Question

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

Answer 1

Here, we are given $\Delta ABC$ is an isosceles right angled triangle.

$\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow {90}^{\circ }+\angle B+\angle B={180}^{\circ }$

$\Rightarrow \angle B={45}^{\circ },\angle C={45}^{\circ }$

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 $(x_1,y_1)$ and making an angle $\alpha$ with the given line $y=mx+c$ are

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

Here,

Equation of the given line is,

$3x+4y=4$

$\Rightarrow 4y=-3x+4$

$\Rightarrow y=-\frac{3}{4}x+1$

Comparing this equation with $y=mx+c$ we get,

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

$x_1=2,y_1=2,\alpha ={45}^{\circ },m=-\frac{3}{4}$

So, the equation of the required lines are

$y-2=\frac{-\frac{3}{4}+tan{45}^{\circ }}{1+\frac{3}{4}tan{45}^{\circ }}(x-2)$ and $y-2$

$=\frac{-\frac{3}{4}-tan{45}^{\circ }}{1-\frac{3}{4}tan{45}^{\circ }}(x-2)$

$\Rightarrow y-2=\frac{-\frac{3}{4}+1}{1+\frac{3}{4}}(x-2)$ and

$y-2=\frac{-\frac{3}{4}-1}{1-\frac{3}{4}}(x-2)$

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

$\Rightarrow x-7y+12=0$ and $7x+y-16=0$