A point P os equi distant from R(-2,4) and S(6,-4) and its x co ordinate is twice its Y co ordinate find co ordinates of P

Answer: [tex]P(-4,-2)[/tex]

Step-by-step explanation: Distance between two points is calculated as

[tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]

Equidistant means the same distance, i.e., the distance between points P and R is equal to the distance between points P and S.

Suppose P's y-coordinate is a. If its x-coordinate is double, then

(x,y) = (2a,a)

Thus

[tex]\sqrt{(2a+2)^{2}+(a-4)^{2}} =\sqrt{(2a-6)^{2}+(a+4)^{2}}[/tex]

Solving

[tex]4a^{2}+8a+4+a^{2}-8a+16=4a^{2}-24a+36+a^{2}+8a+16[/tex]

[tex]5a^{2}+20=5a^{2}-16a+52[/tex]

[tex]-16a=32[/tex]

[tex]a=-2[/tex]

y-coordinate is [tex]-2[/tex], then x-ccordinate is

2a = 2(-2) = [tex]-4[/tex]

Coordinates of point P equidistant for R and S is [tex](-4,-2)[/tex].

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