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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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.
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 $a_i x+b_i y+c_i=0, \mathrm{i}=1,2,3$ are concurrent if $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=0$
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.
Summary
The point of intersection of two lines represents a unique solution satisfied by both lines. We can find this method by substitution, elimination, or graphical method. These points not only represent the point of intersection of lines but also tell about the features of lines and their graphical behavior.
When two lines have a common point they are called intersecting lines. This point of intersection is called the point of intersection.
If $P\left(x_1, y_1\right)$ is a point of intersection of $L_1$ and $L_2$, then
$\left(x_1, y_1\right)=\left(\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}, \frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\right)$
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.
Three lines $a_i x+b_i y+c_i=0, \mathrm{i}=1,2,3$ are concurrent if $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=0$.
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.
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