Straight Line - Equations, Definition, Properties and Examples

A straight line in geometry can be described by its various points in terms of its equation. A line is an infinite length between two points. Adding on to this basic definition we can say that a straight line is a line that does not have any curves on its path and extends between 2 points.

In this article, we will cover the concept 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 eight questions have been asked on JEE MAINS( 2013 to 2023) from this topic including two in 2013, three in 2019, one in 2020, one in 2021, and one in 2023.

What is a Straight Line?

It is a curve such that all points on the line segment joining any two points on it lie on it.

Every equation of the first degree in $x, \text{and} \space y$ represents a straight line. The general equation of a straight line is given as $ax+b y+c=0$ where a, b, and c are real numbers and at least one of a and b is non-zero.

Slope of a Line

The slope of a line tells us the direction in which a line is drawn.

A line in a coordinate plane forms two angles with the x-axis, which are supplementary.

The angle $\theta$ made by the line 'l' with the positive direction of the x-axis and measured anticlockwise is called the inclination of the line. $\theta$ lies in the range $\left[0^{\circ}, 180^{\circ}\right)$

If $\theta$ is the angle at which a straight line is inclined to a positive direction of the x-axis, then the slope (or gradient) of this line is defined by m $=\tan \theta$

Slope will be positive when $\theta$ is an acute angle and it will be negative when $\theta$ is an obtuse angle.

Important points

1) The inclination of a line parallel to the y-axis or vertical line. The slope is not defined for a vertical line, as $\theta$ $=90^{\circ}$, and $\tan \left(90^{\circ}\right)$ is not defined.

2) If a line is equally inclined with the coordinate axes then it will make an angle of $45^{\circ}$ and $135^{\circ}$ in the positive direction of the x-axis. In this case, the slope will be $\tan \left(45^{\circ}\right)$ or $\tan \left(135^{\circ}\right)$. i.e. $m=1$ or $m=-1$.

3) The inclination of a line parallel to the x-axis is 0 degrees. Thus, the slope of a horizontal line is tan(0 ) = 0.

If a is the inclination of the line, then its slope is $\tan (a)$, generally denoted by m.

Thus, $\mathrm{m}=\tan (\mathrm{a})$

If $a$ is acute, then slope, $\tan (a)>0$
If $a$ is obtuse, then slope, $\tan (a)<0$

The slope of the line joining two given points

${ }_{\text {If }} \mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ are two points on a straight line then the slope of the line is
$
\tan \theta=\frac{B C}{A C}=\frac{y_2-y_1}{x_2-x_1}
$

Intercepts of a Line

X-Intercept: The x-coordinate of the point where the straight line cuts the x-axis.

Here straight line cuts the X-axis at $(x, 0)$ so

• the x-intercept is x (it can be positive or negative)
• The length of the x-intercept is $|x|$.

Y-intercept: The y-coordinate of the point where the straight line cuts the y-axis.

Here straight line cuts the Y-axis at $(0, y)$ so

• the y-intercept is y (it can be positive or negative)
• The length of the y-intercept is $|y|$.

The slope of the Line having given equation

The general equation of a straight line is $a x+b y+c=0$.

To get the slope of the line, we require two points on the line.

This line meets axes at points $\mathrm{A}(-\mathrm{c} / \mathrm{a}, 0)$ and $\mathrm{B}(0,-\mathrm{c} / \mathrm{b})$.

Slope of line = $-(a / b)$

Important points

1) If three points A, B, and C are collinear, then the slope of AB= slope of BC= slope of AC

2) If two lines are parallel, then their inclinations and hence, slopes are the same.

The angle between two lines

Let us get the formula for the angle between two lines having slopes m1 and m2.

Let the inclinations of the lines L and M be a and b, respectively.

So, $\mathrm{m} 1=\tan (\mathrm{a})$ and $\mathrm{m} 2=\tan (\mathrm{b})$.

Important points

1) Parallel lines: if $L_1$ and $L_2$ are parallel, then $m_1=m_2$. So, $\tan (a)=0$ or $a=0$ degree.
2) Perpendicular lines: if $L_1$ and $L_2$ are perpendicular , then $m_1, m_2=-1$.

Thus, the product of slopes of two perpendicular lines is -1 . If $m$ is the slope of one line then the slope of the perpendicular line is $-(1 / \mathrm{m})$.

Equation of Straight Line

(a) Slope-Intercept form

The equation of a straight line whose slope is given as $m$ and making $y$ intercept of length c unit is $\mathrm{y}=\mathrm{mx}+\mathrm{c}$.

