Point of Intersection Formula: How to Find and Examples

Q: How do you calculate the intersection point?

If $P\left(x_1, y_1\right)$ is a point of intersection of $L_1$ and $L_2$, then $\left(x_1, y_1\right)=\left(\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}, \frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\right)$

Edited By Komal Miglani | Updated on Feb 15, 2025 02:02 AM IST

In this article, we will cover the concept of the Point of the intersection of two lines. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of six questions have been asked on JEE MAINS( 2013 to 2023) from this topic including two in 2014, three in 2021, and one in 2023.

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This Story also Contains

What is the Point of intersection of two lines?
What is Concurrent Lines?
Condition for Concurrent Lines
Solved Examples Based on the Point of intersection of two lines

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 $P (x_{1}, y_{1})$ 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,
$(x_{1}, y_{1}) = (\frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}, \frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}})$

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

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.
Three lines $a_{i} x + b_{i} y + c_{i} = 0, i = 1, 2, 3$ are concurrent if $| \begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array} | = 0$

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 $y = (m 1 + m 2) x with L lies on.$ [JEE MAINS 2023]

Solution

$\begin{aligned} \to P_{x} = \frac{2 \times 5 + 1 \times 0}{1 + 2} = \frac{10}{3} \\ P_{y} = \frac{0 \times 2 + 9 \times 1}{1 + 2} = 3 \\ P : (\frac{10}{3}, 3) \end{aligned}$

Similarly $\to Q_{x} = \frac{1 \times 5 + 2 \times 0}{1 + 2} = \frac{5}{3}$
$\begin{aligned} Q_{y} = \frac{1 \times 0 + 2 \times 9}{1 + 2} = 6 \\ Q : (\frac{5}{3}, 6) \end{aligned}$

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 $L_{1} : y (m_{1} + m_{2}) x \Rightarrow y = (\frac{9}{2}) x \Rightarrow 9 x = 2 y \dots (2)$

from (1) & (2)

$\begin{aligned} 9 x + 5 y = 45 \\ 9 x - 2 y = 0 \\ - + - \\ 7 y = 45 \Rightarrow y = \frac{45}{7} \\ \Rightarrow x = \frac{10}{7} \end{aligned}$
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 $(\frac{5}{3}, - \frac{4}{3})$ \& $B$ is $(- \frac{2}{3}, \frac{7}{3})$

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

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.

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

so it's an isosceles triangle

Hence, the answer is an isosceles triangle.

Example 4: The number of integral values of $m$ 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

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

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]

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

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

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

Frequently Asked Questions (FAQs)

1. What is the intersection point?

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

2. How do you calculate the intersection point?

If $P (x_{1}, y_{1})$ is a point of intersection of $L_{1}$ and $L_{2}$ , then

$(x_{1}, y_{1}) = (\frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}, \frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}})$

3. What are concurrent lines?

Three lines are said to be concurrent if they pass through a common point, i.e. they meet at a point. Concurrent lines have a common point of interaction.

4. What is the required condition of concurrent lines?

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.