JEE Main 2016 · 9 Apr · Shift 1 · Mathematics Q79

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MathematicsSingle correct+4 / -1Question 79 of 90

The point (2, 1) is translated parallel to the line L : x − y = 4 by $2\sqrt3$ 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 :

x+y=2-\sqrt6

x+y=3-3\sqrt6

x+y=3-2\sqrt6

2x+2y=1-\sqrt6

Answer and worked solution

Official answer: C

Moving $2\sqrt3$ along $\frac1{\sqrt2}(1,1)$ towards the third quadrant gives Q = $(2-\sqrt6,\,1-\sqrt6)$ . The perpendicular to L through Q is $x+y=3-2\sqrt6$ .

Topics:Equations of a line

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JEE Main 2016 · 9 Apr · Shift 1, Mathematics question 79