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).
Let the point on x-axis equidistant from P(2,–5) and Q(–2,9) be \[A\left( x, 0 \right)\].
\[AP = \sqrt{\left( x - 2 \right)^2 + \left( 0 - \left( - 5 \right) \right)^2} = \sqrt{\left( x - 2 \right)^2 + 25}\]
\[QA = \sqrt{\left( x - \left( - 2 \right) \right)^2 + \left( 0 - 9 \right)^2} = \sqrt{\left( x + 2 \right)^2 + 81}\]
\[AP = QA\]
\[ \Rightarrow \sqrt{\left( x - 2 \right)^2 + 25} = \sqrt{\left( x + 2 \right)^2 + 81}\]
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)