Suppose the hypotenuse and its opposite vertex of an isosceles right angled triangle are $3 x+4 y-4=0$ and…

Suppose the hypotenuse and its opposite vertex of an isosceles right angled triangle are $3 x+4 y-4=0$ and $(2,2)$ respectively. Then, which of the following is another side of the triangle?
  1. $x-7 y-12=0$
  2. $x+7 y+12=0$
  3. $7 x+y-16=0$
  4. $7 x+y+16=0$

Solution

Equation of $B C$ is $3 x+4 y-4=0$
$\therefore$ Slope of $A D=\frac{4}{3}$ Let the slope of one of other sides be $m$. $ \begin{aligned} & \text { Then, } \tan 45^{\circ}=\left|\frac{m-\frac{4}{3}}{1+\frac{4 m}{3}}\right| \Rightarrow 1=\left|\frac{3 m-4}{3+4 m}\right| \\ & \Rightarrow \quad \frac{3 m-4}{3+4 m}=1 \text { or } \frac{3 m-4}{3+4 m}=-1 \\ & \Rightarrow \quad m=-7 \text { or } m=\frac{1}{7} \end{aligned} $ and side passes through $(2,2)$. $\therefore$ Equation of side can be $ 7 x+y-16=0 \text { or } x-7 y+12=0 $

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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