Find the point equidistant from the points P(1,3) and Q(-5,-5), and which lies on the line $y = - 1$ .

(-2, -2)

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(-2, -1)

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(-4, -1)

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(3, -1)

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Solution

The correct option is B
(-2, -1)

The required point is equidistant from
the points $P$ and $Q$ , and also lies on
the line $y = - 1$ .

Let the point be $A$ . Since it lies on $y = - 1$ ,
its y coordinate will be -1. Thus, the
coordinates of point $A$ will be $(x, - 1)$ .

Distance $A P$ = Distance $A Q$

$\Rightarrow {A P}^{2} = {A Q}^{2}$

$\Rightarrow (x - 1)^{2} + (- 1 - 3)^{2} = (x - (- 5))^{2} + (- 1 - (- 5))^{2} \Rightarrow (x - 1)^{2} + 16 = (x + 5)^{2} + 16 \Rightarrow x^{2} - 2 x + 1 + 16 = x^{2} + 10 x + 25 + 16 \Rightarrow - 2 x + 17 = 10 x + 41 \Rightarrow x = - 2$

Thus, the coordinates of point $A$ will be (-2,-1).

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Find the point equidistant from the points P(1,3) and Q(-5,-5), and which lies on the line y=−1.

Find the point equidistant from the points P(1,3) and Q(-5,-5), and which lies on the line $y = - 1$ .