Class 10 - Chapter 7: Coordinate Geometry

Exercise 7.1

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

Distance Formula:

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

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

d = √[(4 - 2)² + (1 - 3)²]

= √[2² + (-2)²]

= √(4 + 4)

= √8

= 2√2 units

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

d = √[(-1 + 5)² + (3 - 7)²]

= √[4² + (-4)²]

= √(16 + 16)

= √32

= 4√2 units

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

d = √[(-a - a)² + (-b - b)²]

= √[(-2a)² + (-2b)²]

= √(4a² + 4b²)

= 2√(a² + b²)

Answer: 2√(a² + b²) units

Q2. Find the distance between the points (0, 0) and (36, 15).

d = √[(36 - 0)² + (15 - 0)²]

= √(36² + 15²)

= √(1296 + 225)

= √1521

= 39 units

Therefore, the distance between the two towns A and B is 39 km.

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

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 = √5 + 2√53 = √265

Therefore,

AB + BC = AC

Hence, the three points are collinear.

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

Let A(5,-2), B(6,4), 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,

the triangle has two equal sides.

Therefore, the given points form an isosceles triangle.

Q5. In the classroom, determine whether ABCD is a square.

From the figure:

A(3,4), B(6,7), C(9,4), D(6,1)

Find all sides

AB = √[(6-3)² + (7-4)²]

= √(9 + 9)

= √18 = 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

Find diagonals

AC = √[(9-3)² + (4-4)²]

= √36

= 6

BD = √[(6-6)² + (1-7)²]

= √36

= 6

All four sides are equal and both diagonals are equal.

Therefore, ABCD is a square.

Hence, Champa is correct.

Q6. Name the type of quadrilateral formed by the following points.

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)

Let the vertices be A, B, C and D respectively.

AB = BC = CD = DA = 2√2

Also,

AC = 4 and BD = 4

All sides are equal and diagonals are equal.

Therefore, the quadrilateral is a Square.

(ii) (-3, 5), (3, 1), (0, 3), (-1, -4)

On calculating the side lengths, no pair of opposite sides is equal and the conditions for square, rectangle, rhombus or parallelogram are not satisfied.

Therefore, it forms an irregular quadrilateral.

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

AB = √10

BC = √18

CD = √10

DA = √18

Opposite sides are equal.

Therefore, the quadrilateral is a Parallelogram.

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

Let the required point be P(x, 0).

Since P is equidistant from both points,

Distance from P to (2, -5) = Distance from P to (-2, 9)

√[(x-2)² + (0+5)²] = √[(x+2)² + (0-9)²]

Squaring both sides,

(x-2)² + 25 = (x+2)² + 81

x² - 4x + 4 + 25 = x² + 4x + 4 + 81

29 - 4x = 85 + 4x

-8x = 56

x = -7

Therefore, the required point is:

(-7, 0)

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

Using the distance formula:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

Given:

P(2, −3), Q(10, y)

Distance PQ = 10 units

Therefore,

√[(10 − 2)² + (y + 3)²] = 10

√[8² + (y + 3)²] = 10

√[64 + (y + 3)²] = 10

Squaring both sides,

64 + (y + 3)² = 100

(y + 3)² = 36

y + 3 = ±6

Case 1:

y + 3 = 6

y = 3

Case 2:

y + 3 = −6

y = −9

Answer: y = 3 or y = −9

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

Since Q is equidistant from P and R,

QP = QR

Step 1: Find QP

QP = √[(5 − 0)² + (−3 − 1)²]

= √[25 + 16]

= √41

Step 2: Find QR

QR = √[(x − 0)² + (6 − 1)²]

= √(x² + 25)

Since QP = QR,

√41 = √(x² + 25)

Squaring both sides,

41 = x² + 25

x² = 16

x = ±4

Case 1: x = 4

R = (4, 6)

QR = √41

PR = √[(4 − 5)² + (6 + 3)²]

= √[1 + 81]

= √82

Case 2: x = −4

R = (−4, 6)

QR = √41

PR = √[(−4 − 5)² + (6 + 3)²]

= √[81 + 81]

= √162

= 9√2

Answer:

x = 4 or x = −4

QR = √41 units

PR = √82 units (when x = 4)

PR = 9√2 units (when x = −4)

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

Let P(x, y) be equidistant from A(3, 6) and B(−3, 4).

Therefore,

PA = PB

Using the distance formula,

√[(x − 3)² + (y − 6)²] = √[(x + 3)² + (y − 4)²]

Squaring both sides,

(x − 3)² + (y − 6)² = (x + 3)² + (y − 4)²

x² − 6x + 9 + y² − 12y + 36 = x² + 6x + 9 + y² − 8y + 16

−6x − 12y + 45 = 6x − 8y + 25

−12x − 4y + 20 = 0

Dividing by −4,

3x + y − 5 = 0

Required relation:

3x + y = 5