Question

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

Answer 1

Slope of the line $3x+4y–4=0$ is $m_1=\frac{-3}{4}$

$\tan {45}^{\circ }=\pm \left(\frac{m-m_1}{1+m{m}_1}\right)$

$\Rightarrow 1=\pm \left(\frac{m+\frac{3}{4}}{1-\frac{3}{4}m}\right)$

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

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

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

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

$\Rightarrow \frac{7m}{4}=\frac{1}{4},\frac{m}{4}=\frac{-7}{4}$

$\therefore m_A=\frac{1}{7},m_b=-7$

Hence the required equation of the two lines are

$y-2=m_A(x-2)$ and $y-2=m_B(x-2)$

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

$\Rightarrow 7y-14=x-2$ and $y-2=-7x+14$

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