Homogeneous Equations in Two Variables

The homogenous equation is that equation in which each term has the same degree. A general equation in the second degree represents either a conic section or a pair of straight lines. Based on different criteria we differentiate the pair of straight lines represents a conic section or a pair of straight lines.

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 eleven questions have been asked on JEE MAINS( 2013 to 2023) from this topic including two in 2021 and one in 2023.

Homogeneous Equations in Two Variables

Homogeneous equations are those equations where each term has the same degree of variable terms.
The equation $a x^2+2 h x y+b y^2=0$ is a homogeneous equation of second degree.
It always represents two straight lines passing through the origin.
The given equation is :
$
a x^2+2 h x y+b y^2=0
$

Divide the equation by $x^2$
$
\Rightarrow \quad a+2 h\left(\frac{y}{x}\right)+b\left(\frac{y^2}{x^2}\right)=0
$
now put $\frac{y}{x}=m$
$
\Rightarrow \quad \mathrm{a}+2 \mathrm{~h}(\mathrm{~m})+\mathrm{b}\left(\mathrm{m}^2\right)=0
$

If $\mathrm{m}_1$ and $\mathrm{m}_2$ are roots, then
$
\mathrm{m}_1+\mathrm{m}_2=-\frac{2 \mathrm{~h}}{\mathrm{~b}}
$
amd, $\quad \mathrm{m}_1 \mathrm{~m}_2=\frac{\mathrm{b}}{\mathrm{a}}$
Thus, $y=m_1 x$ and $y=m_2 x$ are two straight lines given by Eq. (i)

Real and Imaginary Lines

For equation (ii), if roots $m_1$ and $m_2$ are real, then the lines are also real, and if $m_1$ and $m_2$ are not real (imaginary), then the lines are also not real (imaginary). This depends on the discriminant of equation (ii)
Discriminant $=4 h^2-4 a b=4\left(h^2-a b\right)$
1. The lines are real and distinct if $D>0: h^2-a b>0$
2. The lines are coincident if $D=0: h^2-a b=0$
3. The lines are imaginary if $D<0: h^2-a b<0$

The angle between a pair of Straight Lines

The general equation of a pair of straight lines is $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$. The angle between the pair of straight lines is
$
\theta=\tan ^{-1}\left\{\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{|\mathrm{a}+\mathrm{b}|}\right\}
$

Homogenization of Second-Degree Equation

Consider the general second-degree equation

$S \equiv a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$

This equation may represent a pair of straight lines, circles, parabolas, ellipses, or hyperbolas.

Now, let a straight line: $L=\mid x+m y+n=0$ (2) intersects the curve at two points $A$ and $B$.

The combined equation of a pair of straight lines $O A$ and $O B$ must be homogenous and must contain only second-degree terms.
In order to make Eq (1) homogeneous using Eq (2), we can use (2) to write
$
1=\frac{\mathrm{lx}+\mathrm{my}}{-\mathrm{n}}
$
now, the equation of the curve is
$
S \equiv\left(a x^2+2 \mathrm{hxy}+\mathrm{by}^2\right)+(2 \mathrm{gx}+2 \mathrm{fy}) \cdot 1+\mathrm{c} \cdot 1=0
$
put the value of 1 in the above equation, we get
$
\left(a x^2+2 h x y+b y^2\right)+(2 g x+2 f y)\left(\frac{l x+m y}{-n}\right)+c\left(\frac{l x+m y}{-n}\right)^2=0
$

Since points A and B satisfy the above equation, it represents the pair of straight lines OA and OB. This process of making a second-degree equation homogeneous is called homogenization through which we get the equation of a pair of straight lines joining the points of intersection of the given line and curve with the origin.

