First slide

Equation of line in Straight lines

Question

Slope of a line $L$ is 1. The points on the line $L$ at a distance of 2 units from the point $(3 \sqrt{2}, 3 \sqrt{2})$ which lies on the line $L$ are

Moderate

A

$(4 \sqrt{2}, 4 \sqrt{2}), (2 \sqrt{2}, 2 \sqrt{2})$
B

$(4 \sqrt{2}, 4 \sqrt{2}), (- 2 \sqrt{2}, - 2 \sqrt{2})$
C

$(- 4 \sqrt{2}, - 4 \sqrt{2}), (2 \sqrt{2}, 2 \sqrt{2})$
D

$(4 \sqrt{2}, - 4 \sqrt{2}), (- 2 \sqrt{2}, 2 \sqrt{2})$

Solution

$The inclination of the line L is θ = \tan^{- 1} (1) = \frac{π}{4}$
$The coordinates of a point on the line L at a distance of r units from (x_{1}, y_{1}) are$
$x = x_{1} \pm r \cos θ and y = y_{1} \pm r \sin θ, here θ is inclination of the line θ$
$x = x_{1} \pm r \cos θ and y = y_{1} \pm r \sin θ, here θ is inclination of the line θ$
Hence the required coordinates are
$\begin{matrix} x = x_{1} \pm r \cos θ \\ = 3 \sqrt{2} \pm 2 (\frac{1}{\sqrt{2}}) \\ = 3 \sqrt{2} \pm \sqrt{2} \\ = 4 \sqrt{2}, 2 \sqrt{2} \end{matrix}$
and
$\begin{matrix} y = y_{1} \pm r \sin θ \\ = 3 \sqrt{2} \pm 2 (\frac{1}{\sqrt{2}}) \\ = 3 \sqrt{2} \pm \sqrt{2} \\ = 4 \sqrt{2}, 2 \sqrt{2} \end{matrix}$
$Therefore, the required points are (4 \sqrt{2}, 4 \sqrt{2}) = (2 \sqrt{2}, 2 \sqrt{2})$

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