The point of intersection of the straight lines given by the equation `3y^2-8xy-3x^2-29x+3y-18=0 is `

Explanation

Comparing the given equation `3y^(2)+2gx+2fy+c=0`
We get, `a=3, 2h=-8,b=-3,2g=-29,2f=3,c=-18`
We find that
`abc+2fgh-af^(2)-bg^(2)-ch^(2)`
`=(-3)(3)(-18)+2((3)/(2))((-29)/(2))(-4)-3((3)/(2))^(2)-(-3)((-29)/(2))^(2)-(-18)(-4)^(2)=0`
Hence two stringht lines are represented.
Now, `3y^(2)-8xy-3x^(2)=(3y+x)(y-3x).`
Hence let `3y^(2)-8xy-3x^(2)-29x+3y-18-=(3y+x+p)(y-3x+q).`
Equating the coefficients of x & y, we get
`-3p+q=-29andp+3q=3`
Whence `p=9and q=-2`
Thus the equation of the represented lines are `3y+x+9=0and y-3x-2=0.`
Solving them, we get the point of interesection as `((-3)/(2),(-5)/(2))`
Let `theta` be the angle between them, then
`tan theta=(2sqrt((-4)^(2)-(3)(-3)))/(-3+3)=oo`
`impliestheta=90^(@)`
or equivalently, we observe that coefficient of `x^(2)+` coefficient of `y^(2)=3+(-3)=0` angle between represented lines is `90^(@)`