Question

In an isosceles right angled triangle whose hypotenuse is given by $3 x+4 y=4$ and the opposite vertex of hypotenuse is $(2,2)$, then the equation of one of the sides other than hypotenuse is
I. $x-7 y+12=0$
II. $7 x+y-16=0$

• A. Both I and II are true
• B. Only I is true
• C. Only II is true
• D. Neither I nor II is true

Answer 1

Answer: (A)

Let $A B C$ be an isosceles right angled triangle, with right angle at $A$ and $A B=A C$.
$\Rightarrow \angle A B C=\angle A C B=45^{\circ}$
The slope of line $B C$ i.e., $3 x+4 y=4$ is
$m_2=-\frac{3}{4}$
Let the slope of the line $A C$ be $m_1=m$.
$\because$ Angle between two lines is
$\Rightarrow \tan \theta =\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$
$\Rightarrow \tan 45^{\circ} =\left|\frac{m-\left(\frac{-3}{4}\right)}{1+m\left(\frac{-3}{4}\right)}\right| $
$\Rightarrow 1 =\left|\frac{m+\frac{3}{4}}{1-\frac{3 m}{4}}\right|$
$\pm =\frac{4 m+3}{4-3 m}$

Taking positive sign,
$1= \frac{4 m+3}{4-3 m}$
$\Rightarrow 4-3 m =4 m+3$
$\Rightarrow 4-3 =4 m+3 m $
$\Rightarrow 7 m=1 \Rightarrow m=\frac{1}{7}$
Taking negative sign,
$ -1 =\frac{4 m+3}{4-3 m}$
$\Rightarrow -4+3 m =4 m+3 $
$\Rightarrow -4-3 =m$
$\Rightarrow m =-7$
Hence, equations of line $A B$ and line $A C$ are
$ y-2 =\frac{1}{7}(x-2)$
and $ y-2 =-7(x-2)\left[\because y-y_1=m\left(x-x_1\right)\right]$
$\Rightarrow 7 y-14 =x-2 $
and $ y-2 =-7 x+14$
$\Rightarrow x-7 y+12 =0 $ and $7 x+y-16=0$