If the equation to the locus of points equidi | vedclass

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(B) Let $P(x, y)$ be the point whose locus is given by $a x + b y + c = 0$. Since $P$ is equidistant from $A(-2, 3)$ and $B(6, -5)$,we have $PA = PB$,

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$c, b, a$

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(B) Let $P(x, y)$ be the point whose locus is given by $a x + b y + c = 0$. Since $P$ is equidistant from $A(-2, 3)$ and $B(6, -5)$,we have $PA = PB$,which implies $PA^2 = PB^2$. $(x + 2)^2 + (y - 3)^2 = (x - 6)^2 + (y + 5)^2$ $x^2 + 4x + 4 + y^2 - 6y + 9 = x^2 - 12x + 36 + y^2 + 10y + 25$ $4x + 4 - 6y + 9 = -12x + 36 + 10y + 25$ $16x - 16y - 48 = 0$ Dividing by $16$,we get $x - y - 3 = 0$. Comparing this with $a x + b y + c = 0$,we find $a = 1$,$b = -1$,and $c = -3$. Since $-3 < -1 < 1$,the ascending order is $c, b, a$.

If the equation to the locus of points equidistant from the points $(-2, 3)$ and $(6, -5)$ is $a x + b y + c = 0$,where $a > 0$,then the ascending order of $a, b, c$ is

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