Question

Find the equation of one of the sides of an isosceles right-angled triangle whose hypotenuse is given by $3x+4y=4$ and the opposite vertex of the hypotenuse is $(2,2)$.

Answer 1

Let the equation of other sides be

$y=mx+c$

It passes through $(2,2)$, so

$2=2m+c$

$\Rightarrow c=2(1-m)$

So, $y=mx+2(1-m)$

Angle between two lines is given by :

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

Equation of hypotenuse is given by $3x+4y=4$, so the slope is

$m_1=-\frac{3}{4}$

Using angle between two lines formula : Angle between the base and hypotenuse is ${45}^{\circ }$, so

$\tan \theta =\left|\frac{-\frac{3}{4}-m}{1-\frac{3m}{4}}\right|$

$\Rightarrow \tan {45}^{\circ }=\left|\frac{-3-4m}{4-3m}\right|$

$\Rightarrow \left|\frac{3+4m}{4-3m}\right|=1$

$\Rightarrow \frac{3+4m}{4-3m}=\pm 1$

$\Rightarrow \frac{3+4m}{4-3m}=1$ or $\Rightarrow \frac{3+4m}{4-3m}=-1$

$\Rightarrow 3+4m=4-3m$ or $3+4m=3m-4$

$\Rightarrow m=\frac{1}{7},-7$

Therefore, the equation of sides are

$y=\frac{1}{7}x+2(1-\frac{1}{7})$

and $y=-7+2(1+7)$

i.e. $x-7y+12=0$ and $7x+y-16=0$