2$ internally.

Exercise Chapter 9 - Question 14

Question: Find the distance of the line $4x + 7y + 5 = 0$ from the point $(1, 2)$ along the line $2x - y = 0$.

Solution:

Let the given point be $A(1, 2)$. The line drawn through $A$ is parallel to the line $2x - y = 0$.

Since the line passes through $(1, 2)$ and has slope $m = 2$ (parallel to $2x - y = 0$), its equation in parametric form is:

$$\frac{x - 1}{\cos\theta} = \frac{y - 2}{\sin\theta} = r$$

where $\tan\theta = 2$. Thus, $\cos\theta = \frac{1}{\sqrt{5}}$ and $\sin\theta = \frac{2}{\sqrt{5}}$ (or their negatives).

Any point on this line is given by $P(1 + r\cos\theta, 2 + r\sin\theta)$.

This point lies on the line $4x + 7y + 5 = 0$. Substituting $x$ and $y$:

$$4(1 + r\cos\theta) + 7(2 + r\sin\theta) + 5 = 0$$

$$4 + 14 + 5 + r(4\cos\theta + 7\sin\theta) = 0$$

$$23 + r\left(4\left(\frac{1}{\sqrt{5}}\right) + 7\left(\frac{2}{\sqrt{5}}\right)\right) = 0$$

$$23 + r\left(\frac{4 + 14}{\sqrt{5}}\right) = 0 \implies 23 + \frac{18r}{\sqrt{5}} = 0$$

$$r = -\frac{23\sqrt{5}}{18}$$

Taking the magnitude of distance, the required distance is $\frac{23\sqrt{5}}{18}$ units.

Exercise Chapter 9 - Question 15

Question: Find the direction in which a straight line must be drawn through the point $(-1, 2)$ so that its point of intersection with the line $x + y = 4$ may be at a distance of $3$ units from this point.

Solution:

Let the line make an angle $\theta$ with the positive direction of the x-axis. The line passes through $(-1, 2)$ and the distance $r = 3$.

The coordinates of the point of intersection on the line $x + y = 4$ are:

$$x = -1 + 3\cos\theta, \quad y = 2 + 3\sin\theta$$

Since this point lies on $x + y = 4$:

$$(-1 + 3\cos\theta) + (2 + 3\sin\theta) = 4$$

$$3(\cos\theta + \sin\theta) + 1 = 4$$

$$3(\cos\theta + \sin\theta) = 3 \implies \cos\theta + \sin\theta = 1$$

Dividing by $\sqrt{2}$:

$$\frac{1}{\sqrt{2}}\cos\theta + \frac{1}{\sqrt{2}}\sin\theta = \frac{1}{\sqrt{2}}$$

$$\cos\left(\theta - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)$$

Thus, $\theta - \frac{\pi}{4} = \pm \frac{\pi}{4}$.

Case 1: $\theta - \frac{\pi}{4} = \frac{\pi}{4} \implies \theta = \frac{\pi}{2}$ ($90^\circ$)

Case 2: $\theta - \frac{\pi}{4} = -\frac{\pi}{4} \implies \theta = 0$ ($0^\circ$)

Therefore, the line must be drawn in the direction of $0^\circ$ or $90^\circ$ (i.e., parallel to the x-axis or y-axis).

Exercise Chapter 9 - Question 16

Question: The hypotenuse of a right angled triangle has its ends at the points $(1, 3)$ and $(-4, 1)$. Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

Solution:

Let the ends of the hypotenuse be $A(1, 3)$ and $B(-4, 1)$.

Since the legs of the right-angled triangle are parallel to the coordinate axes, one leg is parallel to the x-axis (horizontal) and the other is parallel to the y-axis (vertical).

The vertex containing the right angle ($C$) formed by the intersection of these legs must have coordinates combining the x-coordinate of one endpoint and the y-coordinate of the other.

Thus, the possible coordinates for the vertex $C$ are:

1. $C_1 = (1, 1)$

2. $C_2 = (-4, 3)$

Case 1: When the vertex is $C(1, 1)$

- The horizontal leg passes through $(1, 1)$ and has the equation $y = 1$.

- The vertical leg passes through $(1, 1)$ and has the equation $x = 1$.

Case 2: When the vertex is $C(-4, 3)$

- The horizontal leg passes through $(-4, 3)$ and has the equation $y = 3$.

- The vertical leg passes through $(-4, 3)$ and has the equation $x = -4$.

Thus, the equations of the legs are either $x = 1, y = 1$ or $x = -4, y = 3$.

Exercise Chapter 9 - Question 17

Question: Find the image of the point $(3, 8)$ with respect to the line $x + 3y = 7$ assuming the line to be a plane mirror.

