Question

The equation of the lines which passes through the point

(3, – 2) and are inclined at $60 ^ { \circ }$ to the line $\sqrt { 3 } x + y = 1$

• A. $\sqrt { 3 } x - y - 2 - 3 \sqrt { 3 } = 0$
• B. $x - 2 = 0 , \sqrt { 3 } x - y + 2 + 3 \sqrt { 3 } = 0$
• C. $\sqrt { 3 } x - y - 2 - 3 \sqrt { 3 } = 0$
• D. None of these

Answer 1

Answer: (A)

$\sqrt { 3 } x - y - 2 - 3 \sqrt { 3 } = 0$

Answer 2

The equation of lines passing through (3, –2) is

$( y + 2 ) = m ( x - 3 )$ .....(i)

The slope of the given line is $- \sqrt { 3 }$ .

So, $m = 0$ or $\sqrt { 3 }$

Putting the values of m in (i), the required equation is $y + 2 = 0$ and $\sqrt { 3 } x - y - 2 - 3 \sqrt { 3 } = 0$.