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  4. Two equal sides of an isosceles triangle are given by the equations 7 x-y+3=0 and x+y-3=0 and its third side passes through the point (1,-10). The equation of the third side can be

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Q. Two equal sides of an isosceles triangle are given by the equations $7 x-y+3=0$ and $x+y-3=0$ and its third side passes through the point $(1,-10)$. The equation of the third side can be

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Solution:

$m _{ AB }=7$ and $m _{ AC }=-1$
Let slope of $B C$ (third side) be $m$.
Now $\left|\frac{7- m }{1+7 m }\right|=\left|\frac{-1- m }{1- m }\right| \Rightarrow\left(\frac{7- m }{1+7 m }\right)= \pm\left(\frac{-1- m }{1- m }\right)$
image
$\therefore $ Taking ' + ' sign, we get
$m ^2+1=0$ (no real values of $m$ possible)
And taking'-' sign, we get
$3 m ^2+8 m -3=0 \Rightarrow(3 m -1)( m +3)=0 \Rightarrow m =\frac{1}{3},-3$
$\therefore $ Required equation of third side can be $x-3 y-31=0$ or $3 x+y+7=0$ Hence, $a=3, b=1$
$\therefore\left( a ^2- b ^2\right)=9-1=8$.