Chapter 10.6 - Normal Form

In the previous section, we saw the intercept form of the equation of a line. In this section, we will see normal form.

Normal form

Let us first see what a normal is. It can be written in 3 steps:
1. 'Normal' is another word of 'perpendicular line'.
2. Suppose that, we have a line L₁. We can draw another line L₂ in such a way that, L₂ is perpendicular to L₁.
3. Then we say that, L₂ is a normal to L₁

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• Suppose that, we are give the following two items:
    ♦ Length of the normal from the origin to the line
    ♦ Angle which the normal makes with the +ve direction of the x-axis.
• Then we can quickly write the equation of the line. Let us see how it is done.    

Case I
This can be written in 5 steps:
1. In fig.10.29(a) below, OA is the normal to the line L₁.
• The length OA is p
• OA makes an angle of ω with the +ve direction of the x-axis.

Fig.10.29

2. Our first task is to find the slope of OA
• We know that, the slope of any line is the tangent of the inclination (angle with the +ve side of the x-axis).
• So the slope of OA = tan ω
3. Our next task is to find the slope of L₁
• In step (2), we have calculated the slope of OA.
• OA is perpendicular to L₁. So the slope of L₁ will be the -ve reciprocal of the slope of OA.
• Thus we get:  
 Slope of L₁ = $\frac{-1}{\tan \omega}$
4. Next, we want the coordinates of point A
• For that, we drop a perpendicular from A to the x-axis. M is the foot of the perpendicular.
• From the triangle OAM, we have:
    ♦ OM = p cos ω
    ♦ AM = p sin ω
• So the coordinates of A will be (p cos ω , p sin ω)
5. So now we have the slope of L₁. We also have a point on L₁.
• We can use the point-slope form. We get:
$\begin{array}{ll}
{}&{y-y_0}
&{}={}& {m(x-x_0)}
&{} \\

{\Rightarrow}&{y~-~p \sin \omega}
&{}={}& {\frac{-1}{\tan \omega}(x~-~p \cos \omega)}
&{} \\

{\Rightarrow}&{y~-~p \sin \omega}
&{}={}& {\frac{- \cos \omega}{\sin \omega}(x~-~p \cos \omega)}
&{} \\

{\Rightarrow}&{y \sin \omega~-~p \sin^2 \omega}
&{}={}& {-x \cos \omega~+~p \cos^2 \omega}
&{} \\

{\Rightarrow}&{x \cos \omega~+~y \sin \omega}
&{}={}& {p \sin^2 \omega~+~p \cos^2 \omega}
&{} \\

{\Rightarrow}&{x \cos \omega~+~y \sin \omega}
&{}={}& {p (\sin^2 \omega~+~ \cos^2 \omega)}
&{} \\

{\Rightarrow}&{x \cos \omega~+~y \sin \omega}
&{}={}& {p}
&{\color {green}{\text{- - - (a)}}} \\

\end{array}$

◼ Remarks:
• Line marked as (a):
sin²ω + cos²ω = 1

Case II
This can be written in 8 steps:
1. In case I, we derived an equation based on fig.10.29(a) above.
• In this fig.,
    ♦ the line intersects the x-axis at the +ve side.
    ♦ the line intersects the y-axis also at the +ve side.
2. What if
    ♦ the line intersect the x-axis at the -ve side.
    ♦ the line intersect the y-axis at the +ve side.
• Such a situation is shown in fig.10.29(b) above.
3. But this situation should not discourage us. Because here also, we can obtain the slope of the red line. Also we can obtain the coordinates of A.
4. In fig.10.29(b) above, OA is the normal to the line L₂.
• The length OA is p
• OA makes an angle of ω with the +ve direction of the x-axis.  
5. Our first task is to find the slope of OA
• We know that, the slope of any line is the tangent of the inclination (angle with the +ve side of the x-axis).
• So the slope of OA = tan ω
• We see that, OA is sloping downwards. That means, as we move along the x-axis towards the +ve side, the line OA is falling. So OA must have a -ve slope.
• Consider the angle ω in fig.10.29(b). It lies in the second quadrant. In the second quadrant, tan is -ve. So the slope tan ω will be -ve.
• We do not have to worry about the sign of the slope of OA. All we need to write is that: Slope of OA is tan ω. The ω will take care of the sign. 
6. Our next task is to find the slope of L₂
• In step (5), we have calculated the slope of OA.
• OA is perpendicular to L₂. So the slope of L₂ will be the -ve reciprocal of the slope of OA.
• Thus we get:  
Slope of L₂ = $\frac{-1}{\tan \omega}$
7. Next, we have to find the coordinates of point A
• For that, we drop a perpendicular from A to the x-axis. M is the foot of the perpendicular.
• From the triangle OAM, we have:
    ♦ OM = p cos (180 - ω) = -pcos ω
        ✰ See identity 9.c in the list of trigonometric identities.
    ♦ AM = p sin (180 - ω) = p sin ω
        ✰ See identity 9.d in the list of trigonometric identities.
• So the coordinates of A will be (-p cos ω , p sin ω)
• Here also, we do not have to worry about the sign of the x-coordinate. Since ω is in the second quadrant, cos ω will automatically become -ve.
• We can write the coordinates of A as: (p cos ω , p sin ω)
8. So now we have the slope of L₂. We also have a point on L₂.
• Both are same as in case I. So we will get the same equation as in case I:
The equation of line L₂ is: $x \cos \omega~+~y \sin \omega = p$

