We have studied the general equation of first degree in two variables, \(\mathrm{Ax}+\mathrm{B} y+\mathrm{C}=0\), where \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are real constants such that \(\mathrm{A}\) and \(\mathrm{B}\) are not zero simultaneously. Graph of the equation \(\mathrm{Ax}+\mathrm{B} y+\mathrm{C}=0\) is always a straight line.Therefore, any equation of the form \(\mathrm{A} x+\mathrm{B} y+\mathrm{C}=0\), where \(\mathrm{A}\) and \(\mathrm{B}\) are not zero simultaneously is called general linear equation or general equation of a line.

The general equation of a line can be reduced into various forms of the equation of a line, by the following procedures:

Slope-intercept form

If \(\mathrm{B} \neq 0\), then \(\mathrm{A} x+\mathrm{By}+\mathrm{C}=0\) can be written as \(y=-\frac{\mathrm{A}}{\mathrm{B}} x-\frac{\mathrm{C}}{\mathrm{B}}\) or \(y=m x+c \dots(1)\)
where \(\quad m=-\frac{\mathrm{A}}{\mathrm{B}}\) and \(c=-\frac{\mathrm{C}}{\mathrm{B}}\).
We know that Equation (1) is the slope-intercept form of the equation of a line whose slope is \(-\frac{\mathrm{A}}{\mathrm{B}}\), and \(y\)-intercept is \(-\frac{\mathrm{C}}{\mathrm{B}}\).

If \(\mathrm{B}=0\), then \(x=-\frac{\mathrm{C}}{\mathrm{A}}\), which is a vertical line whose slope is undefined and \(x\)-intercept is \(-\frac{\mathrm{C}}{\mathrm{A}}\).

Intercept form

If \(\mathrm{C} \neq 0\), then \(\mathrm{A} x+\mathrm{B} y+\mathrm{C}=0\) can be written as
\(
\frac{x}{-\frac{\mathrm{C}}{\mathrm{A}}}+\frac{y}{-\frac{\mathrm{C}}{\mathrm{B}}}=1 \text { or } \frac{x}{a}+\frac{y}{b}=1 \dots(2)
\)
where \(\quad a=-\frac{\mathrm{C}}{\mathrm{A}}\) and \(b=-\frac{\mathrm{C}}{\mathrm{B}}\).
We know that equation (2) is intercept form of the equation of a line whose \(x\)-intercept is \(-\frac{\mathrm{C}}{\mathrm{A}}\) and \(y\)-intercept is \(-\frac{\mathrm{C}}{\mathrm{B}}\).

If \(\mathrm{C}=0\), then \(\mathrm{Ax}+\mathrm{By}+\mathrm{C}=0\) can be written as \(\mathrm{Ax}+\mathrm{By}=0\), which is a line passing through the origin and, therefore, has zero intercepts on the axes.

Normal form

Let \(x \cos \omega+y \sin \omega=p\) be the normal form of the line represented by the equation \(\mathrm{Ax}+\mathrm{B} y+\mathrm{C}=0\) or \(\mathrm{A} x+\mathrm{B} y=-\mathrm{C}\). Thus, both the equations are same and therefore, \(\quad \frac{\mathrm{A}}{\cos \omega}=\frac{\mathrm{B}}{\sin \omega}=-\frac{\mathrm{C}}{p}\)
which gives \(\quad \cos \omega=-\frac{\mathrm{A} p}{\mathrm{C}}\) and \(\sin \omega=-\frac{\mathrm{B} p}{\mathrm{C}}\).
Now
\(
\begin{aligned}
& \sin ^2 \omega+\cos ^2 \omega=\left(-\frac{\mathrm{A} p}{\mathrm{C}}\right)^2+\left(-\frac{\mathrm{B} p}{\mathrm{C}}\right)^2=1 \\
& p^2=\frac{\mathrm{C}^2}{\mathrm{~A}^2+\mathrm{B}^2} \text { or } p=\pm \frac{\mathrm{C}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}} \\
& \cos \omega=\pm \frac{\mathrm{A}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}} \text { and } \sin \omega=\pm \frac{\mathrm{B}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}} .
\end{aligned}
\)
Thus, the normal form of the equation \(\mathrm{A} x+\mathrm{B} y+\mathrm{C}=0\) is
\(
x \cos \omega+y \sin \omega=p \text {, }
\)
where \(\cos \omega=\pm \frac{A}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}, \sin \omega=\pm \frac{\mathrm{B}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\) and \(p=\pm \frac{\mathrm{C}}{\sqrt{\mathrm{A}^2+\mathrm{B}^2}}\).
Proper choice of signs is made so that \(p\) should be positive.

