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(d)  Let the equation of perpendicular bisector $FN$ of $AB$ is $x - y + 5 = 0$ ......$(i)$

The middle point $F$ of $AB$ is $\left( {\frac{{{x

Vedclass · Accepted Answer

$14x + 23y - 40 = 0$

Vedclass · Answer

(d)  Let the equation of perpendicular bisector $FN$ of $AB$ is $x - y + 5 = 0$ ......$(i)$

The middle point $F$ of $AB$ is $\left( {\frac{{{x_1} + 1}}{2},\frac{{{y_1} - 2}}{2}} ight)$ lies on line $(i)$. Therefore ${x_1} - {y_1} = - 13$&hellip;..$(ii)$

Also $AB$ is perpendicular to $FN$. So the product of their slopes is $-1$.

i.e. $\frac{{{y_1} + 2}}{{{x_1} - 1}} 	imes 1 = - 1$or ${x_1} + {y_1} = - 1$&hellip;&hellip;$(iii)$

On solving $(ii)$ and $(iii)$, we get $B( - 7,6)$.

Similarly $C{m{ }}\left( {\frac{{11}}{5},\frac{2}{5}} ight)$.

Hence the equation of $BC$ is $14x + 23y - 40 = 0$.

The equation of perpendicular bisectors of the sides $AB$ and $AC$ of a triangle $ABC$ are $x - y + 5 = 0$ and $x + 2y = 0$ respectively. If the point $A$ is $(1,\; - \;2)$, then the equation of line $BC$ is

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