Question

If $A\equiv (-6,0),B\equiv (3,-3)$ and $C\equiv (5,3)$ are three points, then the locus of the point $P$ such that $|\overline{AP}{|}^2+|\overline{BP}{|}^2=2|\overline{CP}{|}^2$ is

• A. $y=-\frac{7}{9}x-\frac{13}{9}$
• B. $y=-\frac{13}{9}x+\frac{7}{9}$
• C. $y=\frac{13}{9}x+\frac{7}{9}$
• D. $y=\frac{7}{9}x-\frac{13}{9}$

Answer 1

The correct option is B $y=-\frac{13}{9}x+\frac{7}{9}$

Let $P(x,y)$ be an arbitrary point.
Now, $|\overline{AP}{|}^2+|\overline{BP}{|}^2=2|\overline{CP}{|}^2$
$\Rightarrow [(x+6{)}^2+y^2]+[(x-3{)}^2+(y+3{)}^2]=2[(x-5{)}^2+(y-3{)}^2]\Rightarrow 26x+18y=14\Rightarrow y=-\frac{13}{9}x+\frac{7}{9}$

Hence, the locus of the point $P$ is $y=-\frac{13}{9}x+\frac{7}{9}$