1.

Find the joint equation 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`.

Answer» The slope of the line `3x+2y-11=0` is `m_(1)=-(3)/(2)`
Let m be the slope of the lines making an angle of `30^(@)`
`tan 30^(@)=|(m-(-(3)/(2)))/(1+m(-(3)/(2)))|`
`rArr" "(1)/(sqrt3)=|(2m+3)/(2-3m)|`
On squaring both sides, we get
`(1)/(3)=((2m+3)^(2))/((2-3m)^(2))`
`rArr" "4+9m^(2)-12m=3(4m^(2)+9+12m)`
`rArr" "4+9m^(2)-12m=12m^(@)+27+36m`
`rArr" "3m^(2)+48m+23=0" ...(i)"`
For the joint equation of the pair of the lines is obtained by putting `m=y//x.`
`therefore " The joint equaiton of the two lines is : "`
`3((y)/(x))^(2)+48((y)/(x))+23=0`
`rArr" "3y^(2)+48xy+23x^(2)`
`rArr" "23x^(2)+48xy+3y^(2)=0`


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