Question

Find the joint equation of the pair of lines through the origin each of which is making an angle of 30° with the line 3x + 2y - 11 = 0

Answer 1

The slope of the line3x + 2y -11 = 0 is m₁=3/2

Let m be the slope of one of the lines making an angle of 300 with the 3x + 2y -11= 0. The angle between the lines having slopes m and m₁ is 30°

${\displaystyle {\tan 30}^{\circ }=\left|\frac{m-m_1}{1+m{m}_1}\right|,where{\tan 30}^{\circ }=\frac{1}{\sqrt{3}}}$

${\displaystyle \therefore \frac{1}{\sqrt{3}}=\left|\frac{m-(-\frac{3}{2})}{1+m(-\frac{3}{2})}\right|}$

On squaring both the sides, we get,

${\displaystyle \frac{1}{3}=\frac{{(2m+3)}^2}{{(2-3m)}^2}}$

${\displaystyle \therefore {(2-3m)}^2=3{(2m+3)}^2}$

${\displaystyle 4-12m+9m^2=3(4m^2+12m+9)}$

${\displaystyle 9m^2-12m+4=12m^2+36m+27}$

${\displaystyle 3m^2+48m+23=0}$

This is the auxiliary equation of the two lines and their joint equation is obtained by putting m=y/x

∴ the joint equation of the two lines is

${\displaystyle 3{\left(\frac{y}{x}\right)}^2+48\left(\frac{y}{x}\right)+23=0}$

${\displaystyle \frac{{\left(3y\right)}^2}{x^2}+\frac{48y}{x}+23=0}$

3

${\displaystyle 3y^2+48xy+23x^2=0}$