JEE Main & Advanced Mathematics Pair of Straight Lines Point of Intersection of Lines Represented by \[\mathbf{a}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+ 2hxy+b}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{+ 2gx+ 2fy+c= 0}\]

Let \[\varphi \equiv a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\]

\[\frac{\partial \varphi }{\partial x}=2ax+2hy+2g=0\]                          (Keeping \[y\] as constant)  

and \[\frac{\partial \varphi }{\partial y}=2hx+2by+2f=0\]    (Keeping \[x\] as constant)

For point of intersection \[\frac{\partial \varphi }{\partial x}=0\] and \[\frac{\partial \varphi }{\partial y}=0\]

We obtain, \[ax+hy+g=0\] and \[hx+by+f=0\]

On solving these equations, we get

\[\frac{x}{fh-bg}=\frac{y}{gh-af}=\frac{1}{ab-{{h}^{2}}}\]i.e.,\[(x,y)=\left( \frac{bg-fh}{{{h}^{2}}-ab},\frac{af-gh}{{{h}^{2}}-ab} \right)\].

(3) Separate equations from joint equation: The general equation of second degree be \[a{{x}^{2}}+2hxy+b{{y}^{2}}+2gx+2fy+c=0\]

To find the lines represented by this equation we proceed as follows :

Step I : Factorize the homogeneous part \[a{{x}^{2}}+2hxy+b{{y}^{2}}\] into two linear factors. Let the linear factors be \[a'x+b'y\] and \[a''x+b''y\].

Step II : Add constants \[c'\]and \[c''\] in the factors obtained in step I to obtain \[a'x+b'y+c'\] and \[a''x+b''y+c''\]. Let the lines be \[a'x+b'y+c'=0\] and \[a''x+b''y+c''=0\].

Step III : Obtain the joint equation of the lines in step II and compare the coefficients of \[x,\,\,y\] and constant terms to obtain equations in \[c'\]and \[c''\].

Step IV : Solve the equations in \[c'\]and \[c''\] to obtain the values of \[c'\]and \[c''\].

Step V : Substitute the values of \[c'\]and \[c''\] in lines in step II to obtain the required lines.