If the straight line passes through the origin, then the equation of the straight line becomes $y=m x$
(b) Point-Slope form

The equation of a straight line whose slope is given as ' $m$ ' and passes through the point $\left(x_1, y_1\right)$ is $\mathbf{y}-\mathbf{y}_1=\mathbf{m}\left(\mathbf{x}-\mathbf{x}_1\right)$.
(c) Two-point form

The equation of a straight line passing through the two given points $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\left(x_2, y_2\right)$ is given by
$
\mathbf{y}-\mathbf{y}_1=\left(\frac{\mathbf{y}_2-\mathbf{y}_1}{\mathbf{x}_2-\mathbf{x}_1}\right)\left(\mathbf{x}-\mathbf{x}_1\right)
$
(d) Intercept form of line

The equation of a straight line which makes intercepts ' $a$ ' and ' $b$ ' on the $X$ axis and $Y$-axis respectively is given by
$
\frac{x}{a}+\frac{y}{b}=1
$

e) Normal form of a line

The equation of a straight line on which the length of the perpendicular from the origin is $p$ and this normal makes an angle $\theta$ with the positive direction of the X -axis is given by
$
\mathbf{x} \cos \theta+\mathbf{y} \sin \theta=\mathbf{p}
$
f) Parametric form of a line

The equation of a straight line passing through the point $\left(x_1, y_1\right)$ and making an angle $\theta$ with the positive direction of the X -axis is
$
\frac{\mathrm{x}-\mathrm{x}_1}{\cos \theta}=\frac{\mathrm{y}-\mathrm{y}_1}{\sin \theta}=\mathrm{r}
$

Angle between two straight line
Two lines are given with the slope $m_1$ and $m_2$, then acute angle $\theta$ between the lines is given by
$
\theta=\tan ^{-1}\left|\frac{\mathrm{m}_1-\mathrm{m}_2}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|
$

If the angle between the two lines is $0^{\circ}$ or $\pi$ then lines are parallel two each other. In this case, $m_1=m_2$ where $m_1$ and $m_2$ are slopes of two lines.

If the angle between the two lines is $\frac{\pi}{2}$ or $-\frac{\pi}{2}$ then lines are perpendicular two each other. Then in this case $m_1 \cdot m_2=-1$ where $m_1$ and $m_2$ are slopes of two lines.

Position of two points with respect to a line

Two given points $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ and $\mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)$ lies on the same side of a line $a x+b y+c=0$ when $\frac{a x_1+b y_1+c}{a x_2+b y_2+c}>0$ and points lie on the opposite side when $\frac{\mathrm{ax}_1+\mathrm{by}_1+\mathrm{c}}{\mathrm{ax}_2+\mathrm{by}_2+\mathrm{c}}<0$
1. A point $(p, q)$ will lie on the origin side of the line $a x+b y+c=0$ if $\frac{a p+b q+c}{a .0+b .0+c}>0$, meaning ap $+\mathrm{bq}+\mathrm{c}$ and c will have the same sign.
2. A point $(p, q)$ will lie on the non-origin side of the line $a x+b y+c$ $=0$, if $\frac{a p+b q+c}{a .0+b .0+c}<0$, meaning ap $+\mathrm{bq}+\mathrm{c}$, and c will have the opposite sign.

Distance of a point from a line

The distance of a point from a line is the perpendicular distance between a point and the line.

Perpendicular length from a point $\left(x_1, y_1\right)$ to the line $L$ : $A x+B y+C=0$ is
$
\frac{\left|\mathrm{Ax}_1++\mathrm{By}_1+\mathrm{C}\right|}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}
$

Distance between two parallel lines

The equation of two parallel lines is $a x+b y+c=0$ and $a x+b y+d=0$, then the distance between them is the perpendicular distance of any point on one line from the other line.

If $\left(x_1, y_1\right)$ is any point on the line $a x+b y+c=0$
Then, $a x_1+b y_1+c=0$
Now, the perpendicular distance of the point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ from the line $\mathrm{ax}+$ by + $d=0$ is
$
\frac{\left|\mathrm{ax}_1+\mathrm{by}_1+\mathrm{d}\right|}{\sqrt{\mathbf{a}^2+\mathbf{b}^2}}=\frac{|d-\mathbf{d}|}{\sqrt{\mathbf{a}^2+\mathbf{b}^2}}
$

Point of 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)
$

Concurrent Lines

If three straight lines meet in a point then three given lines are called concurrent.

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 $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|=0$.

Family of Lines

Any equation of the line through the point of intersection of the lines $\mathrm{L}_1=\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{L}_2=\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$ can be represented as
$
\begin{aligned}
& a_1 x+b_1 y+c_1+\lambda\left(a_2 x+b_2 y+c_2\right)=0 \\
& \text { or, } L_1+\lambda L_2=0
\end{aligned}
$

Where $\lambda$ is a parameter.

Angle Bisectors

The line which bisects the angle in two equal halves is called Angle Bisectors.

The equation of the angle bisectors between the two lines

$\mathrm{L}_1=\mathrm{a}_1 \mathrm{x}+\mathrm{b}_1 \mathrm{y}+\mathrm{c}_1=0$ and $\mathrm{L}_2=\mathrm{a}_2 \mathrm{x}+\mathrm{b}_2 \mathrm{y}+\mathrm{c}_2=0$ is $\frac{\left(a_1 x+b_1 y+c_1\right)}{\sqrt{a_1^2+b_1^2}}= \pm \frac{\left(a_2 x+b_2 y+c_2\right)}{\sqrt{a_2^2+b_2^2}}$

Pair of Straight Lines

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
$
\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 $y$.