Example 1: The equation of a line is \(3 x-4 y+10=0\). Find its (i) slope, (ii) \(x\) – and \(y\)-intercepts.

Solution:

(i) Given equation \(3 x-4 y+10=0\) can be written as
\(
y=\frac{3}{4} x+\frac{5}{2} \dots(1)
\)
Comparing (1) with \(y=m x+c\), we have the slope of the given line as \(m=\frac{3}{4}\).
(ii) Equation \(3 x-4 y+10=0\) can be written as
\(
3 x-4 y=-10 \text { or } \frac{x}{-\frac{10}{3}}+\frac{y}{\frac{5}{2}}=1 \dots(2)
\)
Comparing (2) with \(\frac{x}{a}+\frac{y}{b}=1\), we have \(x\)-intercept as \(a=-\frac{10}{3}\) and \(y\)-intercept as \(b=\frac{5}{2}\).

Example 2: Reduce the equation \(\sqrt{3} x+y-8=0\) into normal form. Find the values of \(p\) and \(\omega\).

Solution:

Given equation is
\(
\sqrt{3} x+y-8=0 \dots(1)
\)
Dividing (1) by \(\sqrt{(\sqrt{3})^2+(1)^2}=2\), we get
\(
\frac{\sqrt{3}}{2} x+\frac{1}{2} y=4 \text { or } \cos 30^{\circ} x+\sin 30^{\circ} y=4 \dots(2)
\)
Comparing (2) with \(x \cos \omega+y \sin \omega=p\), we get \(p=4\) and \(\omega=30^{\circ}\).

Example 3: \(\text { Find the angle between the lines } y-\sqrt{3} x-5=0 \text { and } \sqrt{3} y-x+6=0 \text {. }\)

Solution:

Given lines are
and
\(
\begin{aligned}
& y-\sqrt{3} x-5=0 \text { or } y=\sqrt{3} x+5 \dots(1)\\
& \sqrt{3} y-x+6=0 \text { or } y=\frac{1}{\sqrt{3}} x-2 \sqrt{3} \dots(2)
\end{aligned}
\)
Slope of line (1) is \(m_1=\sqrt{3}\) and slope of line (2) is \(m_2=\frac{1}{\sqrt{3}}\).
The acute angle (say) \(\theta\) between two lines is given by
\(
\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right| \dots(3)
\)
Putting the values of \(m_1\) and \(m_2\) in (3), we get
\(
\tan \theta=\left|\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}\right|=\left|\frac{1-3}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}}
\)
which gives \(\theta=30^{\circ}\). Hence, angle between two lines is either \(30^{\circ}\) or \(180^{\circ}-30^{\circ}=150^{\circ}\).

Example 4: Show that two lines \(a_1 x+b_1 y+c_1=0\) and \(a_2 x+b_2 y+c_2=0\), where \(b_1, b_2 \neq 0\) are:
\(\text { (i) Parallel if } \frac{a_1}{b_1}=\frac{a_2}{b_2} \text {, and (ii) Perpendicular if } a_1 a_2+b_1 b_2=0 \text {. }\)

Solution:

Given lines can be written as
and
\(
\begin{gathered}
y=-\frac{a_1}{b_1} x-\frac{c_1}{b_1} \dots(1) \\
y=-\frac{a_2}{b_2} x-\frac{c_2}{b_2} \dots(2)
\end{gathered}
\)
Slopes of the lines (1) and (2) are \(m_1=-\frac{a_1}{b_1}\) and \(m_2=-\frac{a_2}{b_2}\), respectively. Now
(i) Lines are parallel, if \(m_1=m_2\), which gives
\(
-\frac{a_1}{b_1}=-\frac{a_2}{b_2} \text { or } \frac{a_1}{b_1}=\frac{a_2}{b_2} .
\)
(ii) Lines are perpendicular, if \(m_1 \cdot m_2=-1\), which gives \(\frac{a_1}{b_1} \cdot \frac{a_2}{b_2}=-1\) or \(a_1 a_2+b_1 b_2=0\)

Example 5: Find the equation of a line perpendicular to the line \(x-2 y-3=0\) and passing through the point \((1,-2)\).

Solution:

Given line \(x-2 y+3=0\) can be written as
\(
y=\frac{1}{2} x+\frac{3}{2} \dots(1)
\)
The slope of the line (1) is \(m_1=\frac{1}{2}\). Therefore, slope of the line perpendicular to line (1) is
\(
m_2=-\frac{1}{m_1}=-2
\)
Equation of the line with slope \(-2\) and passing through the point \((1,-2)\) is
\(
y-(-2)=-2(x-1) \text { or } y=-2 x
\)
which is the required equation.