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M(-5,3) and N(2,4) are two fixed points . Find the locus of a point P so that PM:PN=3:2

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Answer:

Step-by-step explanation:

let (x,y) be coordinates of P.

[tex]PM=\sqrt{(-5-x)^2+(3-y)^2} \\PN=\sqrt{(2-x)^2+(4-y)^2} \\\frac{PM}{PN} =\frac{\sqrt{(-5 -x)^2 +(3-y)^2} }{\sqrt{(2-x)^2+(4-y)^2 } } =\frac{3}{2}\\4[(-5-x)^2+(3-y)^2]=9[(2-x)^2+(4-y)^2]\\4[25+x^2+10x+9+y^2-6y]=9[4+x^2-4x+16+y^2-8y] \\4x^2+4y^2+40x-24y+100+36=9x^2+9y^2-36x-72y+36+144\\4x^2-9x^2+4y^2-9y^2+40x+36x-24y+72y+100+36-36-144=0\\-5x^2-5y^2+76x+48y-44=0\\5x^2+5y^2-76x-48y+44=0[/tex]