Pair of Straight Lines
7.0 Point of intersection of lines
7.0 Point of intersection of lines
Partial differentiation is used to find the point of intersection of lines represented by $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$
Let $$\begin{equation} \begin{aligned} \phi (x,y) \equiv a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0 \\ \therefore \frac{{\partial \phi }}{{\partial x}} = 2ax + 2hy + 2g{\text{ (take }}y{\text{ as constant)}} \\ \frac{{\partial \phi }}{{\partial y}} = 2hx + 2by + 2f{\text{ (take }}y{\text{ as constant)}} \\\end{aligned} \end{equation} $$
For point of intersection $$\frac{{\partial \phi }}{{\partial x}} = 0{\text{ and }}\frac{{\partial \phi }}{{\partial y}} = 0$$
We get, $$ax + hy + g = 0{\text{ and }}hx + by + f = 0$$
Solving them, we get $$\begin{equation} \begin{aligned} \frac{x}{{fh - bg}} = \frac{y}{{gh - af}} = \frac{1}{{ab - {h^2}}} \\ (x,y) = \left( {\frac{{bg - fh}}{{{h^2} - ab}},\frac{{af - gh}}{{{h^2} - ab}}} \right) \\\end{aligned} \end{equation} $$
