Question

Two sides of a triangle have the joint equation $(x-3y+2)(x+y-2)=0$, the third side which is variable always passes through the point $(-5,-1)$, then the possible values of slope of third side such that origin is an interior point of the triangle is/are

• A. $-\frac{4}{3}$
• B. $-\frac{2}{3}$
• C. $-\frac{1}{3}$
• D. $\frac{1}{6}$

Answer 1

The correct option is A $-\frac{4}{3}$

Let line be $y=mx+c$

So, $-1=-5m+c$

and for origin to lie inside triangle,

$\frac{-m\left(0\right)+\left(0\right)-c}{1}>0$

$c<0$

$c=5m-1<0$

$m<\frac{1}{5}$
So, all options are correct