First slide

Equation of line in Straight lines

Question

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

Moderate

A

$\begin{aligned} y - 3 x + 9 = 0, 3 y + x - 3 = 0 \end{aligned}$
B

$\begin{array}{r} y + 3 x + 9 = 0, 3 y + x - 3 = 0 \end{array}$
C

$\begin{array}{r} y - 3 x + 9 = 0, 3 y - x + 3 = 0 \end{array}$
D

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

Solution

Clearly the point (3, 0) does not lie on the diagonal $x = 2 y .$
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
$π / 4 .$ Therefore, angle between $x = 2 y$ and $y - 0 = m (x - 3)$ is
also $π / 4 .$ Consequently, we have
$\tan \frac{π}{4} = \pm \frac{m - 1 / 2}{1 + m / 2} \Rightarrow m = 3, - \frac{1}{3}$
Substituting the values of min (i), we obtain
$y - 3 x + 9 = 0$ and $3 y + x - 3 = 0$ as the required sides.

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