Solved Examples Based on Homogeneous Equations in Two Variables

Example 1: The combined equation of the two lines $a x+b y+c=0$ and $a^{\prime} x+b^{\prime} y+c^{\prime}=0$ can be written as $(a x+b y+c)\left(a^{\prime} x+b^{\prime} y+c^{\prime}\right)=0$. The equation of the angle bisectors of the lines represented by the equation $2 x^2+x y-3 y^2=0$ is
[JEE MAINS 2023]

Solution: For a pair of straight lines in the form
$
a x^2+b y^2+2 h x y+2 g x+2 f y+c=0
$
equation of angle bisector is
$
\begin{aligned}
& \frac{x^2-y^2}{a-b}=\frac{x y}{h} \\
& \text { for } 2 \mathrm{x}^2+\mathrm{xy}-3 \mathrm{y}^2=0 \\
& \mathrm{a}=2, \mathrm{~b}=-3, \mathrm{~h}=\frac{1}{2}
\end{aligned}
$
equation of angle bisector is
$
\begin{aligned}
& \frac{x^2-y^2}{5}=\frac{x y}{1 / 2} \\
& \Rightarrow \mathrm{x}^2-\mathrm{y}^2-10 \mathrm{xy}=0
\end{aligned}
$

Hence, the answer is $x^2-y^2-10 x y=0$

Example 2: Let the equation of the pair of lines, $y=p x$ and $y=q x$, can be written as $(y-p x)(y-q x)=0$. Then the equation of the pair of the angle bisectors of the lines $x^2-4 x y-5 y^2=0$ is:
[JEE MAINS 2021]

Solution: Now $x^2-4 x y-5 y^2=0$
$
a=1,2 h=-4 \Rightarrow h=-2, b=-5
$

The equation of bisectors is
$
\begin{aligned}
& \frac{x^2-y^2}{1+5}=\frac{x y}{-2} \\
& \Rightarrow-2 x^2+2 y^2=6 x y \\
& \Rightarrow x^2+3 x y^2-y^2=0
\end{aligned}
$

Hence, the answer is $x^2+3 x y-y^2=0$

Example 3: If $4 x^2+4 x y=0$ is an equation of a pair of straight lines and $\tan ^{-1} m$ is is angle between two lines then find out the value of $m$.
Solution: The angle between the pair of straight lines is
$
\begin{gathered}
\theta=\tan ^{-1}\left\{\frac{2 \sqrt{h^2-a b}}{|a+b|}\right\} \\
4 x^2+4 x y=0
\end{gathered}
$

Angle between the pair of the straight line is
$
\begin{aligned}
\theta & =\tan ^{-1}\left\{\frac{2 \sqrt{h^2-a b}}{|a+b|}\right\} \\
a & =4, b=0, h=2 \\
\theta & =\tan ^{-1}\left\{\frac{2 \sqrt{2^2-4 \times 0}}{|4+0|}\right\} \\
\theta & =\tan ^{-1}(1) \\
m & =1
\end{aligned}
$

Hence, the answer is 1.

Example 4: The angle between the lines $x^2+4 x y+y^2=0$, is
Solution: The angle between a pair of lines
$
\tan \theta=2\left|\frac{\sqrt{h^2-a b}}{a+b}\right|
$

Here $a=1, b=1, h=2$
$
\begin{aligned}
& \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\frac{2 \sqrt{4-1}}{2}=\sqrt{3} \\
& \therefore \quad \theta=60^{\circ}
\end{aligned}
$

Hence, the answer is 60 degrees.

Example 5: If the component lines whose combined equation is $a x^2-b x y-y^2=0$ makes angle $\phi$ and $\theta$ with the x -axis, then find the value of $\tan (\phi+\theta)$.
Solution
Given Equation is $a x^2-b x y-y^2=0$
Let $m_1(=\tan \phi), m_2(=\tan \theta)$ be the slopes of the lines
$
\tan (\phi+\theta)=\frac{m_1+m_2}{1-m_1 m_2}
$

As we know, for $a x^2+2 h x y+b y^2=0$
$
\begin{aligned}
& \mathrm{m}_1+\mathrm{m}_2=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \\
& \mathrm{~m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{\mathrm{b}}
\end{aligned}
$
$
\begin{aligned}
& \text { So, } \mathrm{m}_1+\mathrm{m}_2=-\frac{-b}{-1}=-b \\
& \mathrm{~m}_1 \mathrm{~m}_2=\frac{\mathrm{a}}{-1} \\
& \tan (\phi+\theta)=\frac{-b}{1+a}
\end{aligned}
$

Summary

Homogenous equation plays a crucial role in coordinate geometry and differential equations. It makes our calculation as all the variables are of the same degree.