Question

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

• A. $y-3x+7=0,x-3y-3=0$
• B. $y-3x+9=0,x-3y-3=0$
• C. $y+3x-9=0,x+3y-3=0$
• D. $y-3x+9=0,x+3y-3=0$

Answer 1

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

Scope of $AC=m_1=\frac{1}{2}$

Let scope of AB is $m$

As $tan\theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$

$\Rightarrow tan45=\left|\frac{m-\frac{1}{2}}{1+\frac{m}{2}}\right|$

$\Rightarrow 1=\left|\frac{2m-1}{2+m}\right|$

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

or

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

$\Rightarrow 3m=-1\Rightarrow m=1\frac{1}{3}$

As both AB and AC are at 45

to $AC\Rightarrow$ slope of $AB=3,BC=-\frac{1}{3}$

And both pass through $B(3,0)$

$\Rightarrow AB=y-o=3(x-3)=y=3x-9\Rightarrow y-3x+9=0$

$BC=y-0=-\frac{1}{3}(x-3)\Rightarrow 3y=-x+3$

$\Rightarrow x+3y-3=0$

$\Rightarrow \left(D\right)$