Find the point on X-axis which is equidistant from P(2,-5) and Q(-2,9).

Find the point on X-axis which is equidistant from P(2,-5) and Q(-2,9).

Answer:-

Let the point on x-axis equidistant from P(2,–5) and Q(–2,9) be [Aleft( x, 0 right)].

[AP = sqrt{left( x – 2 right)^2 + left( 0 – left( – 5 right) right)^2} = sqrt{left( x – 2 right)^2 + 25}]

[QA = sqrt{left( x – left( – 2 right) right)^2 + left( 0 – 9 right)^2} = sqrt{left( x + 2 right)^2 + 81}]

[AP = QA]

[ Rightarrow sqrt{left( x – 2 right)^2 + 25} = sqrt{left( x + 2 right)^2 + 81}]

Squaring both sides

[left( x – 2 right)^2 + 25 = left( x + 2 right)^2 + 81]

[ Rightarrow x^2 + 4 – 4x + 25 = x^2 + 4 + 4x + 81]

[ Rightarrow – 8x = 56]

[ Rightarrow x = – 7]

Thus, the required point is [left( – 7, 0 right)].

Explanation:-

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

Then, using the distance formula, we have:

distance from P to (x, 0) = distance from Q to (x, 0)

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

Simplifying this equation, we get:

(x – 2)² + 25 = (x + 2)² + 81

Expanding and simplifying, we get:

x² – 4x – 56 = 0

Factoring this quadratic equation, we get:

(x – 8)(x + 7) = 0

Therefore, the solutions are:

x = 8 or x = -7.

The point on the x-axis which is equidistant from P and Q is either (8, 0) or (-7, 0).

Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)