Solved Examples Based on Straight Lines

Example 1: A triangle is formed by the X -axis, Y -axis, and the line $3 \mathrm{x}+4 \mathrm{y}=$ $4 y=60$. Then the number of points $P(a, b)$ which lie strictly inside the triangle, where $a$ is an integer and $b$ is a multiple of $a$, is [JEE MAINS 2023]

Solution
$
\begin{aligned}
& 3 x+4 y=60 \\
& x=1,4 y=57, y=14.2 \\
& x=1, y=1,2,3, \ldots \ldots 14 \rightarrow 14 \text { points } \\
& x=2,4 y=54, y=13.5 \\
& x=2, y=2,4,6,8,10,12 \rightarrow 6 \text { points } \\
& x=3, y=3,6,9,12 \rightarrow 4 \text { points } \\
& x=4, y=4,8 \rightarrow 2 \text { points } \\
& x=5, y=5,10 \rightarrow 2 \text { points } \\
& x=6, y=6 \rightarrow 1 \text { points } \\
& x=7, y=7 \rightarrow 1 \text { points } \\
& x=8, y=8 \rightarrow 1 \text { points } \\
& x=9,4 y=23, y=5.7 \quad \times \text { no point }
\end{aligned}
$
$
\text { Total points }=14+6+4+2+2+1+1+1=31
$

Hence, the answer is 31 .

Example 2: If $\alpha, \beta$ are natural numbers such that $100^\alpha-199 \beta=(100)(100)+(99)(101)+(98)(102)+\ldots \ldots+(1)(199)$, then the slope of the line passing through $(\alpha, \beta)$ and the origin is $\quad[\mathrm{JEE}$ MAINS 2021]

Solution
$
\begin{aligned}
& \text { RHS }=\sum_{\mathrm{r}=0}^{99}(100-\mathrm{r})(100+\mathrm{r})=\sum_{\mathrm{r}=0}^{99}\left(100^2-r^2\right) \\
& =(100)^3-\frac{99 \times 100 \times 199}{6}=(100)^3-(1650) 199 \\
& \text { LHS }=(100)^\alpha-(199) \beta
\end{aligned}
$

So, $\alpha=3, \beta=1650$
Slope $=\tan \theta=\frac{\beta-0}{\alpha-0}=\frac{1650}{3}$
$
\tan \theta=550
$

Hence, the answer is 550

Example 3: If the perpendicular bisector of the line segment joining the points $P(1,4)$ and $Q(k, 3)$ has a $y$-intercept equal to -4 then a value of $k$ is:
[JEE MAINS 2020]

Solution

Slope $=m=\frac{1}{1-k}$ equation of perpendicular bisector is
$
\begin{aligned}
& y+4=(k-1)(x-0) \\
& \Rightarrow y+4=x(k-1) \\
& \Rightarrow \frac{7}{2}+4=\frac{k+1}{2}(k-1) \\
& \Rightarrow \frac{15}{2}=\frac{\mathrm{k}^2-1}{2} \Rightarrow \mathrm{k}^2=16 \Rightarrow \mathrm{k}=4,-4
\end{aligned}
$

Hence, the answer is -4.

Example 4: If the line $3 x+4 y-24=0$ intersects the x -axis at the point A and the y -axis at the point B , then the incentre of the triangle OAB , where O is the origin, is:
[JEE MAINS 2019]

Solution:

From the fig $(r, r)$ is the center of the circle.
Distance from point $(\mathrm{r}, \mathrm{r})$ to the line $3 x+4 y-24=0$
$
\begin{aligned}
& \left|\frac{3 r+4 r-24}{\sqrt{9+16}}\right|=r \\
& 7 r-24= \pm 5 r \\
& r=14 \text { or } r=2
\end{aligned}
$

Center is $(2,2)$
Hence, the answer is $(2,2)$

Example 5: If a straight line passing through the point $P(-3,4)$ is such that its intercepted portion between the coordinates axes is bisected at P , then its equation is :
[JEE MAINS 2019]

Solution: $x$-intercept -The distance on the x -axis from the origin where the straight line cuts it.

Intercept form of a straight line -
$
\frac{x}{a}+\frac{y}{b}=1
$
$a$ and $b$ are the $x$-intercept and $y$-intercept respectively.
So that line says intersect the $y$-axis at ( $0, \mathrm{~b}$ ) and x -axis at ( $a, 0$ ).
since it is bisected at $p(-3,4)$
$
\begin{aligned}
\Rightarrow a & =2 \times(-3) \\
& =-6
\end{aligned}
$
$
\begin{aligned}
& b=2 \times 4 \\
& =8
\end{aligned}
$

Equation of line -
$
\begin{aligned}
& \frac{x}{a}+\frac{y}{b}=1 \\
& \frac{x}{-6}+\frac{y}{8}=1 \\
& \Rightarrow 4 x-3 y+24=0
\end{aligned}
$

Hence, the answer is $4 x-3 y+24=0$

Summary

The straight line is a fundamental concept of coordinate geometry. Its properties provide a wide variety of geometrical applications. Mastering Straight lines helps us to solve complex problems.