# Point of Intersection Formula

Point of intersection means the point at which two lines intersect. These two lines are represented by the equation a1x+ b1x + c1= 0  and a2x+ b2x + c2 = 0 respectively. Given figure illustrate the point of intersection of two lines.
We can find the point of intersection of three or more lines also. By solving the two equations, we can find the solution for point of intersection of two lines.

### Solved Examples

Question 1: Find out the point of intersection of two lines $x^{2}$ + 2x + 1 = 0 and 2$x^{2}$ + 3x + 5 = 0 ?

Solution:

Given straight line equations are: $x^{2}$ + 2x + 1 = 0 and 2$x^{2}$ + 3x + 5 = 0

Here,
a1 = 1
b1 = 2
c1 = 1

a2 = 2, b2 = 3, c2 = 5

Intersection point can be calculated using this formula,

x = $\frac{b_{1}c_{2}-b_{2}c_{1}}{a_{1}b_{2}-a_{2}b_{1}}$; y = $\frac{a_{2}c_{1}-a_{1}c_{2}}{a_{1}b_{2}-a_{2}b_{1}}$

(x,y) = ($\frac{2\times5-3\times1}{1\times3-2\times2}$,$\frac{2\times1-1\times5}{1\times3-2\times2}$)

(x,y) = ($\frac{10-3}{3-4}$,$\frac{2-5}{3-4}$)

(x,y) = (-7,3)

#### Practise This Question

Two charges of equal magnitude 'q'  but of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at the point P  is given by  (r2a) (Where p=2qa )