Question

Find the point on x-axis which is equidistant from p(2,-5) and q(-2,9) solution

Answer 1

Final Answer : (-7,0)

Logic Used :
1) Distance between two points :
$\sqrt{ {(x1 - x2)}^{2} + {(y1 - y2)}^{2} }$
where x, y are respective points.

Steps:
1) Let the point on x-axis be A (a, 0) .
Then, According to the Question

AP = AQ
=> AP^2 = AQ^2

$= > {(a - 2)}^{2} + {5}^{2} = {(a + 2)}^{2} + {9}^{2} \\ = > - 4 \times 2a = 4 \times 14 \\ = > a = - 7$

Hence, The Required point on x -axis is (-7,0) .

Answer 2

Answer:

Final Answer : (-7,0)

Logic Used :

1) Distance between two points :

where x, y are respective points.

Steps:

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

Then, According to the Question

AP = AQ

=> AP^2 = AQ^2

Hence, The Required point on x -axis is (-7,0) .

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