Pair of Straight Lines: Formulas & Solved Examples

A straight line, also called just a line, is a never-ending, one-dimensional shape that only has length and no width. It consists of an endless series of points. We use a linear equation with two variables to represent a straight line. Sometimes, a quadratic equation in two variables can describe a pair of straight lines.

This Story also Contains

1. What is a Pair of Straight Lines?
2. Derivation of Joint Equation of a Pair of Straight Lines
3. Condition for a general two-degree equation to represent a pair of straight lines
4. Derivation of a general two-degree equation to represent a pair of straight lines
5. Point of Intersection of Pair of Straight Lines
6. Derivation of Point of Intersection of Pair of Straight Lines
7. Solved Examples Based on a Pair of Straight Line

In this article, we will cover the concept of Pair of straight 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 thirty questions have been asked on JEE MAINS( 2013 to 2023) from this topic.

What is a Pair of Straight Lines?

A pair of straight lines can be represented as a product of two linear equations in $x$ and $y$, representing a straight line.

If the equation of two straight lines is $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=$ 0, then the combined or joint equation of these two lines is
$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$

Derivation of Joint Equation of a Pair of Straight Lines

$
\left(a_1 x+b_1 y+c_1\right)\left(a_2 x+b_2 y+c_2\right)=0
$

Multiplying the brackets we get
$
\begin{aligned}
& \Rightarrow a_1 a_2 x^2+\left(a_1 b_2+a_2 b_1\right) x y+b_1 b_2 y^2+\left(a_1 c_2+c_1 a_2\right) x+\left(b_1 c_2+c_1 b_2\right) y+ \\
& c_1 c_2=0
\end{aligned}
$

Here the coefficients can be re-named and this equation can be re-written as
$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$

So, a pair of straight lines is represented by a two-degree equation in $x$ and $y$.

Condition for a general two-degree equation to represent a pair of straight lines

A general two degree equation $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ may represent $a$

• Pair of straight lines, or
• Circle, or
• Parabola, or
• Ellipse, or
• Hyperbola

To identify which of the following curves is represented by a given equation, we have a determinant

$\Delta=\left|\begin{array}{lll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right|$

If this determinant equals zero, then the given equation represents a pair of straight lines.

Derivation of a general two-degree equation to represent a pair of straight lines

If we write the equation as a quadratic in x , where $\mathrm{a} \neq 0$
$
a x^2+2 x(h y+g)+\left(b y^2+2 f y+c\right)=0
$

Solving for ' $x$ ', we get
$
\begin{aligned}
x & =\frac{-2(h y+g) \pm \sqrt{4(h y+g)^2-4 a\left(b^2+2 f y+c\right)}}{2 a} \\
& =\frac{-(h y+g) \pm \sqrt{\left(h^2-a b\right) y^2+2(g h-a f) y+\left(g^2-a c\right)}}{a}
\end{aligned}
$

For this equation to represent two straight lines, we must have ' $x$ ' as a linear expression in ' $y$ '. For that, the value under the square root must be a perfect square of some linear expression in ' $y$ '. For this to happen, the Discriminant of the quadratic equation in $y$ (the equation inside the root) must be zero.

So,
Discriminant of $\left(h^2-a b\right) y^2+2(g h-a f) y+\left(g^2-a c\right)$ is 0
$
\begin{array}{ll}
\Rightarrow & 4(g h-a f)^2-4\left(h^2-a b\right)\left(g^2-a c\right)=0 \\
\Rightarrow & a b c+2 f g h-a f^2-b g^2-c h^2=0
\end{array}
$

This is the condition for which $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ represents a pair of straight lines.
This condition can also be written in a determinant form as
$
\left|\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right|=0
$

Point of Intersection of Pair of Straight Lines

To get the point of intersection, first find the separate equations of straight lines and solve them simultaneously.
Or
The point of intersection of a pair of straight lines can also be determined with the help of partial differentiation
Point of Intersection of Pair of Straight Lines is
$
\therefore \quad(x, y)=\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}\right)
$

Derivation of Point of Intersection of Pair of Straight Lines

Let $\phi \equiv \mathrm{ax}^2+2 \mathrm{hxy}+\mathrm{by}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$
Differentiating $\phi$ with respect to $x$, keeping $y$ constant, we get,
$
\frac{\partial \phi}{\partial \mathrm{x}}=2 \mathrm{ax}+2 \mathrm{hy}+2 \mathrm{~g}
$

