Chapter 10.7 - General Equation of A Line

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:

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.

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• 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 $-\frac{C}{A}$.

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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.

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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: $\sin \omega~=~\pm \frac{B}{\sqrt{A^2~+~B^2}}$

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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.

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• 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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