Q1. Find the distance between the following pairs of points:

(i) (2, 3), (4, 1)

Solution: By Distance Formula, d = √[(4 - 2)² + (1 - 3)²] = √[2² + (-2)²] = √[4 + 4] = √8 = 2√2 units.

(ii) (-5, 7), (-1, 3)

Solution: d = √[(-1 - (-5))² + (3 - 7)²] = √[4² + (-4)²] = √[16 + 16] = √32 = 4√2 units.

(iii) (a, b), (-a, -b)

Solution: d = √[(-a - a)² + (-b - b)²] = √[(-2a)² + (-2b)²] = √[4a² + 4b²] = 2√(a² + b²) units.

Q2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2?

Solution: Distance between (0, 0) and (36, 15):
d = √[(36 - 0)² + (15 - 0)²] = √[1296 + 225] = √1521 = 39 units.
Yes, the distance between town A and town B is 39 km.

Q3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.

Solution: Let A(1, 5), B(2, 3) and C(-2, -11).
AB = √[(2 - 1)² + (3 - 5)²] = √[1 + 4] = √5
BC = √[(-2 - 2)² + (-11 - 3)²] = √[16 + 196] = √212 = 2√53
AC = √[(-2 - 1)² + (-11 - 5)²] = √[9 + 256] = √265
Since AB + BC ≠ AC, the given points are not collinear.

Q4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

Solution: Let A(5, -2), B(6, 4), and C(7, -2).
AB = √[(6 - 5)² + (4 - (-2))²] = √[1 + 36] = √37
BC = √[(7 - 6)² + (-2 - 4)²] = √[1 + 36] = √37
AC = √[(7 - 5)² + (-2 - (-2))²] = √[4 + 0] = 2
Since AB = BC = √37, ABC is an isosceles triangle.

Q5. In a classroom, 4 friends are seated at A(3, 4), B(6, 7), C(9, 4), D(6, 1). Is ABCD a square?

Solution: Calculating side lengths:
AB = √[(6 - 3)² + (7 - 4)²] = √[9 + 9] = 3√2
BC = √[(9 - 6)² + (4 - 7)²] = √[9 + 9] = 3√2
CD = √[(6 - 9)² + (1 - 4)²] = √[9 + 9] = 3√2
DA = √[(3 - 6)² + (4 - 1)²] = √[9 + 9] = 3√2
Diagonals:
AC = √[(9 - 3)² + (4 - 4)²] = √36 = 6
BD = √[(6 - 6)² + (1 - 7)²] = √36 = 6
Since all four sides are equal and diagonals are equal, ABCD is a square. Champa is correct.

Q6. Name the type of quadrilateral formed, if any, by the following points:

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
Solution: All sides equal √8 = 2√2 and diagonals equal 4. It forms a Square.

(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)
Solution: Distance AB = √52, BC = √13, AC = √13. Since AC + BC = AB, points A, B, C are collinear. No quadrilateral is formed.

(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution: Opposite sides equal (√10 and √18 = 3√2), but diagonals unequal (2 and √52). It forms a Parallelogram.

Q7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).

Solution: Any point on x-axis is P(x, 0).
Distance from (2, -5) = Distance from (-2, 9)
(x - 2)² + (0 - (-5))² = (x - (-2))² + (0 - 9)²
x² - 4x + 4 + 25 = x² + 4x + 4 + 81
-8x = 56 ⇒ x = -7.
The required point is (-7, 0).

Q8. Find the values of y for which the distance between P(2, -3) and Q(10, y) is 10 units.

Solution: PQ = 10 ⇒ PQ² = 100
(10 - 2)² + (y - (-3))² = 100
64 + (y + 3)² = 100
(y + 3)² = 36 ⇒ y + 3 = ±6
y = 3 or y = -9.

Q9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.

Solution: QP² = QR² ⇒ (5 - 0)² + (-3 - 1)² = (x - 0)² + (6 - 1)²
25 + 16 = x² + 25 ⇒ x² = 16 ⇒ x = ±4.
For x = 4: QR = √41, PR = √82.
For x = -4: QR = √41, PR = 9√2.

