Question

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

Answer 1

Given equations two sides of isosceles triangle are $7x-y+3=0$ and $x+y-3=0$

Slope of the given lines are 7 and -1, erspectively,

Then, equation of line passing through(1,10)is

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

Since it makes equal angle $\theta$ with the give lines.

$\therefore tan\theta =\frac{m-7}{1+7m}=\frac{(-1)-m}{1+m(-1)}$

$\Rightarrow \frac{m-7}{1+7m}=\frac{(m+1)}{1-m}$

$\Rightarrow (m-7)(1-m)=-(m+1)(1+7m)$

$\Rightarrow -m^2+8m-7=-(7m^2+8m+1$

$\Rightarrow 6m^2+16m-6=0\Rightarrow 3m^2+8m-3=0$

$\Rightarrow 3m^2+9m-m-3=0\Rightarrow 3m(m+3)-1(m+3)=0$

$\Rightarrow (3m-1)(m+3)=0\Rightarrow m=\frac{1}{3}orm=-3$

Hence , the equation of third side are $y+10=\frac{1}{3}(x-1)ory+10=-3(x-1)$

i.e. $x-3y-31=0or3x+y+7=0$