Question

The equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of hypotenuse is (2, 2) is

• A. $7x-y=16$
• B. $7x+y=16$
• C. $x+7y=16$
• D. $x-7y+12=0$

Answer 1

The correct option is D $x-7y+12=0$

The equation of the hypotenuse is $3x+4y=4$

Slope of the line is $-\frac{3}{4}$

Since it is an isoceles right angled triangle the angles should be ${45}^o,{45}^o$ and ${90}^o$

$\tan {45}^{\circ }=\left|\frac{m-(-\frac{3}{4})}{1+m(-\frac{3}{4})}\right|$

$1=\left|\frac{m+\frac{3}{y}}{1-\frac{3m}{y}}\right|$

$1-\frac{3m}{y}=\pm (m+\frac{m}{y})$

$1-\frac{3m}{y}=m+\frac{3}{y}$

$m+\frac{3m}{y}=1-\frac{3}{y}$

$\frac{7m}{y}=\frac{1}{y}$

$m=\frac{1}{7}$

The coordinates of B is $(2,2)$

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

$7y-1y=x=2$

$x-7y+12=0$ is the req equation