In the previous section, we saw the normal form of the equation of a line. In this section, we will see the general equation of a line.
Some basics about the general equation can be written in 4 steps:
1. Consider the equation: Ax + By + C = 0
• It is a first degree equation in two variables.
2. Let us put some random values for A, B and C:
A = 2, B = 3 and C = 5. Now the equation becomes: 2x + 3y + 5 = 0
• In this equation, if we put some random values for x, we will get corresponding values for y.
• For example:
♦ If x = 2, then y will be -3
♦ If x = 3, then y will be -3.67
♦ If x = 6, then y will be -5.67
♦ If x = 7, then y will be -6.33
3. So we get some coordinates: (2,-3), (3,-3.67), (6,-5.67), (7,-6.33)
• If we plot these coordinates, they will lie on a straight line. This is shown in fig.10.33 below:
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| Fig.10.33 |
4. So we can write:
• Any equation of the form Ax + By + C = 0 is called general linear equation.
• It is also called general equation of a line.
◼ But there is one condition:
A and B should not be zero simultaneously.
• This can be explained in 3 steps:
(i) If A = 0, the general equation will become: By + C = 0
♦ This is OK because, By + C = 0 will give a straight line.
(ii) If B = 0, the general equation will become: Ax + C = 0
♦ This is OK because, Ax + C = 0 will give a straight line.
(iii) If A = 0 and B = 0, the general equation will become: C = 0
♦ This is not OK because, C = 0 will not give a straight line.
• Now we will see the different forms of Ax + By + C = 0
• We have seen:
♦ Slope-intercept form
♦ Intercept form
♦ Normal form.
• The general equation can be written in each of these three forms.
First we will see the slope-intercept form. It can be written in 11 steps:
1. The slope-intercept form is: y = mx + c
2. The general form is: Ax + By + C = 0
3. Let the equations in both (1) and (2) represent the same line.
4. The point at which (1) intersects the x-axis can be obtained by putting y = 0.
So we get:
$\begin{array}{ll}
{}&{y}
&{}={}& {mx+c}
&{} \\
{\Rightarrow}&{0}
&{}={}& {mx+c}
&{} \\
{\Rightarrow}&{x}
&{}={}& {-\frac{c}{m}}
&{} \\
\end{array}$
5. So the line in (1) intersects the x-axis at $\left(-\frac{c}{m},~0 \right)$
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.
• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{A × -\frac{c}{m} ~+~B × 0~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{-\frac{Ac}{m}~+~C}
&{}={}& {0}
&{} \\
\end{array}$
6. The point at which (1) intersects the y-axis can be obtained by putting x = 0.
So we get:
$\begin{array}{ll}
{}&{y}
&{}={}& {mx+c}
&{} \\
{\Rightarrow}&{y}
&{}={}& {m × 0+c}
&{} \\
{\Rightarrow}&{y}
&{}={}& {c}
&{} \\
\end{array}$
7. So the line in (1) intersects the y-axis at (0,c)
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.
• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{A × 0 ~+~B × c~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{B × c~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{c}
&{}={}& {-\frac{C}{B}}
&{} \\
\end{array}$
8. Substituting this value of c in (5), we get:
$\begin{array}{ll}
{}&{-\frac{Ac}{m}~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{-\frac{A}{m} × -\frac{C}{B}~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{-\frac{A}{m} × -\frac{1}{B}~+~1}
&{}={}& {0}
&{} \\
{\Rightarrow}&{\frac{A}{B m}~+~1}
&{}={}& {0}
&{} \\
{\Rightarrow}&{\frac{A}{Bm}}
&{}={}& {-1}
&{} \\
{\Rightarrow}&{m}
&{}={}& {-\frac{A}{B}}
&{} \\
\end{array}$
9. So we get the following results:
♦ From (7), we get: $c~=~-\frac{C}{B}$
♦ From (8), we get: $m~=~-\frac{A}{B}$
10. Now we can write:
• If we are given the equation of a line in the general form Ax + By + C = 0, then:
♦ Slope of that line can be obtained as: $m~=~-\frac{A}{B}$
♦ y-intercept of that line can be obtained as: $c~=~-\frac{C}{B}$
11. If B = 0, then the general equation becomes: $x~=~-\frac{C}{A}$
♦ This is an equation of vertical line.
