Question

Find the equations to the sides of an isosceles right angled triangle the equation of whose hypotenues is 3x + 4y = 4 and the opposite vertex is the point (2, 2).

Answer 1

$$\begin{gathered} \mathrm{Here},\mathrm{we}\mathrm{are}\mathrm{given}△ABC\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{right}\mathrm{angled}\mathrm{triangle}. \\ \angle A+\angle B+\angle C=180° \\ \Rightarrow 90°+\angle B+\angle B=180° \\ \Rightarrow \angle B=45°,\angle C=45° \end{gathered}$$


Now, we have to find the equations of the sides AB and AC, where 3x + 4y = 4 is the equation of the hypotenuse BC.

We know that the equations of two lines passing through a point $\left(x_1,y_1\right)$ and making an angle $\mathrm{\alpha }$ with the given line y = mx + c are

$y-y_1=\frac{m\pm \tan \alpha }{1\mp m\tan \alpha }\left(x-x_1\right)$

Here,
$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{line}\mathrm{is}, \\ 3x+4y=4 \\ \Rightarrow 4y=-3x+4 \\ \Rightarrow y=-\frac{3}{4}x+1 \\ \mathrm{Comparing}\mathrm{this}\mathrm{equation}\mathrm{with}\mathit{}y=mx+c \\ \mathrm{we}\mathrm{get}, \\ m=-\frac{3}{4} \end{gathered}$$
$x_1=2,y_1=2,\mathrm{\alpha }={45}^{\circ },m=-\frac{3}{4}$

So, the equations of the required lines are

$$\begin{gathered} y-2=\frac{{\displaystyle -\frac{{\displaystyle 3}}{4}+\tan {45}^{\circ }}}{{\displaystyle 1+\frac{{\displaystyle 3}}{4}\tan {45}^{\circ }}}\left(x-2\right)\mathrm{and}y-2=\frac{{\displaystyle -\frac{{\displaystyle 3}}{4}-\tan {45}^{\circ }}}{{\displaystyle 1-\frac{{\displaystyle 3}}{4}\tan {45}^{\circ }}}\left(x-2\right) \\ \Rightarrow y-2=\frac{{\displaystyle -\frac{{\displaystyle 3}}{4}+1}}{{\displaystyle 1+\frac{{\displaystyle 3}}{4}}}\left(x-2\right)\mathrm{and}y-2=\frac{{\displaystyle -\frac{{\displaystyle 3}}{4}-1}}{{\displaystyle 1-\frac{{\displaystyle 3}}{4}}}\left(x-2\right) \\ \Rightarrow y-2=\frac{1}{{\displaystyle 7}}\left(x-2\right)\mathrm{and}y-2=\frac{-7}{1}\left(x-2\right) \\ \Rightarrow x-7y+12=0\mathrm{and}7x+y-16=0 \end{gathered}$$