Question

Two 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

Let ABC be the isosceles traingle, where $7x-y+3=0$ and $x+y-3=0$ represent the sides AB and AC, respectively.

Let AB = BC

$\because$ AB = BC

$\therefore tanB=tanC$

Here,

Slope of AB = 7

Slope of AC = -1

Let m be the slope of BC

Then, $\left|\frac{m-7}{1+7m}\right|=\left|\frac{m+1}{1-m}\right|=\left|\frac{m+1}{m-1}\right|$

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

Taking the positive sign, we get:

$m^2-8m+7=7m^2+8m+1$

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

$\Rightarrow m=-3,\frac{1}{3}$

Now, taking the negative sign, we get:

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

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

$\Rightarrow m^2=-1$ (not possible)

Equations of the third side is

$y+10=-3(x-1)$ and $y+10=\frac{1}{3}(x-1)$

$\Rightarrow 3x+y+7=0$ and $x-3y-31=0$