QJEEM16Ma9ade: JEE Main 2016 Mathematics - Class 11 JEEM16Ma9ade Solution: The point (2, 1) is translated parallel to the line L : x− y =…

Question

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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 newpoint Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is :

Accepted Answer

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}

=

\frac{| 2 + 1 - c |}{\sqrt{2}}

2

\sqrt{6}

=

| 3 - c |

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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