JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 17)
The point (2, 1) is translated parallel to the line L : x− y = 4 by $$2\sqrt 3 $$ units. If the newpoint Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :
x + y = 2 $$-$$ $$\sqrt 6 $$
x + y = 3 $$-$$ 3$$\sqrt 6 $$
x + y = 3 $$-$$ 2$$\sqrt 6 $$
2x + 2y = 1 $$-$$ $$\sqrt 6 $$
Explanation
x $$-$$ y = 4
To find equation of R
slope of L = 0 is 1
$$ \Rightarrow $$ slope of QR = $$-$$ 1
Let QR is y = mx + c
y = $$-$$ x + c
x + y $$-$$ c = 0
distance of QR from (2, 1) is 2$$\sqrt 3 $$
2$$\sqrt 3 $$ = $${{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}$$
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2$$\sqrt 6 $$ = $$\left| {3 - c} \right|$$
c $$-$$ 3 = $$ \pm 2\sqrt 6 $$ c = 3 $$ \pm $$ 2$$\sqrt 6 $$
Line can be x + y = 3 $$ \pm $$ 2$$\sqrt 6 $$
x + y = 3 $$-$$ 2$$\sqrt 6 $$
To find equation of R
slope of L = 0 is 1
$$ \Rightarrow $$ slope of QR = $$-$$ 1
Let QR is y = mx + c
y = $$-$$ x + c
x + y $$-$$ c = 0
distance of QR from (2, 1) is 2$$\sqrt 3 $$
2$$\sqrt 3 $$ = $${{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}$$
2$$\sqrt 6 $$ = $$\left| {3 - c} \right|$$
c $$-$$ 3 = $$ \pm 2\sqrt 6 $$ c = 3 $$ \pm $$ 2$$\sqrt 6 $$
Line can be x + y = 3 $$ \pm $$ 2$$\sqrt 6 $$
x + y = 3 $$-$$ 2$$\sqrt 6 $$
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