If the equation to the locus of points equidistant from the points $(-2,3),(6,-5)$ is $a x+b y+c=0$, where…
If the equation to the locus of points equidistant from the points $(-2,3),(6,-5)$ is $a x+b y+c=0$, where $a>0$, then the ascending order of $a, b, c$ is
$a,b,c$
$c,b,a$
$b,c,a$
$a,c,b$
Solution
Let $P(x, y)$ be the required point whose locus is given by $a x+b y+c=0$
Also given that $P$ is equidistant from $A(-2,3)$ and $B(6,-5)$. Then,
$\begin{aligned}
& P A=P B \text { or } P A^2=P B^2 \\
& (x+2)^2+(y-3)^2=(x-6)^2+(y+5)^2 \\
& x^2+4 x+4+y^2-6 y+9 \\
& =x^2-12 x+36+y^2+10 y+25 \\
& 16 x-16 y-48=0 \\
& x-y-3=0
\end{aligned}$
On comparing with $a x+b y+c=0$, we get $a=1, b=-1, c=-3$
Hence, the ascending order is $c, b, a$.