The point (2,1) is translated parallel to the line L: x-y=4 by 2√3 units.

Question

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:

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$$\sqrt 3 $$ = $${{\left| {2 + 1 - c} \right|} \over {\sqrt 2 }}$$

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$$\sqrt 6 $$

Solutions · 1