Question

If one of the diagonals 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

Answer: (A)

Clearly the point (3, 0) does not lie on the diagonal x = 2y. Let m be the slope of a side passing through (3, 0). Then its equation is
$y-0=m(x-3)....(i)$
Since the angle between a diagonal and a side of a square is $\frac{\pi }{4}$. Therefore angle between
$x=2y\&y-0=m\left(x-3\right)$ is also $\frac{\pi }{4}$
Consequently, $tan\frac{\pi }{4}=\pm \frac{m-\frac{1}{2}}{1+\frac{m}{2}}\Rightarrow m=3,-\frac{1}{3}$
$\therefore$ from $\left(i\right)$, we get the required equations
$y=3\left(x-3\right)\Rightarrow y-3x+9=0$ or
$y=-\frac{1}{3}\left(x-3\right)\Rightarrow 3y+x-3=0$