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

Question

If one diagonal of a square is along the line x =2 y and one of its vertex is (3,0), then its sides through this vertex are given by the equations - (A) y-3 x+9=0, x-3 y-3=0 . (B) y-3 x+9=0, x-3 y-3=0 (C) y+3 x-9=0, x+3 y-3=0 . (D) y-3 x+9=0, x+3 y-3=0

Tardigrade Admin · Accepted Answer

Let slope of required line is m <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/7cbf8ccd589d33f5c431b66b1493acc2-.png" /> tan 45°=|((1/2)- m /1+( m )2)| ⇒ ± 1=((1/2)- m /1+( m )2) ⇒ m =-(1/3), 3 equation of lines y=3(x-3) y=-(1/3)(x-3)

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

$y - 3 x + 9 = 0, x - 3 y - 3 = 0$ .

$y - 3 x + 9 = 0, x - 3 y - 3 = 0$

$y + 3 x - 9 = 0, x + 3 y - 3 = 0$ .

$y - 3 x + 9 = 0, x + 3 y - 3 = 0$

Let slope of required line is $m$

$tan 4 5^{\circ} = ∣ ∣ \frac{\frac{1}{2} - m}{1 + \frac{m}{2}} ∣ ∣$
$\Rightarrow \pm 1 = \frac{\frac{1}{2} - m}{1 + \frac{m}{2}} \Rightarrow m = - \frac{1}{3}, 3$
equation of lines $y = 3 (x - 3) & y = - \frac{1}{3} (x - 3)$

Q. 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 -

y−3x+9=0,x−3y−3=0 .

y−3x+9=0,x−3y−3=0

y+3x−9=0,x+3y−3=0 .

y−3x+9=0,x+3y−3=0