The point equidistant from P (1, 3) and Q (-5, -5) and lying on the line $y = - 1$ is .

(-2 , -1)

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(-4 , -1)

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(-3 , -1)

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Solution

The correct option is A (-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 AP = Distance AQ

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

$\Rightarrow (x - 1)^{2} + (- 1 - 3)^{2} = (x - (- 5))^{2} + (- 1 - (- 5))^{2} \Rightarrow (x - 1)^{2} + 16 = (x + 5)^{2} + 16 \Rightarrow x^{2} - 2 x + 1 + 16 = x^{2} + 10 x + 25 + 16 \Rightarrow - 2 x + 17 = 10 x + 41 \Rightarrow - 12 x = 24 \Rightarrow x = - 2$

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

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Similar questions

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The point equidistant from P (1, 3) and Q (-5, -5) and lying on the line y=−1 is .

The point equidistant from P (1, 3) and Q (-5, -5) and lying on the line $y = - 1$ is .