Q10. Find a relation between x and y such that the point (x, y) is equidistant from (3, 6) and (-3, 4).

Solution: (x - 3)² + (y - 6)² = (x + 3)² + (y - 4)²
x² - 6x + 9 + y² - 12y + 36 = x² + 6x + 9 + y² - 8y + 16
-12x - 4y + 20 = 0 ⇒ 3x + y - 5 = 0.

Q1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2 : 3.

Solution: By Section Formula:
x = (2 × 4 + 3 × (-1)) / (2 + 3) = (8 - 3)/5 = 1
y = (2 × (-3) + 3 × 7) / (2 + 3) = (-6 + 21)/5 = 3
The point is (1, 3).

Q2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Solution: Point P divides in ratio 1 : 2:
P = ((1(-2) + 2(4))/3, (1(-3) + 2(-1))/3) = (2, -5/3)
Point Q divides in ratio 2 : 1:
Q = ((2(-2) + 1(4))/3, (2(-3) + 1(-1))/3) = (0, -7/3).

Q3. Sports Day Activity: Niharika runs 1/4th distance AD on 2nd line, Preet runs 1/5th distance AD on 8th line. Total AD = 100m.

Solution: Niharika's position = (2, 25). Preet's position = (8, 20).
Distance = √[(8 - 2)² + (20 - 25)²] = √[36 + 25] = √61 m.
Rashmi's flag position (Midpoint) = ((2 + 8)/2, (25 + 20)/2) = (5, 22.5) i.e. 5th line at 22.5m distance.

Q4. Find the ratio in which the line segment joining (-3, 10) and (6, -8) is divided by (-1, 6).

Solution: Let ratio be k : 1.
-1 = (6k - 3)/(k + 1) ⇒ -k - 1 = 6k - 3 ⇒ 7k = 2 ⇒ k = 2/7.
The ratio is 2 : 7.

Q5. Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution: Point on x-axis has y = 0. Let ratio be k : 1.
0 = (5k - 5)/(k + 1) ⇒ 5k = 5 ⇒ k = 1. Ratio is 1 : 1.
Coordinates of division point = ((1(-4) + 1(1))/2, 0) = (-3/2, 0).

Q6. If (1, 2), (4, y), (x, 6) and (3, 5) are vertices of a parallelogram taken in order, find x and y.

Solution: Diagonals of parallelogram bisect each other.
Midpoint of AC = Midpoint of BD
((1 + x)/2, (2 + 6)/2) = ((4 + 3)/2, (y + 5)/2)
(1 + x)/2 = 7/2 ⇒ x = 6.
8/2 = (y + 5)/2 ⇒ y = 3.

Q7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Solution: Centre is the midpoint of AB. Let A = (x, y).
(x + 1)/2 = 2 ⇒ x = 3
(y + 4)/2 = -3 ⇒ y = -10
Coordinates of A are (3, -10).

Q8. If A(-2, -2) and B(2, -4), find coordinates of P such that AP = 3/7 AB and P lies on AB.

Solution: AP : PB = 3 : 4.
x = (3(2) + 4(-2))/(3 + 4) = -2/7
y = (3(-4) + 4(-2))/(3 + 4) = -20/7
Coordinates of P are (-2/7, -20/7).

Q9. Find coordinates of points dividing the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Solution: Midpoint of AB (P₂) = (( -2 + 2)/2, (2 + 8)/2) = (0, 5).
Midpoint of AP₂ (P₁) = ((-2 + 0)/2, (2 + 5)/2) = (-1, 7/2).
Midpoint of P₂B (P₃) = ((0 + 2)/2, (5 + 8)/2) = (1, 13/2).

Q10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

Solution: Diagonal d₁ (AC) = √[(-1 - 3)² + (4 - 0)²] = √[16 + 16] = √32 = 4√2 units.
Diagonal d₂ (BD) = √[(-2 - 4)² + (-1 - 5)²] = √[36 + 36] = √72 = 6√2 units.
Area = 1/2 × d₁ × d₂ = 1/2 × 4√2 × 6√2 = 24 sq units.