Solution:

Let the given point be $P(x_1, y_1) = (3, 8)$ and let its image with respect to the line mirror $L: x + 3y - 7 = 0$ be $Q(x_2, y_2)$.

Using the properties of reflection across a line:

1. The midpoint $M$ of line segment $PQ$ lies on the mirror line $x + 3y = 7$. $$M = \left(\frac{3 + x_2}{2}, \frac{8 + y_2}{2}\right)$$ Substituting $M$ into the line equation: $$\frac{3 + x_2}{2} + 3\left(\frac{8 + y_2}{2}\right) = 7 \implies (3 + x_2) + 3(8 + y_2) = 14$$ $$x_2 + 3y_2 + 27 = 14 \implies x_2 + 3y_2 = -13 \quad \text{--- (1)}$$
2. The line segment $PQ$ is perpendicular to the mirror line $x + 3y = 7$.
   Slope of mirror line $= -\frac{1}{3}$. Thus, the slope of $PQ$ is $\frac{y_2 - 8}{x_2 - 3} = 3$. $$y_2 - 8 = 3(x_2 - 3) \implies 3x_2 - y_2 = 1 \quad \text{--- (2)}$$

Solving equations (1) and (2) simultaneously:

From (2), $y_2 = 3x_2 - 1$. Substituting into (1): $$x_2 + 3(3x_2 - 1) = -13 \gets 10x_2 - 3 = -13 \implies 10x_2 = -10 \implies x_2 = -1$$ Then, $y_2 = 3(-1) - 1 = -4$.

Therefore, the image point is $(-1, -4)$.

Exercise - Question 18

Question: If the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$, find the value of $m$.

Solution:

The slopes of the given lines are:

• Line 1: $y = 3x + 1 \implies m_1 = 3$
• Line 2: $2y = x + 3 \implies y = \frac{1}{2}x + \frac{3}{2} \implies m_2 = \frac{1}{2}$
• Given line: $y = mx + 4 \implies m_3 = m$

Since the first two lines are equally inclined to the line $y = mx + 4$, the angle between the first line and the given line is equal to the angle between the second line and the given line. Using the angle between two lines formula $\tan \theta = \left|\frac{m_a - m_b}{1 + m_a m_b}\right|$:

$$\left|\frac{3 - m}{1 + 3m}\right| = \left|\frac{\frac{1}{2} - m}{1 + \frac{1}{2}m}\right|$$ $$\left|\frac{3 - m}{1 + 3m}\right| = \left|\frac{1 - 2m}{2 + m}\right|$$

Removing the absolute value gives two cases:

Case 1: $$\frac{3 - m}{1 + 3m} = \frac{1 - 2m}{2 + m}$$ $$(3 - m)(2 + m) = (1 - 2m)(1 + 3m)$$ $$6 + 3m - 2m - m^2 = 1 + 3m - 2m - 6m^2$$ $$6 + m - m^2 = 1 + m - 6m^2$$ $$5m^2 + 5 = 0 \implies m^2 = -1 \quad (\text{No real solution})$$

Case 2: $$\frac{3 - m}{1 + 3m} = -\left(\frac{1 - 2m}{2 + m}\right) = \frac{2m - 1}{2 + m}$$ $$(3 - m)(2 + m) = (2m - 1)(1 + 3m)$$ $$6 + m - m^2 = 2m + 6m^2 - 1 - 3m$$ $$6 + m - m^2 = 6m^2 - m - 1$$ $$7m^2 - 2m - 7 = 0$$

Using the quadratic formula to solve for $m$:

$$m = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(7)(-7)}}{2(7)} = \frac{2 \pm \sqrt{4 + 196}}{14} = \frac{2 \pm \sqrt{200}}{14} = \frac{2 \pm 10\sqrt{2}}{14} = \frac{1 \pm 5\sqrt{2}}{7}$$

Thus, $m = \frac{1 + 5\sqrt{2}}{7}$ or $m = \frac{1 - 5\sqrt{2}}{7}$.

Exercise - Question 19

Question: If sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always $10$. Show that $P$ must move on a line.