♦ It’s slope is undefined.
✰ Indeed "$m~=~-\frac{A}{B}$" is undefined if B = 0
♦ It intersects the x axis at $-\frac{C}{A}$.
Next we will see the intercept form. It can be written in 10 steps:
1. The intercept form is: $\frac{x}{a}~+~\frac{y}{b}~=~1$
2. The general form is: Ax + By + C = 0
3. Let the equations in both (1) and (2) represent the same line.
4. The point at which (1) intersects the x-axis can be obtained by putting y = 0.
So we get:
$\begin{array}{ll}
{}&{\frac{x}{a}~+~\frac{y}{b}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{\frac{x}{a}~+~\frac{0}{b}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{\frac{x}{a}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{x}
&{}={}& {a}
&{} \\
\end{array}$
5. So the line in (1) intersects the x-axis at (a,0)
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.
• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{A × a ~+~B × 0~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{Aa~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{a}
&{}={}& {-\frac{C}{A}}
&{} \\
\end{array}$
6. The point at which (1) intersects the y-axis can be obtained by putting x = 0.
So we get:
$\begin{array}{ll}
{}&{\frac{x}{a}~+~\frac{y}{b}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{\frac{0}{a}~+~\frac{y}{b}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{\frac{y}{b}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{y}
&{}={}& {b}
&{} \\
\end{array}$
7. So the line in (1) intersects the x-axis at (0,b)
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.
• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{A × 0 ~+~B × b~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{B × b~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{b}
&{}={}& {-\frac{C}{B}}
&{} \\
\end{array}$
8. So we get the following results:
♦ From (5), we get: $a~=~-\frac{C}{A}$
♦ From (7), we get: $b~=~-\frac{C}{B}$
9. Now we can write:
• If we are given the equation of a line in the general form Ax + By + C = 0, then:
♦ x-intercept of that line can be obtained as: $a~=~-\frac{C}{A}$
♦ y-intercept of that line can be obtained as: $b~=~-\frac{C}{B}$
✰ Note that, we obtained the same y-intercept in slope-intercept form also.
10.If C = 0, then:
♦ x-intercept = $-\frac{0}{A}$ = 0
♦ y-intercept = $-\frac{0}{B}$ = 0
• That means, the line will pass through the origin.
◼ We can write:
In the general equation of a line, if "term with no variable" is absent, then that line will pass through the origin.
Next we will see the normal form. It can be written in 12 steps:
1. The normal form is: $x \cos \omega~+~y \sin \omega~=~p$
2. The general form is: Ax + By + C = 0
3. Let the equations in both (1) and (2) represent the same line.
4. The point at which (1) intersects the x-axis can be obtained by putting y = 0.
So we get:
$\begin{array}{ll}
{}&{x \cos \omega~+~y \sin \omega}
&{}={}& {p}
&{} \\
{\Rightarrow}&{x \cos \omega~+~0 × \sin \omega}
&{}={}& {p}
&{} \\
{\Rightarrow}&{x \cos \omega}
&{}={}& {p}
&{} \\
{\Rightarrow}&{x}
&{}={}& {\frac{p}{\cos \omega}}
&{} \\
\end{array}$
5. So the line in (1) intersects the x-axis at $\left(\frac{p}{\cos \omega},~0 \right)$
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.
• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{A × \frac{p}{\cos \omega} ~+~B × 0~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{\frac{A p}{\cos \omega}~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{\cos \omega}
&{}={}& {-\frac{Ap}{C}}
&{} \\
\end{array}$
6. The point at which (1) intersects the y-axis can be obtained by putting x = 0.
So we get:
$\begin{array}{ll}
{}&{x \cos \omega~+~y \sin \omega}
&{}={}& {p}
&{} \\
{\Rightarrow}&{0 × \cos \omega~+~ y \sin \omega}
&{}={}& {p}
&{} \\
{\Rightarrow}&{y \sin \omega}
&{}={}& {p}
&{} \\
{\Rightarrow}&{y}
&{}={}& {\frac{p}{\sin \omega}}
&{} \\
\end{array}$
7. So the line in (1) intersects the x-axis at $\left(0,~\frac{p}{\sin \omega} \right)$
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.
• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{A × 0 ~+~B × \frac{p}{\sin \omega}~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{\frac{B p}{\sin \omega}~+~C}
&{}={}& {0}
&{} \\
{\Rightarrow}&{\sin \omega}
&{}={}& {-\frac{Bp}{C}}
&{} \\
\end{array}$
8. So we get the following results:
♦ From (5), we get: $\cos \omega~=~-\frac{Ap}{C}$
♦ From (7), we get: $\sin \omega~=~-\frac{Bp}{C}$
9. But $\sin^2 \omega~+~\cos^2 \omega~=~1$
So we get:
$\begin{array}{ll}
{}&{\left(-\frac{Bp}{C} \right)^2~+~\left(-\frac{Ap}{C} \right)^2}
&{}={}& {1}
&{} \\
{\Rightarrow}&{\frac{B^2 p^2}{C^2}~+~\frac{A^2 p^2}{C^2}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{\frac{ p^2(A^2~+~B^2)}{C^2}}
&{}={}& {1}
&{} \\
{\Rightarrow}&{p^2}
&{}={}& {\frac{C^2}{A^2~+~B^2}}
&{} \\
{\Rightarrow}&{p}
&{}={}& {\pm \frac{C}{\sqrt{A^2~+~B^2}}}
&{} \\
\end{array}$
10. We can substitute this value of p in (5). We get:
$\begin{array}{ll}
{}&{\cos \omega}
&{}={}& {-\frac{Ap}{C}}
&{} \\
{}&{}
&{}={}& {-\frac{A}{C}~ × ~\pm \frac{C}{\sqrt{A^2~+~B^2}}}
&{} \\
{}&{}
&{}={}& {\pm \frac{A}{\sqrt{A^2~+~B^2}}}
&{} \\
\end{array}$
11. Similarly, we can substitute this value of p in (7). We get:
$\begin{array}{ll}
{}&{\sin \omega}
&{}={}& {-\frac{Bp}{C}}
&{} \\
{}&{}
&{}={}& {-\frac{B}{C}~ × ~\pm \frac{C}{\sqrt{A^2~+~B^2}}}
&{} \\
{}&{}
&{}={}& {\pm \frac{B}{\sqrt{A^2~+~B^2}}}
&{} \\
\end{array}$
12. Thus we get three results:
♦ From (9), we get: $p~=~\pm \frac{C}{\sqrt{A^2~+~B^2}}$
♦ From (10), we get: $\cos \omega~=~\pm \frac{A}{\sqrt{A^2~+~B^2}}$
♦ From (11), we get: $\sin \omega~=~\pm \frac{B}{\sqrt{A^2~+~B^2}}$
Let us compile the above results. It can be done in 5 steps:
1. Slope of a given line can be calculated as: $m~=~-\frac{A}{B}$
• This result is available from slope-intercept form.
2. x-intercept of a given line can be calculated as: $c~=~-\frac{C}{A}$
• This result is available from intercept form.
3. y-intercept of a given line can be calculated as: $c~=~-\frac{C}{B}$
• This result is available from both slope-intercept form and intercept form.
4. For writing the normal form, we can use three results:
♦ $p~=~\pm \frac{C}{\sqrt{A^2~+~B^2}}$
♦ $\cos \omega~=~\pm \frac{A}{\sqrt{A^2~+~B^2}}$
♦ $\sin \omega~=~\pm \frac{B}{\sqrt{A^2~+~B^2}}$
5. These results are easy to remember because, they follow a pattern.
• In the above results, we have $\pm$ sign for p, $\cos \omega$ and $\sin \omega$.
• So two questions arise:
♦ When do we use the '+' sign ?
♦ When do we use the '-' sign ?
• We will see the answers in the next section.
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