Question

The triangle formed by the lines $x-7y+12=0,7x+y-16=0$ and $3x+4y=0$ is right-angled triangle

• A. True
• B. False

Answer 1

The correct option is A True

Let

$L_1:x-7y+12=0$ and slope of $L_1$ is $m_1=\frac{1}{7}$

$L_2:7x+y-16=0$ and slope of $L_2$ is $m_2=-7$

$L_3:3x+4y=0$

Now product of $m_1$ and $m_2$ is:

$⟹m_1\times m_2=7\times \frac{-1}{7}=-1$

Hence product of slope is $-1$. So $L_1\perp L_2$.

Hence triangle form by these lines are right angled triangle.