First slide

Equation of line in Straight lines

Question

If one of the diagonals of a square is along he line $x = 2 y$ and one of its vertices is (3, 0), then its side through this vertex nearer to the origin is given by the equation

Moderate

A

$y - 3 x + 9 = 0$
B

$3 y + x - 3 = 0$
C

$x = 3 y - 3 = 0$
D

$3 x + y - 9 = 0$

Solution

The point (3, 0) does not lie on the diagonal $x = 2 y$ . Let the equation of a side through the vertex (3, 0) be
$y - 0 = m (x - 3)$
Since the angle between a side and a diagonal of a square is $π / 4$ we have
$\pm \tan \frac{π}{4} = \frac{m - 1 / 2}{1 + m (1 / 2)} = \frac{2 m - 1}{2 + m}$
$\Rightarrow m = 3, - 1 / 3$
Thus the equation of a side through (3, 0) is
$y = 3 (x - 3) or y = (- \frac{1}{3}) (x - 3)$ and the one nearer to the origin is $3 y + x - 3 = 0$

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