Question

Let $\Delta ABC$ be an isosceles right-angled triangle where $AB\perp BC$, if the equation of hypotenuse is $2x+3y=4$, then the sum of slopes of $AB$ and $AC$ can be equal to

• A. $0$
• B. $-\frac{7}{15}$
• C. $-\frac{17}{3}$
• D. $-\frac{24}{5}$

Answer 1

Answer: (B), (C)

$-\frac{17}{3}$

Let slope of $AB=m_{AB}$
$\Delta ABC$ is an right angled isosceles triangle so it makes equal angle of ${45}^{\circ }$.
$\therefore \tan {45}^{\circ }=\left|\frac{m_{AC}-m_{AB}}{1+m_{AC}.m_{AB}}\right|...\left(1\right)$
Now, equation of $AC$ is $2x+3y=4$
$\therefore m_{AC}=-\frac{2}{3}$
From equation $\left(1\right)$
$1=\left|\frac{-\frac{2}{3}-m_{AB}}{1+(-\frac{2}{3}.m_{AB})}\right|$
$\Rightarrow 1=\left|\frac{-2-3m_{AB}}{3-2m_{AB}}\right|$
$\Rightarrow \frac{-2-3m_{AB}}{3-2m_{AB}}=\pm 1$

Case I
$\frac{-2-3m_{AB}}{3-2m_{AB}}=1$
$\Rightarrow -2-3m_{AB}=3-2m_{AB}$
$m_{AB}=-5$

Case II
$\frac{-2-3m_{AB}}{3-2m_{AB}}=-1$
$\Rightarrow 2+3m_{AB}=3-2m_{AB}$
$\Rightarrow m_{AB}=\frac{1}{5}$
Hence
Sum of slopes can be $=-5+\frac{-2}{3}=-\frac{17}{3}$
or
$=\frac{1}{5}+\frac{-2}{3}=-\frac{7}{15}$