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# The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from $Q\left( {2, - 5} \right)$ and $R\left( { - 3,6} \right)$, then find the coordinates of P.  Verified
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Hint: Use the distance formula $D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$ to compare the distance of P from two points and find the coordinates.

Let the coordinate of point P is $\left( {x,y} \right)$.

According to the question, the x-coordinate of P is twice its y-coordinate. Then we have:
$\Rightarrow x = 2y$

So, the coordinates of P will be $\left( {2y,y} \right)$.

We know that the distance between two point $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ can be found using distance formula:
$D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
Using this, the distance between points $P\left( {2y,y} \right)$ and point $Q\left( {2, - 5} \right)$ is:
$\Rightarrow PQ = \sqrt {{{\left( {2y - 2} \right)}^2} + {{\left( {y + 5} \right)}^2}}$
And the distance between points $P\left( {2y,y} \right)$ and point $R\left( { - 3,6} \right)$ is:
$\Rightarrow PR = \sqrt {{{\left( {2y - 3} \right)}^2} + {{\left( {y + 6} \right)}^2}}$

Given in the question, P is equidistant from Q and R. So, we have:
$\Rightarrow PQ = PR \\ \Rightarrow \sqrt {{{\left( {2y - 2} \right)}^2} + {{\left( {y + 5} \right)}^2}} = \sqrt {{{\left( {2y - 3} \right)}^2} + {{\left( {y + 6} \right)}^2}} \\ \Rightarrow 4{y^2} + 4 - 8y + {y^2} + 25 + 10y = 4{y^2} + 9 - 12y + {y^2} + 36 + 12y \\ \Rightarrow 2y + 29 = 45 \\ \Rightarrow 2y = 16 \\ \Rightarrow y = 8 \\$

So, the coordinate of point P is $\left( {16,8} \right)$.

Note: In the above question, if in such cases the slope PQ and PR also comes out to be equal then, P, Q and R will lie on the same straight line i.e. they’ll be collinear and P will be the mid-point of Q and R.