The equation of the sides of an Isosceles right angled triangle whose hypotenuse is  $7x+y-8=0$ and opposite vertex to the hypotenuse is $\left(1,1\right)$  are

$\text{ The equation of any line which makes an angle }\alpha \text{ with the line }ax+by+c=0\text{ and passing }$

$\text{ through the point }\left(x_1,y_1\right)\text{ can be taken as }y-y_1=\tan (\theta \pm \alpha )\left(x-x_1\right),\text{ here }\theta \text{ is the }$

$\text{ inclination of the line }ax+by+c=0$

$\text{ The slope of the line }ax+by+c=0\text{ is }m=-\frac{a}{b}=\tan \theta$

$\text{ Since the hypotenuse of right angled isosceles triangle is }7x+y-8=0\text{ and opposite vertex is }$

$\text{ (1,1) then the other two sides makes equal angles }{45}^{\circ }\text{ each with the hypotenuse and passing }$

$\text{ through the point }(1,1)$

$\text{ Hence, }\tan \theta =-7,\alpha ={45}^{\circ }$

$\text{ Therefore, the equation of one side is }$

$\begin{array}{c}y-1=\tan \left(\theta +45°\right)\left(x-1\right)\\ =\frac{1+\tan \theta }{1-\tan \theta }\left(x-1\right)\\ =-\frac{3}{4}\left(x-1\right)\end{array}$

It implies, 

$\begin{array}{c}4y-4=-3x+3\\ 3x+4y-7=0\end{array}$

And the equation of the other side is 

$\begin{array}{c}y-1=\tan \left(\theta -45°\right)\left(x-1\right)\\ =\frac{\tan \theta -1}{\tan \theta +1}\left(x-1\right)\\ =\frac{4}{3}\left(x-1\right)\end{array}$

It implies, 

$\begin{array}{c}3y-3=4x-4\\ 4x-3y-1=0\end{array}$

$\text{ Therefore, two sides of given triangle are }3x+4y-7=0,4x-3y-1=0$