Class 10 Maths Chapter 7 Question Answers - Coordinate Geometry

Coordinate Geometry

Q1. Find a point on the y-axis equidistant from (− 5, 2) and (9, − 2).

Let the required point on the y-axis be P (0, y)

∴ PA = PB

⇒ √((0 + 5)2 + (y − 2)2) = √((0 − 9)2 + (y + 2)2)
⇒ √(52 + y2 − 4y + 4 − 4y) = √(92 + y2 + 4 + 4y + 4y)

⇒ 25 + y2 − 4y = 81 + y2 + 4 + 4y

⇒ y² − y² − 4y − 4y = 81 + 4 − 4 − 25
⇒ − 8y = 85 − 29
⇒ − 8y = 56

⇒ y-8 = 56-8 = −7

∴ The required point is (0, −7).

Q2. Find a point on x-axis at a distance of 4 units from point A (2, 1).

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

∴ PA = 4

⇒ √((x − 2)2 + (0 − 1)2) = 4

⇒ x² − 4x + 4 + 1 = 4² = 16
⇒ x² − 4x + 1 + 4 − 16 = 0
⇒ x²− 4x − 11 = 0

⇒ x = 2 ± √15

Q3. Find the distance of the point (3, − 4) from the origin.

The coordinates of origin (0, 0).
∴ Distance of (3, − 4) from the origin

= √((3 − 0)2 + (−4 − 0)2)
= √((3)2 + (−4)2)
= √(9 + 16) = √25 = 5

Q4. For what value of x is the distance between the points A (− 3, 2) and B (x, 10) 10 units?

The distance between A (− 3, 2) and B (x, 10)

= √((x + 3)2 + (10 − 2)2)
∴ √((x + 3)2 + (8)2) = 10

⇒ (x + 3)² + (8)² = 10²
⇒ (x + 3)² = 10² − 8²
⇒ (x + 3)²= (10 − 8) (10 + 8) = 36

⇒ x + 3 = ±√36 = ±6

For +ve sign, x =6 − 3 = 3
For −ve sign, x = − 6 − 3 = − 9

Q5. Find a point on the x-axis which is equidistant from points A (5, 2) and B (1, − 2).

The given points are: A (5, 2) and B (1, − 2) Let the required point on the x-axis be C (x, 0).
Since, C is equidistant from A and B.
∴ AC = BC

∴ √((x − 5)2 + (0 − 2)2) = √((x − 1)2 + (0 + 2)2)
⇒ (x − 5)2 + (−2)2 = (x − 1)2 + (2)2
⇒ x2 + 25 − 10x + 4 = x2 + 1 − 2x + 4
⇒ −10x + 2x = 5 − 29
⇒ −8x = −24
⇒ x = −24−8 = 3

∴ The required point is (0, 3).

Q6. Establish the relation between x and y when P (x, y) is equidistant from the points A (− 1, 2) and B (2, − 1).

As P is equidistant from A and B
∴ PA = PB

√((x + 1)2 + (y − 2)2) = √((x − 2)2 + (y + 1)2)
⇒ (x + 1)2 + (y − 2)2 = (x − 2)2 + (y + 1)2
⇒ x2 + 1 + 2x + y2 − 4y + 4 = x2 + 4 − 4x + y2 + 1 + 2y
⇒ 2x − 4x + 5 = −4x + 2y + 5
⇒ 2x + 4x + 5 = 2y + 4y + 5
⇒ 6x = 6y
⇒ x = y

which is the required relation.

Q7. Find a relation between x and y such that the point P (x, y) is equidistant from the points A (−5, 3) and B (7, 2) 

Since, P (x, y) is equidistant from A (−5, 3) and B (7, 2)
∴ AP = BP ⇒ √((x + 5)2 + (y − 3)2) = √((x − 7)2 + (y − 2)2)
⇒ (x + 5)2 + (y − 3)2 = (x − 7)2 + (y − 2)2
⇒ x2 + 10x + 25 + y2 − 6y + 9 = x2 − 14x + 49 + y2 − 4y + 4
⇒ 10x − 6y + 34 = −14x − 4y + 53
⇒ 10x + 14x − 6y + 4y = 53 − 34
⇒ 24x − 2y = 19
Thus, 24x − 2y = 19 is the required relation

Q8. In the given figure, ABC is a triangle. D and E are the mid points of the sides BC and AC respectively. Find the length of DE. Prove that  DE = 12 AB

Coordinates of the midpoint of BC are:

= ( -6 + 22, -1 + (-2)2 )

= ( -2, -32 ) ⇒ E ( -2, -32 )

Coordinates of the midpoint of AC are:

= ( -6 + 42, -1 + (-2)2 )

= ( -1, -32 ) ⇒ D ( -1, -32 )

Now,

DE = √( -2 - 12 )² + ( -32 + 32 )²

= √( -11 )² + 0 = 1

AB = √( 4 - 22 )² + ( -2 + 22 )² = 2

Hence, DE = 12 AB

Q9. Find the distance between the points   ( -85 , 2 ) and ( 25 , 2 )

Distance between ( -85 , 2 ) and ( 25 , 2 ) is given by:

√[ ( 25 + 85 )² + (2 - 2)²]

= √(2² - 0²) = 2 units.

Q10. If the mid point of the line joining the points P (6, b − 2) and Q (− 2, 4) is (2, − 3), find the value of b.

Here, P (6, b − 2) and Q (− 2, 4) are the given points.
 ∴ Mid point of PQ is given by:

[ 6 + (-2)2 , 4 + b - 22 ]

or [ 6 - 22 , 4 - 2 + b2 ]

or [ 2, 2 + b2 ]

∴ 2 + b2 = -3 ⇒ 2 + b = -6

⇒ b = -6 - 2

⇒ b = -8