Find the joint equation the pair of line through the origin and making angle $\frac{π}{6}$ with the line $3 x - y = 7$

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Solution

Solution -

Let slope of one line be $m_{1}$ & another be $m_{2}$
slope of $3 x - y = 7$ is 3.

$t a n 30^{\circ} = \frac{m_{1} - 3}{1 + 3 m_{1}}$
$\frac{1}{\sqrt{3}} = \frac{m_{1} - 3}{1 + 3 m_{1}}$
$1 + 3 m_{1} = \sqrt{3} m_{1} - 3 \sqrt{3}$
$(3 - \sqrt{3}) m_{1} = - 3 \sqrt{3} - 1$
$m_{1} = \frac{3 \sqrt{3} + 1}{\sqrt{3} - 3}$

Also, $t a n 30^{\circ} = \frac{3 - m_{2}}{1 + 3 m_{2}}$

$\frac{1}{\sqrt{3}} = \frac{3 - m_{2}}{1 + 3 m_{2}}$

$1 + 3 m_{2} = 3 \sqrt{3} - \sqrt{3} m_{2}$
$(3 + \sqrt{3}) m_{2} = 3 \sqrt{3} - 1$
$m_{2} = \frac{3 \sqrt{3} - 1}{3 + \sqrt{3}}$

pair of lines equation - $(y - m_{1} x) (y - m_{2} x) = 0$

$y^{2} - (m_{1} + m_{2}) x y + m_{1} m_{2} x^{2} = 0$
$m_{1} + m_{2} = \frac{10 \sqrt{3} + 12 + 9 - \sqrt{3} - 9 \sqrt{3} + 3}{(\sqrt{3} - 3) (\sqrt{3} + 3)} = \frac{24}{3 - 9} = - 4$
$m_{1} m_{2} = \frac{(3 \sqrt{3} + 1) (3 \sqrt{3} - 1)}{(\sqrt{3} - 3) (\sqrt{3} + 3)} = \frac{27 - 1}{3 - 9} = \frac{- 26}{6} = \frac{- 13}{3}$

$y^{2} + 4 x y - \frac{13}{3} x^{2} = 0$

$3 y^{2} + 12 x y - 13 x^{2} = 0$

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