Similarly, differentiating $\phi$ with respect to $y$, keeping $x$ constant, we get
$
\frac{\partial \phi}{\partial \mathrm{y}}=2 \mathrm{hx}+2 \mathrm{by}+2 \mathrm{f}
$

For point of intersection, we get
$
\frac{\partial \phi}{\partial \mathrm{x}}=0 \text { and } \frac{\partial \phi}{\partial \mathrm{y}}=0
$

Thus, we have $a x+h y+g=0$ and $h x+b y+f=0$.
Solving the two equations, we get
$
\begin{aligned}
& \frac{x}{f h-b g}=\frac{y}{g h-a f}=\frac{1}{a b-h^2} \\
\therefore & (x, y)=\left(\frac{b g-f h}{h^2-a b}, \frac{a f-g h}{h^2-a b}\right)
\end{aligned}
$

IMPORTANT NOTE:

In the above equations, $2 a x+2 h y+2 g=0$ and $2 h x+2 b y+2 f=0$ are NOT the equations of components of the straight line.

Recommended Video Based on Pair of Straight Lines

Solved Examples Based on a Pair of Straight Line

Example 1: The combined equation of the lines $x+y+2=0, y=x+2$ is
Solution:

Let us first re-write the given equations in standard form (with 0 on one side)
$
x+y+2=0 \text { and } x-y+2=0
$

Joint Equation $=(x+y+2)(x-y+2)=0 \Rightarrow(x+2)^2-y^2=0 \Rightarrow x^2-y^2+4 x+4=0$
Hence, the answer is $x^2-y^2+4 x+4=0$

Example 2: Find the separate equations of the lines represented by $x^2-4 x y+3 y^2=0$.
Solution:

In this question, we can use the method learned.
But as we do not have any term containing $x$, $y$, or constant term, we can easily factorize such expressions using middle-term split and get the desired lines
$
\begin{aligned}
& x^2-4 x y+3 y^2=0 \\
& \Rightarrow \quad x^2-x y-3 x y+3 y^2=0 \\
& \Rightarrow \quad x(x-y)-3 y(x-y)=0 \\
& \Rightarrow \quad(x-y)(x-3 y)=0
\end{aligned}
$

As combined equation is $(x-y)(x-3 y)=0$
so, separate equations are $x-y=0, x-3 y=0$
Hence, the answer is $x-y=0, x-3 y=0$

Example 3: If the sum of the slopes of the lines given by $x^2-2 c x y-7 y^2=0$ is four times their product, then $c$ has the value
Solution:

$x^2=2 c x y-7 y^2=0$
Divide equation by
$
\begin{gathered}
x^2 ; \frac{y}{x}=m \\
1-2 c m-7 m^2=0 \\
7 m^2+2 c m-1=0 \\
m_1+m_2=4 m_1 m_2
\end{gathered}
$
$
-\frac{2 c}{7}=4 \times \frac{-1}{7} \quad C=2
$

Hence, the answer is 2

Example 4: The absolute value of the difference of the slopes of the lines $x^2\left(\sec ^2 \theta-\sin ^2 \theta\right)-2 x y \tan \theta+y^2 \sin ^2 \theta=0$ is :
Solution:

The angle between a pair of lines -
$
\tan \theta=2\left|\frac{\sqrt{h^2-a b}}{a+b}\right|
$

Equation $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$
$
\begin{aligned}
& x^2\left(\sec ^2 \theta-\sin ^2 \theta\right)-2 x y \tan \theta+y^2 \sin ^2 \theta=0 \\
& \left|m_1-m_2\right|=\sqrt{\left(m_1+m_2\right)^2-4 m_1 m_2} \\
& \sqrt{\left(\frac{2 \tan \theta}{\sin ^2 \theta}\right)^2-4\left(\frac{\sec ^2 \theta-\sin ^2 \theta}{\sin ^2 \theta}\right)}=2
\end{aligned}
$

Hence, the answer is 2.

Example 5: If the pairs of straight lines $x^2-2 p x y-y^2=0$ and $x^2-2 q x y-y^2=0$ be such that each pair bisects the angle between the other pair, then
Solution:

$x^2+2 p x y-y^2=0$
and $x^2-2 p x y-y^2=0$
They bisect the angle between the other pair
The equation of bisector is given by
$
\frac{x^2-y^2}{1-(-1)}=\frac{x y}{-p} \Rightarrow x^2+\frac{2}{p} x y-y^2=0
$

So $\frac{2}{p}=-2 q \Rightarrow p q=-1$
Hence, the answer is $p q=-1$.