Question

If one diagonal of a square is along the line $x=2y$ and one of its vertices is $(3,0)$, then its sides through this vertex are given by the equations

• A. $y-3x+9=0,3y+x-3=0$
• B. $y+3x+9=0,3y+x-3=0$
• C. $y+3x+9=0,3y-x+3=0$
• D. $y-3x+3=0,3y+x+9=0$

Answer 1

The correct option is A $y-3x+9=0,3y+x-3=0$

Let slope of required line is $m$

$tan{45}^o=\left|\frac{1}{2}-m1+\frac{m}{2}\right|$

$\Rightarrow \pm 1=\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\Rightarrow m=-\frac{1}{3},3$

The equation of lines $y=3(x-3)$ & $y=-\frac{1}{3}(x-3)$