Question

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

• 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 $ABC$ be an isosceles right angled triangle, with right angle at $A$ and $AB=AC$.
$\Rightarrow \angle ABC=\angle ACB=45^{\circ }$
The slope of line $BC$ i.e., $3x+4y=4$ is
$m_2=-\frac{3}{4}$
Let the slope of the line $AC$ be $m_1=m$.
$\because$ Angle between two lines is
$\Rightarrow \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$\Rightarrow \tan 45^{\circ }=\left|\frac{m-\left(\frac{-3}{4}\right)}{1+m\left(\frac{-3}{4}\right)}\right|$
$\Rightarrow 1=\left|\frac{m+\frac{3}{4}}{1-\frac{3m}{4}}\right|$
$\pm =\frac{4m+3}{4-3m}$

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