Solution:

The perpendicular distance $d_1$ from point $P(x, y)$ to the line $x + y - 5 = 0$ is given by:

$$d_1 = \frac{|x + y - 5|}{\sqrt{1^2 + 1^2}} = \frac{|x + y - 5|}{\sqrt{2}}$$

The perpendicular distance $d_2$ from point $P(x, y)$ to the line $3x - 2y + 7 = 0$ is given by:

$$d_2 = \frac{|3x - 2y + 7|}{\sqrt{3^2 + (-2)^2}} = \frac{|3x - 2y + 7|}{\sqrt{13}}$$

According to the given condition, the sum of these perpendicular distances is $10$:

$$\frac{|x + y - 5|}{\sqrt{2}} + \frac{|3x - 2y + 7|}{\sqrt{13}} = 10$$

Depending on the quadrant/region in which point $P$ lies, the signs of the expressions inside the absolute values can be removed or adjusted to constants. In any specific region bounded by these lines, the equation simplifies to a linear equation of the general form $Ax + By + C = 0$ (since removing absolute value bars yields linear terms in $x$ and $y$).

Since the locus of point $P$ satisfies a linear equation in $x$ and $y$ of the first degree in each region, $P$ must move on a line (specifically, a set of line segments forming straight paths).

Exercise - Question 20

Question: Find equation of the line which is equidistant from parallel lines $9x + 6y - 7 = 0$ and $3x + 2y + 6 = 0$.

Solution:

First, let us rewrite the given parallel lines so that the coefficients of $x$ and $y$ are identical.

Line 1: $9x + 6y - 7 = 0 \implies 3x + 2y - \frac{7}{3} = 0$

Line 2: $3x + 2y + 6 = 0$

A line parallel to both of these lines will have the same coefficients for $x$ and $y$. Let the equation of the required equidistant line be:

$$3x + 2y + k = 0$$

The perpendicular distance from any point on this line to Line 1 must equal the perpendicular distance to Line 2. Alternatively, the line equidistant from two parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ has a constant term which is the exact average of $C_1$ and $C_2$ (provided the $x, y$ coefficients are scaled identically).

Here, Line 1 is $3x + 2y - \frac{7}{3} = 0$ and Line 2 is $3x + 2y + 6 = 0$.

Thus, the constant term $k$ is:

$$k = \frac{-\frac{7}{3} + 6}{2} = \frac{\frac{-7 + 18}{3}}{2} = \frac{11}{6}$$

Therefore, the equation of the required line is:

$$3x + 2y + \frac{11}{6} = 0$$

Multiplying the entire equation by $6$ to clear fractions:

$$18x + 12y + 11 = 0$$

Exercise - Question 21

Question: A ray of light passing through the point $(1, 2)$ reflects on the $x$-axis at point $A$ and the reflected ray passes through the point $(5, 3)$. Find the coordinates of $A$.

Solution:

Let the coordinates of point $A$ on the $x$-axis be $(x, 0)$.

Let $P = (1, 2)$ be the point through which the incident ray passes, and $Q = (5, 3)$ be the point through which the reflected ray passes.

By the laws of reflection, the angle of incidence equals the angle of reflection. This implies that the path of the reflected ray, if produced backwards, passes through the reflection (image) of point $P$ across the $x$-axis.

Let $P' = (1, -2)$ be the reflection of $P(1, 2)$ across the $x$-axis.

Thus, points $P'(1, -2)$, $A(x, 0)$, and $Q(5, 3)$ are collinear. The slope of the line segment $P'A$ must equal the slope of the line segment $AQ$ (or slope of $P'Q$).

Using the points $P'(1, -2)$ and $Q(5, 3)$, the slope $m$ of the line $P'Q$ is given by:

$$m = \frac{3 - (-2)}{5 - 1} = \frac{3 + 2}{4} = \frac{5}{4}$$

Now, using the points $P'(1, -2)$ and $A(x, 0)$, the slope is:

$$\frac{0 - (-2)}{x - 1} = \frac{2}{x - 1}$$

Equating the two slopes:

$$\frac{2}{x - 1} = \frac{5}{4}$$ $$5(x - 1) = 8$$ $$5x - 5 = 8 \implies 5x = 13 \implies x = \frac{13}{5}$$

Therefore, the coordinates of the point $A$ are $\left(\frac{13}{5}, 0\right)$.

Exercise - Chapter 9 - Question 22

Question: Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^2 - b^2}, 0)$ and $(-\sqrt{a^2 - b^2}, 0)$ to the line $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$ is $b^2$.

Solution:

Let the given line equation be rewritten in standard form:

$$\frac{\cos\theta}{a}x + \frac{\sin\theta}{b}y - 1 = 0$$

The length of the perpendicular from a point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by:

$$p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Here, $A = \frac{\cos\theta}{a}$, $B = \frac{\sin\theta}{b}$, and $C = -1$.

