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 1,1  are

The equation of any line which makes an angle α with the line ax+by+c=0 and passing  through the point x1,y1 can be taken as y−y1=tan⁡(θ±α)x−x1, here θ is the  inclination of the line ax+by+c=0 The slope of the line ax+by+c=0 is m=−ab=tan⁡θ Since the hypotenuse of right angled isosceles triangle is 7x+y−8=0 and opposite vertex is  (1,1) then the other two sides makes equal angles 45∘ each with the hypotenuse and passing  through the point (1,1) Hence, tan⁡θ=−7,α=45∘ Therefore, the equation of one side is y−1=tanθ+45°x−1=1+tanθ1−tanθx−1=−34x−1It implies, 4y−4=−3x+33x+4y−7=0And the equation of the other side is y−1=tanθ−45°x−1=tanθ−1tanθ+1x−1=43x−1It implies, 3y−3=4x−44x−3y−1=0 Therefore, two sides of given triangle are 3x+4y−7=0,4x−3y−1=0