Find the joint equaiton of the pair of the lines through the origin each of which is making an angle of `30^(@)` with the line `3x+2y-11=0`.

Explanation

LeT OA and OB be the two lines through the origin, each making an angle of `30^(@)` with the line `3x + 2y -11 = 0` whose slope is `m_(1) = -(3)/(2).`
Let OA ( or OB ) has slope m. `therefore` its equation is y = mx
`therefore m = (y)/(x) " "...(1)`
Also `tan 30^(@) = |(m-m_(1))/(1+m*m_(1))| , " where " tan 30^(@) = (1)/(sqrt(3))`
`therefore (1)/(sqrt(3)) = |(m-(-(3)/(2)))/(1+m(-(3)/(2)))| " "therefore (1)/(sqrt(3)) = |(2m+3)/(2-3m)|`
On squaring both sides, we get,
`(1)/(3) = ((2m+3)^(2))/((2-3m)^(2))`
`therefore (2-3m)^(2) = 3(2m+3)^(2)`
`therefore 4-12m + 9m^(2) = 3(4m^(2) + 12m +9)`
` therefore 4 - 12m + 9m^(2) = = 12 m^(2) + 36m + 27`
`therefore 3m^(2) + 48m + 23 = 0`
`therefore 3((y)/(x))^(2) + 48((y)/(x)) + 23 = 0 " "`...[By(1)]
`therefore 3y^(2) + 48xy + 23x^(2) = 0`
`therefore 23x^(2) + 38xy + 3y^(2) = 0`
This is the joint equation of the two lines through the origin making an angle of `30^(@)` with the line `3x + 2y -11 = 0` .