Step 1: Length of perpendicular $p_1$ from point $(\sqrt{a^2 - b^2}, 0)$

Let $c = \sqrt{a^2 - b^2}$. Then $x_1 = c, y_1 = 0$:

$$p_1 = \frac{\left|\frac{\cos\theta}{a}(c) + \frac{\sin\theta}{b}(0) - 1\right|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}} = \frac{\left|\frac{c\cos\theta}{a} - 1\right|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}}$$

Step 2: Length of perpendicular $p_2$ from point $(-\sqrt{a^2 - b^2}, 0)$

Here, $x_1 = -c, y_1 = 0$:

$$p_2 = \frac{\left|\frac{\cos\theta}{a}(-c) + \frac{\sin\theta}{b}(0) - 1\right|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}} = \frac{\left|-\frac{c\cos\theta}{a} - 1\right|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}} = \frac{\left|\frac{c\cos\theta}{a} + 1\right|}{\sqrt{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}}$$

Step 3: Product of the perpendiculars $p_1 p_2$

$$p_1 p_2 = \frac{\left(\frac{c\cos\theta}{a} - 1\right)\left(\frac{c\cos\theta}{a} + 1\right)}{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}} = \frac{\frac{c^2\cos^2\theta}{a^2} - 1}{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}}$$

Since $c = \sqrt{a^2 - b^2}$, we have $c^2 = a^2 - b^2$. Substituting $c^2$ into the numerator:

$$\frac{c^2\cos^2\theta}{a^2} - 1 = \frac{(a^2 - b^2)\cos^2\theta}{a^2} - 1 = \left(1 - \frac{b^2}{a^2}\right)\cos^2\theta - 1 = \cos^2\theta - \frac{b^2}{a^2}\cos^2\theta - 1$$

Using $\cos^2\theta - 1 = -\sin^2\theta$:

$$= -\sin^2\theta - \frac{b^2}{a^2}\cos^2\theta = -\left(\sin^2\theta + \frac{b^2}{a^2}\cos^2\theta\right)$$

Now, substitute this back into the product expression:

$$p_1 p_2 = \frac{-\left(\sin^2\theta + \frac{b^2}{a^2}\cos^2\theta\right)}{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}} = \frac{-b^2\left(\frac{\sin^2\theta}{b^2} + \frac{\cos^2\theta}{a^2}\right)}{\frac{\cos^2\theta}{a^2} + \frac{\sin^2\theta}{b^2}} = b^2$$

Hence proved that the product of the lengths of the perpendiculars is $b^2$.

Exercise - Chapter 9 - Question 23

Question: A person standing at the junction (crossing) of two straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$ wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. Find equation of the path that he should follow.

Solution:

Step 1: Find the coordinates of the junction point.

The junction is the point of intersection of the two paths:

$$2x - 3y + 4 = 0 \quad \text{--- (1)}$$ $$3x + 4y - 5 = 0 \quad \text{--- (2)}$$

Solving equations (1) and (2) simultaneously:

Multiply (1) by 4 and (2) by 3:

$$8x - 12y + 16 = 0$$ $$9x + 12y - 15 = 0$$

Adding the two equations:

$$17x + 1 = 0 \implies x = -\frac{1}{17}$$

Substitute $x = -\frac{1}{17}$ into equation (1):

$$2\left(-\frac{1}{17}\right) - 3y + 4 = 0 \implies -\frac{2}{17} - 3y + 4 = 0 \implies 3y = 4 - \frac{2}{17} = \frac{66}{17} \implies y = \frac{22}{17}$$

Thus, the junction point $P$ is $\left(-\frac{1}{17}, \frac{22}{17}\right)$.

Step 2: Find the path of least time.

To reach the third path $6x - 7y + 8 = 0$ in the least time, the person must travel along the perpendicular distance from the junction point $P$ to the third line.

The slope of the line $6x - 7y + 8 = 0$ is $m_1 = \frac{6}{7}$.

The path to be followed is perpendicular to this line, so its slope $m_2$ is the negative reciprocal of $m_1$:

$$m_2 = -\frac{1}{m_1} = -\frac{7}{6}$$

Step 3: Equation of the required path.

Using the point-slope form with point $P\left(-\frac{1}{17}, \frac{22}{17}\right)$ and slope $m = -\frac{7}{6}$:

$$y - \frac{22}{17} = -\frac{7}{6}\left(x - \left(-\frac{1}{17}\right)\right)$$ $$6\left(\frac{17y - 22}{17}\right) = -7\left(\frac{17x + 1}{17}\right)$$ $$6(17y - 22) = -7(17x + 1)$$ $$102y - 132 = -119x - 7$$ $$119x + 102y - 132 + 7 = 0$$ $$119x + 102y - 125 = 0$$

Therefore, the equation of the path that he should follow is $119x + 102y - 125 = 0$.