Question

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

• A. $3x+y+7=0$
• B. $x-3y+29=0$
• C. $3x+y+3=0$
• D. $3x+y-3=0$

Answer 1

The correct option is A $3x+y+7=0$

Given Sides

$7x-y+3=0$------(1)

$x+y-3=0$-----(2)

Let slope of $7x-y+3=0$ is $m_1$

$m_1=7$

and slope of $x+y-3=0$ is $m_2$

$m_2=-1$

Let the eq of third side from point $(1,-10)$

$y+10=m(x-1)--------\left(3\right)$

since the triangle is isosceles

$\therefore$ angle between (1) and (3) should be equal to angle between (2) and (3)

$\frac{m_1-m}{1+m_{1}m}=\frac{m-m_2}{1+m{m}_2}$

$\frac{7-m}{1+7m}=\frac{m+1}{1-m}$

Solving above eqn

$3m^2+8m-3=0$

on solving above eq we get

$m=-3$ or $\frac{1}{3}$

from eq (3)

$y+10=-3(x-1)$

$y+10=-3x+3$

$3x+y+7=0$