We will discuss the equation of a line under different conditions. A line can be in different forms like horizontal, vertical, point-slope, two-point, etc.

Horizontal and vertical lines

If a horizontal line $\mathrm{L}$ is at distance $a$ from the $x$-axis then ordinate of every point lying on the line is either $a$ or $-a$ [Fig (a)]. Therefore, equation of the line $\mathrm{L}$ is either
$y=a$ or $y=-a$.
The choice of sign will depend upon the position of the line according as the line is above or below the $y$-axis.

Similarly, the equation of a vertical line at a distance $b$ from the $y$-axis is either (Fig (b))

$x=b$ or $x=-b$

Example 1: Find the equation of the lines parallel to axes and passing through (-2,3).

Solution:

The position of the lines is shown in the figure above. The $y$-coordinate of every point on the line parallel to $x$-axis is 3, therefore, the equation of the line parallel to $x$-axis and passing through $(-2,3)$ is
$y=3$.
Similarly, the equation of the line parallel to $y$-axis and passing through $(-2,3)$ is
$x=-2$.

Point-slope form

Suppose that $\mathrm{P}_0\left(x_0, y_0\right)$ is a fixed point on a non-vertical line $\mathrm{L}$, whose slope is $m$.
Let $\mathrm{P}(x, y)$ be an arbitrary point on L (Figure above). Then, by the definition, the slope of $\mathrm{L}$ is given by
$m=\frac{y-y_0}{x-x_0} \text {, i.e., } y-y_0=m\left(x-x_0\right) \dots(1)$
Since the point $\mathrm{P}_0\left(x_0, y_0\right)$ along with all points $(x, y)$ on L satisfies (1) and no other point in the plane satisfies (1). Equation (1) is indeed the equation for the given line L.
Thus, the point $(x, y)$ lies on the line with slope $m$ through the fixed point $\left(x_0, y_0\right)$, if and only if, its coordinates satisfy the equation
$y-y_0=m\left(x-x_0\right)$

Example 2: $\text { Find the equation of the line through }(-2,3) \text { with slope }-4 \text {. }$

Solution:

Here $m=-4$ and given point $\left(x_0, y_0\right)$ is $(-2,3)$.
By slope-intercept form formula
$y-y_0=m\left(x-x_0\right)$
Using this formula the equation of the given line is
$y-3=-4(x+2) \text { or }$
$4 x+y+5=0$, which is the required equation.

Two-point form

Let the line $\mathrm{L}$ passes through two given points $P_1\left(x_1, y_1\right)$ and $P_2\left(x_2, y_2\right)$. Let $\mathrm{P}(x, y)$ be a general point on L (Figure above).

The three points $\mathrm{P}_1, \mathrm{P}_2$ and $\mathrm{P}$ are collinear, therefore, we have slope of $P_1 P=$ slope of $P_1 P_2$
$\text { i.e., } \quad \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}, \quad \text { or } y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \text {. }$
Thus, the equation of the line passing through the points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is given by
$y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \dots(2)$

Example 3: Write the equation of the line through the points (1, –1) and (3, 5).

Solution:

Here $x_1=1, y_1=-1, x_2=3$ and $y_2=5$. Using the two-point form (2) above for the equation of the line, we have
$y-(-1)=\frac{5-(-1)}{3-1}(x-1)$
or
$-3 x+y+4=0$, which is the required equation.

Slope-intercept form

Sometimes a line is known to us with its slope and an intercept on one of the axes. We will now find equations of such lines.

Case I: Suppose a line L with slope $m$ cuts the $y$-axis at a distance $c$ from the origin (Figure above). The distance $c$ is called the $y$ intercept of the line L. Obviously, coordinates of the point where the line meet the $y$-axis are $(0, c)$. Thus, L has slope $m$ and passes through a fixed point $(0, c)$. Therefore, by point-slope form, the equation of $\mathrm{L}$ is
$y-c=m(x-0) \text { or } y=m x+c$
Thus, the point $(x, y)$ on the line with slope $m$ and $y$-intercept $c$ lies on the line if and
only if
$y=m x+c \dots(3)$
Note that the value of $c$ will be positive or negative according as the intercept is made on the positive or negative side of the $y$-axis, respectively.

