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The point (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 :

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

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

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

Option 4)
$2x + 2y = 1 - \sqrt{6}$

Answers (1)

best_answer

Parametric form -

$x=x_{1}+r\cos \Theta$

$y=y_{1}+r\sin \Theta$

- wherein

Where $\Theta$ is the inclination of the line and $r$ is the distance between $(x,y)$ and $(x_{1},y_{1})$

Slope – point from of a straight line -

$y-y_{1}=m(x-x_{1})$

- wherein

$m\rightarrow$ slope

$\left ( x_{1},y_{1} \right )\rightarrow$ point through which line passes

Condition for perpendicular lines -

$m_{1}m_{2}= -1$

- wherein

Here $m_{1},m_{2}$ are the slope of perpendicular lines.

So, $Q=(2-2\sqrt3\cos 45^{\circ},1-2\sqrt3\sin 45^{\circ})$

$Q=(2-\sqrt6,1-\sqrt6)$

Line Required

$\left ( y-1+\sqrt{6} \right )= \left ( -1 \right )\left ( x-2+\sqrt6 \right )$

$\left ( y-1+\sqrt{6} \right )= \left ( -x+2-\sqrt6 \right )$

$x+y=3-2\sqrt6$

Option 1)

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

Option 2)

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

Option 3)

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

Option 4)

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

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