First slide

Equation of line in Straight lines

Question

The point P (2,1) is translated parallel to the line L: x-y = 4 by 2 $\sqrt{3}$ units. If the new point Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is

Moderate

A

$x + y = 2 - \sqrt{6}$
B

$x + y = 3 - 3 \sqrt{6}$
C

$x + y = 3 - 2 \sqrt{6}$
D

$2 x + 2 y = 1 - \sqrt{6}$

Solution

$Q = (2 + 2 \sqrt{3} \frac{1}{2}, 1 + 2 \sqrt{3} . \frac{1}{\sqrt{2}}); (2 - \sqrt{6}, 1 - \sqrt{6})$
required line equation
$y - (1 - \sqrt{6}) = - 1 (x - 2 - \sqrt{6})$
$x + y = 1 - \sqrt{6} + 2 - \sqrt{6} = 3 - 2 \sqrt{6}$

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