Question

Find the equation to the straight line which passes through the point (3, -2) and inclined at ${60}^o$ to the line $\sqrt{3}x+y=1$

Answer 1

Any line through ( 3 , - 2) is
y + 2 = m (x - 3)
slope of $\sqrt{3}x+y=1is-\sqrt{3}$
Angle between the two lines is $6o^{\circ }$
$\therefore tan(\pm {60}^{\circ })=\frac{m-(-\sqrt{3}}{1,+m(-\sqrt{3}}$
or $\pm \sqrt{3}(1-m\sqrt{3})=m+\sqrt{3}$
+ ive sign
$-\sqrt{3}-3m=m+\sqrt{3}$ or 4m = 0
$\therefore$ m = 0
-ive sign
$-\sqrt{3}+3m=m+\sqrt{3}$ 2m = $2\sqrt{3}$
$\therefore m=\sqrt{3}$
putting in (1) the required lines are
y + 2 = 0 and $\sqrt{3}x-y=2+3\sqrt{3}$
Alternative direct methods
slope of given lines is - $\sqrt{3}=m$ say
t = $tan\alpha ={60}^{\circ }=\sqrt{3}$
$\therefore m_1=\frac{m-t}{1+mt},m_2=\frac{m+t}{1-mt}$
$\therefore m_1=\frac{-2\sqrt{3}}{1-3}=\sqrt{3}and{m}_2=0$
Hence the lines on putting m = $m_{1}or{m}_2$ in (1) are as found above