If `theta` is the angle between the lines given by the equation `6x^2+5x y-4y^2+7x+13 y-3=0` , then find the equation of the line passing through the point of intersection of these lines and making an angle `theta` with the positive x-axis.

Answer

`11x-2y+13=0`

Explanation

Writing the given equation as quadratic in x , we have
`6x^(2)+(5y+7)x-(4y^(2)-13y+3)=0`
or `x=(-(5y+7)+-sqrt((5y+7)^(2)+24(4y^(2)-13y+3)*))/(12)`
`=(-(5y+7)+-sqrt(121y^(2)-242y+121))/(12)`
`=(-(y+7)+-11(y-1))/(12)=(6y-18)/(12),(-16y+4)/(12)`
or `2x-y+3=0and3x+4y-1=0`
which are the lines represented by the given equation . the point of intersection is (-1,1) that is obtained by solving these equations .
Also , `tan theta=(2sqrt(h^(2)-ab))/(a+b)`
`=(2sqrt((5//2)^(2)-6(-4)))/(6-4)=sqrt(121/(4))=(11)/(2)` (`:'a=6,b=-4,h=5//2)`
So , the equation of the required line is
`y-1=(11)/(2)(x+1)`
or `11x-2y+13=0`