The x-coordinate of a point $P$ is twice its y-coordinate. If $P$ is equidistant from $Q (2, - 5)$ and $R (- 3, 6),$ then find the coordinates of $P$ .

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Solution

Let the coordinates of P be (x,y). It is given that x=2y
Also,
PQ=PR
$\Rightarrow \sqrt{{(x - 2)}^{2} + {(y + 5)}^{2}} = \sqrt{{(x + 3)}^{2} + {(y - 6)}^{2}}$
$\Rightarrow \sqrt{{(2 y - 2)}^{2} + {(y + 5)}^{2}} = \sqrt{{(2 y + 3)}^{2} + {(y - 6)}^{2}}$
$\Rightarrow \sqrt{{5 y}^{2} + 2 y + 29} = \sqrt{{5 y}^{2} + 45}$
$\Rightarrow {5 y}^{2} + 2 y + 29 = {5 y}^{2} + 45$
$\Rightarrow 2 y = 16$
$\Rightarrow y = 8$
Hence, the coordinates of P are $(16, 8)$ .

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The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2,−5) and R(−3,6), then find the coordinates of P.

Let the coordinates of P be (x,y). It is given that x=2yAlso, PQ=PR⇒√(x−2)2+(y+5)2=√(x+3)2+(y−6)2⇒√(2y−2)2+(y+5)2=√(2y+3)2+(y−6)2⇒√5y2+2y+29=√5y2+45⇒5y2+2y+29=5y2+45⇒2y=16⇒y=8Hence, the coordinates of P are (16,8).

The x-coordinate of a point $P$ is twice its y-coordinate. If $P$ is equidistant from $Q (2, - 5)$ and $R (- 3, 6),$ then find the coordinates of $P$ .