Point of Intersection Formula: How to Find and Examples

What is the Point of intersection of two lines?

When two lines have a common point they are called intersecting lines. This point of intersection is called the point of intersection.

The formula of the Point of the intersection of two lines

If the equations of two non-parallel lines are

$\begin{aligned} & L_1=a_1 x+b_1 y+c_1=0 \\ & L_2=a_2 x+b_2 y+c_2=0\end{aligned}$

If $P\left(x_1, y_1\right)$ is a point of intersection of $L_1$ and $L_2$, then solving these two equations of the line by cross multiplication
$
\frac{x_1}{b_1 c_2-c_1 b_2}=\frac{y_1}{c_1 a_2-a_1 c_2}=\frac{1}{a_1 b_2-b_1 a_2}
$

We get,
$
\left(x_1, y_1\right)=\left(\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}, \frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\right)
$

What is Concurrent Lines?

Three lines are said to be concurrent if they pass through a common point, i.e. they meet at a point. Thus, if three lines are concurrent, then the point of intersection of two lines lies on the third line. This is the required condition of concurrency of three lines.

Condition for Concurrent Lines

To check if three lines are concurrent or not

1. First, find the point of intersection of any two straight lines by solving them simultaneously. If this point satisfies the third equation then three lines are concurrent.

2. Three lines $a_i x+b_i y+c_i=0, \mathrm{i}=1,2,3$ are concurrent if $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=0$

Recommended Video Based on the Point of Intersection of Two Lines

Solved Examples Based on the Point of intersection of two lines

Example 1: The straight lines $1_1$ and $1_2$ pass through the origin and trisect the line segment of the line $L: 9 x+5 y=45$ between the axes. If $m_1$ and $m_2$ are the slopes of the lines $1_1$ and $1_2$, then the point of intersection of the line $y=(m 1+m 2) x {\text {with } \mathrm{L} \text { lies on. }}$ [JEE MAINS 2023]

Solution

$
\begin{aligned}
& \rightarrow P_x=\frac{2 \times 5+1 \times 0}{1+2}=\frac{10}{3} \\
& P_y=\frac{0 \times 2+9 \times 1}{1+2}=3 \\
& P:\left(\frac{10}{3}, 3\right)
\end{aligned}
$

Similarly $\rightarrow Q_x=\frac{1 \times 5+2 \times 0}{1+2}=\frac{5}{3}$
$
\begin{aligned}
& Q_y=\frac{1 \times 0+2 \times 9}{1+2}=6 \\
& Q:\left(\frac{5}{3}, 6\right)
\end{aligned}
$

Now $m_1=\frac{3-0}{\frac{10}{3}-0}=\frac{9}{10}$
$
m_2=\frac{6-0}{\frac{5}{3}-0}=\frac{18}{5}
$

Now $L_1: y\left(m_1+m_2\right) x \Rightarrow y=\left(\frac{9}{2}\right) x \Rightarrow 9 x=2 y \ldots(2)$

from (1) & (2)

$
\begin{aligned}
& 9 x+5 y=45 \\
& 9 \mathrm{x}-2 \mathrm{y}=0 \\
& -\quad+- \\
& 7 \mathrm{y}=45 \Rightarrow \mathrm{y}=\frac{45}{7} \\
& \Rightarrow x=\frac{10}{7}
\end{aligned}
$
which satisfies, $y-x=5$
Hence, the answer is $y-x=5$

Example 2: Two sides of a parallelogram are along the lines $4 x+5 y=0$ and $7 x+2 y=0$. If the equation of one of the diagonals of the parallelogram is $11 x+7 y=9$, then the other diagonal passes through the point [JEE MAINS 2021]

Solution

Clearly point of intersection $A$ is $(0,0)$ $D$ is $\left(\frac{5}{3},-\frac{4}{3}\right)$ \& $B$ is $\left(-\frac{2}{3}, \frac{7}{3}\right)$

As diagonals bisect each other, so other diagonal passes through the midpoint of $B D$ (i.e.E)
E is $\left(\frac{1}{2}, \frac{1}{2}\right)$
Equation of $B C$ is
$
\begin{aligned}
& y-0=\left(\frac{\frac{1}{2}-0}{\frac{1}{2}-0}\right)(x-0) \\
& \Rightarrow y=x
\end{aligned}
$

It passes through $(2,2)$
Hence, the answer is $(2,2)$

Example 3: The intersection of three lines $x-y=0, x+2 y=3$ and $2 x+y=6$ is a: [JEE MAINS 2021]

Solution

Let
$
\begin{aligned}
& L_1: x-y=0 \\
& L_2: x+2 y=3 \\
& L_3: 2 x+y=6
\end{aligned}
$

Let point A be the point of intersection of L₁ and L₂, point B be the point of intersection of L₁ and L₃, and point C be the point of intersection of L₃ and L₂.

$\begin{aligned} & \mathrm{A}=(1,1) \\ & \mathrm{B}=(2,2) \\ & \mathrm{C}=(3,0)\end{aligned}$
$\begin{aligned} & \mathrm{AC}=\sqrt{(1-3)^2+(1-0)^2}=\sqrt{4+1}=\sqrt{5} \\ & \mathrm{BC}=\sqrt{(2-3)^2+(2-0)^2}=\sqrt{1+4}=\sqrt{5} \\ & \mathrm{AB}=\sqrt{(1-2)^2+(1-2)^2}=\sqrt{1+1}=\sqrt{2}\end{aligned}$

so it's an isosceles triangle

Hence, the answer is an isosceles triangle.

Example 4: The number of integral values of $\mathbf{m}$ so that the abscissa of the point of intersection of lines $3 x+4 y=9$ and $y=m x+1$ is also an integer, is
[JEE MAINS 2021]

Solution: Given, the equation of the lines

$
\begin{aligned}
& 3 x+4 y=9 \text { and } y=m x+1 \\
& 3 x+4(m x+1)=9 \\
& x(3+4 m)=5 \\
& x=\frac{5}{(3+4 m)} \\
& (3+4 m)= \pm 1, \pm 5 \\
& 4 \mathrm{~m}=-3 \pm 1,-3 \pm 5 \\
& 4 \mathrm{~m}=-4,-2,-8,2 \\
& \mathrm{~m}=-1,-\frac{1}{2},-2, \frac{1}{2}
\end{aligned}
$

Two integral values of $m$ are possible.

Hence, the answer is 2.

Example 5: If the three distinct lines $x+2 a y+a=0, x+3 b y+b=0$ and $x+4 a y+a=0$ are concurrent, then the point ( a , b) lies on a
[JEE MAINS 2014]

$
\begin{aligned}
&\left|\begin{array}{lll}
1 & 2 a & a \\
1 & 3 b & b \\
1 & 4 a & a
\end{array}\right| \\
& \Rightarrow(3 a b-2 a b)-1\left(2 a^2\right)+1(a b) \\
& \Rightarrow 2 a b-2 a^2=0 \\
& a^2-a b=0 \\
& a(a-b)=0
\end{aligned}
$

Replace $(a, b)$ by $(x, y)$
$
x(x-y)=0
$

So (a.b) lies on a straight line either $\mathrm{x}=0$ or $\mathrm{x}=\mathrm{y}$
Hence, the answer is that points ( $\mathrm{a}, \mathrm{b}$ ) lie in a straight line.