ISI2018-DCG-23

Question

Let $A$ be the point of intersection of the lines $3x-y=1$ and $y=1$. Let $B$ be the point of reflection of the point $A$ with respect to the $y$-axis. Then the equation of the straight line through $B$ that produces a right angled triangle $ABC$ with $\angle ABC=90^{\circ}$, and $C$ lies on the line $3x-y=1$, is

• A. $3x-3y=2$
• B. $2x+3=0$
• C. $3x+2=0$
• D. $3y-2=0$

Answer 1

The intersection point of $y=1$  and  $3x-y=1$  is the point $A=(\frac {2}{3},1)$  

The reflection of the same w.r.t to $y-axis$  would be the point $B=(\frac {-2}{3},1)$

Since the triangle $ABC$ is right angled at $B$,  the equation of the required line would $3x+2=0$  as it is perpendicular to $y=1$

So, option C is correct