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

#### Solution

The slope of the line3x + 2y -11 = 0 is m_{1}=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 m_{1} is 30°

`tan 30^@=|(m-m_1)/(1+mm_1)|,where tan 30^@=1/sqrt3`

`therefore 1/sqrt3=|(m-(-3/2))/(1+m(-3/2))|`

On squaring both the sides, we get,

`1/3=(2m+3)^2/(2-3m)^2`

`therefore (2-3m)^2=3(2m+3)^2`

`4-12m+9m^2=3(4m^2+12m+9)`

`9m^2-12m+4=12m^2+36m+27`

`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

`3(y/x)^2+48(y/x)+23=0`

`(3y)^2/x^2+(48y)/x+23=0`

3

`3y^2+48xy+23x^2=0`