Case II: Suppose line L with slope $m$ makes $x$-intercept $d$. Then equation of $\mathrm{L}$ is
$y-0=m(x-d)$
$y=m(x-d) \dots(4)$

Example 4: Write the equation of the lines for which $\tan \theta=\frac{1}{2}$, where $\theta$ is the inclination of the line and

(i) $y$-intercept is $-\frac{3}{2}$
(ii) $x$-intercept is 4.

Solution:

(i) Here, the slope of the line is $m=\tan \theta=\frac{1}{2}$ and $y$ – intercept $c=-\frac{3}{2}$. Therefore, by slope-intercept form (3) above, the equation of the line is
$y=\frac{1}{2} x-\frac{3}{2} \text { or } 2 y-x+3=0,$
which is the required equation.

(ii) Here, we have $m=\tan \theta=\frac{1}{2}$ and $d=4$.
Therefore, by slope-intercept form (4) above, the equation of the line is
$y=\frac{1}{2}(x-4) \text { or } 2 y-x+4=0,$
which is the required equation.

Intercept – form

Suppose a line L makes $x$-intercept $a$ and $y$-intercept $b$ on the axes. Obviously L meets $x$-ax is at the point $(a, 0)$ and $y$-axis at the point $(0, b)$ (Figure below). By two-point form of the equation of the line, we have

$y-b=\frac{0-b}{a-0}(x-0) \quad$ or $\quad a y=-b x+a b$,
i.e., $\quad \frac{x}{a}+\frac{y}{b}=1$.
Thus, the equation of the line-making intercepts $a$ and $b$ on $x$-and $y$-axis, respectively, is
$\frac{x}{a}+\frac{y}{b}=1 \dots(5)$

Example 5: Find the equation of the line, which makes intercepts $-3$ and 2 on the $x$ – and $y$-axes respectively.

Solution:

Here $a=-3$ and $b=2$. By intercept form (5) above, the equation of the line is
$\frac{x}{-3}+\frac{y}{2}=1 \quad \text { or } \quad 2 x-3 y+6=0 .$

Normal form

Suppose a non-vertical line is known to us with the following data:
(i) Length of the perpendicular (normal) from the origin to the line.
(ii) Angle which normal makes with the positive direction of x-axis.

Let $\mathrm{L}$ be the line, whose perpendicular distance from origin $\mathrm{O}$ be $\mathrm{OA}=p$ and the angle between the positive $x$-axis and $\mathrm{OA}$ be $\angle \mathrm{XOA}=\omega$. The possible positions of line $\mathrm{L}$ in the Cartesian plane are shown in the Figure below. Now, our purpose is to find slope of $\mathrm{L}$ and a point on it. Draw perpendicular AM on the $x$-axis in each case.

In each case, we have $\mathrm{OM}=p \cos \omega$ and $\mathrm{MA}=p \sin \omega$, so that the coordinates of the point $\mathrm{A}$ are $(p \cos \omega, p \sin \omega)$.
Further, line $\mathrm{L}$ is perpendicular to $\mathrm{OA}$. Therefore
The slope of the line $L=-\frac{1}{\text { slope of OA }}=-\frac{1}{\tan \omega}=-\frac{\cos \omega}{\sin \omega}$.
Thus, the line $\mathrm{L}$ has slope $-\frac{\cos \omega}{\sin \omega}$ and point $\mathrm{A}(p \cos \omega, p \sin \omega)$ on it. Therefore, by point-slope form, the equation of the line $\mathrm{L}$ is
$y-p \sin \omega=-\frac{\cos \omega}{\sin \omega}(x-p \cos \omega) \quad \text { or } \quad x \cos \omega+y \sin \omega=p\left(\sin ^2 \omega+\cos ^2 \omega\right)$
or $\quad x \cos \omega+y \sin \omega=p$.
Hence, the equation of the line having normal distance $p$ from the origin and angle $\omega$ which the normal makes with the positive direction of $x$-axis is given by
$x \cos \omega+y \sin \omega=p \dots(6)$

Example 6: Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with the positive direction of $x$-axis is $15^{\circ}$.

Solution:

Here, we are given $p=4$ and $\omega=15^0($ Figure above).
Now $\quad \cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
and $\quad \sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
By the normal form (6) above, the equation of the line is
$x \cos 15^0+y \sin 15^{\circ}=4$ or $\frac{\sqrt{3}+1}{2 \sqrt{2}} x+\frac{\sqrt{3}-1}{2 \sqrt{2}} y=4$ or $(\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}$.
This is the required equation.