# Point of Intersection Formula

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## Introduction to Point of Intersection Formula

In mathematics, we refer to the point of intersection where a point meets two lines or curves. The intersection of lines may be an empty set, a point, or a line in Euclidean geometry. A required criterion for two lines to intersect is that they should be in the same plane and are not skew lines. The intersection formula gives the point at which these lines are meeting.

### Point of Intersection of Two Lines Formula

Consider 2 straight lines $a_1x+b_1y+c_1$ and $a_2x+b_2y+c_2$ which are intersecting at point $(x,y)$ as shown in figure. So, we have to find a line intersection formula to find these points of intersection $(x,y)$. These points satisfy both the equations.

By solving these two equations we can find the intersection of two lines formula.

The formula for the point of intersection of two lines will be as follows:

$x=\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1}$

$y=\dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}$

$(x,y)=\left(\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1} , \dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\right)$

So, this is the point of intersection formula of 2 lines when intersecting at one point.

### A Intersection B Intersection C Formula

If we have 3 sets says $A, B$ and $C$. $A$ intersection $B$ intersection $C$ formula will be as follows:

$n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(B \cap C)-n(C \cap A)+n(A \cap B \cap C)$

### The Formula of A Intersection B

If we have 2 sets say $A$ and $B$. The formula of $A$ intersection $B$ will be as follows:

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Where

$n(A\cup B)$ is the number of elements present in either set $A$ or set $B$.

$n(A \cap B)$ is the number of elements present in both set $A$ and set $B$.

### Problems on Point of Intersection of Two Lines Formula:

1. Find the Point of Intersection of Two Lines $2x + 4y + 6 = 0$ and $2x + 3y + 4 = 0$.

Ans: The two line equations are given as $2x + 4y + 6 = 0$ and $2x + 3y + 4 = 0$.

Comparing these two equations with $a_1x+b_1y+c_1$ and $a_2x+b_2y+c_2$

We get $a_1=2, b_1=4, c_1=6$

$a_2=2, b_2=3, c_2=4$

Point of intersection formula is given as

$(x,y)=\left(\dfrac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1} , \dfrac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\right)$

Substituting the values of $a_1, b_1, c_1, a_2, b_2, c_2$ in the intersection formula

$(x,y)=\left(\dfrac{4 \times 4 - 3 \times 6}{2 \times 3 - 2 \times 4} , \dfrac{6 \times 2 - 4 \times 2}{2 \times 3 - 2 \times 4}\right)$

$(x,y)=\left(\dfrac{16 -18}{6 - 8} , \dfrac{12 - 8}{6 - 8}\right)$

$(x,y)=\left(\dfrac{-2}{-2} , \dfrac{4}{-2}\right)$

$(x,y) = (1,-2)$

Therefore the point of intersection of two lines $2x + 4y + 6 = 0$ and $2x + 3y + 4 = 0$ is $(x,y) = (1,-2)$.