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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+2a−5a2=45−29
⇒2a=16
⇒a=8
Thus, the coordinates of the point P are (16, 8), i.e. (2 × 8, 8).
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