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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), find the coordinates of P.

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Question

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

 

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Solution

 

Let the y-coordinate of the point P be a.

Then, its x-coordinate will be 2a.

Thus, the coordinates of the point P are (2a, a).

t is given that the point P (2a, a) is equidistant from Q (2, −5) and R (−3, 6).

Thus, we have

`sqrt((2a-2)^2+(a-(-5)^2))=sqrt((2a-(-3)^2)+(a-6)^2)`

`=>sqrt((2a-2)^2+(a+5)^2)=sqrt((2a+3)^2+(a-6)^2)`

`sqrt(4a^2+4-8a+a^2+25+10a)=sqrt(4a^2+9+12a+a^2+36-12a)`

 `=>sqrt(5a^2+2a+29)=sqrt(5a^2+45)`

 Squaring both sides, we get

5a2+2a+29=5a2+45

5a2+2a5a2=4529

2a=16

a=8

Thus, the coordinates of the point P are (16, 8), i.e. (2 × 8, 8).

 
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2015-2016 (March) Delhi Set 1

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