Case III
This can be written in 8 steps:
1. In case I, we derived an equation based on fig.10.29(a) above.
• In this fig.,
    ♦ the line intersects the x-axis at the +ve side.
    ♦ the line intersects the y-axis also at the +ve side.
2. What if
    ♦ the line intersect the x-axis at the -ve side.
    ♦ the line intersect the y-axis also  at the -ve side.
• Such a situation is shown in fig.10.30(a) below:

Fig.10.30

3. But this situation should not discourage us. Because here also, we can obtain the slope of the red line. Also we can obtain the coordinates of A.
4. In fig.10.30(a) above, OA is the normal to the line L₃.
• The length OA is p
• OA makes an angle of ω with the +ve direction of the x-axis.  
5. Our first task is to find the slope of OA
• We know that, the slope of any line is the tangent of the inclination (angle with the +ve side of the x-axis).
• So the slope of OA = tan ω
• We see that, OA is sloping upwards. That means, as we move along the x-axis towards the +ve side, the line OA is rising. So OA must have a +ve slope.
• Consider the angle ω in fig.10.30(a). It lies in the third quadrant. In the third quadrant, tan is +ve. So the slope tan ω will be +ve.
• We do not have to worry about the sign of the slope of OA. All we need to write is that: Slope of OA is tan ω. The ω will take care of the sign. 
6. Our next task is to find the slope of L₃
• In step (5), we have calculated the slope of OA.
• OA is perpendicular to L₃. So the slope of L₃ will be the -ve reciprocal of the slope of OA.
• Thus we get:  
Slope of L₃ = $\frac{-1}{\tan \omega}$
7. Next, we have to find the coordinates of point A
• For that, we drop a perpendicular from A to the x-axis. M is the foot of the perpendicular.
• From the triangle OAM, we have:
    ♦ OM = p cos (ω - 180) = pcos [-(180 - ω)] = pcos (180 - ω) = -p cos ω
        ✰ See identities 2 and 9.c in the list of trigonometric identities.
    ♦ AM = p sin (ω - 180) = p sin [-(180 - ω)] = -p sin (180 - ω) = -p sin ω
        ✰ See identities 1 and 9.d in the list of trigonometric identities.
• So the coordinates of A will be (-p cos ω , -p sin ω)
• Here also, we do not have to worry about the sign of the x and y coordinates. Since ω is in the third quadrant, both sin ω and cos ω will automatically become -ve.
• We can write the coordinates of A as: (p cos ω , p sin ω)
8. So now we have the slope of L₃. We also have a point on L₃.
• Both are same as in case I. So we will get the same equation as in case I:
The equation of line L₃ is: $x \cos \omega~+~y \sin \omega = p$

Case IV
This can be written in 8 steps:
1. In case I, we derived an equation based on fig.10.29(a) above.
• In this fig.,
    ♦ the line intersects the x-axis at the +ve side.
    ♦ the line intersects the y-axis also at the +ve side.
2. What if
    ♦ the line intersect the x-axis at the +ve side.
    ♦ the line intersect the y-axis at the -ve side.
• Such a situation is shown in fig.10.30(b) above.

3. But this situation should not discourage us. Because here also, we can obtain the slope of the red line. Also we can obtain the coordinates of A.
4. In fig.10.30(b) above, OA is the normal to the line L₄.
• The length OA is p
• OA makes an angle of ω with the +ve direction of the x-axis.  
5. Our first task is to find the slope of OA
• We know that, the slope of any line is the tangent of the inclination (angle with the +ve side of the x-axis).
• So the slope of OA = tan ω
• We see that, OA is sloping downwards. That means, as we move along the x-axis towards the +ve side, the line OA is falling. So OA must have a -ve slope.
• Consider the angle ω in fig.10.30(b). It lies in the fourth quadrant. In the fourth quadrant, tan is -ve. So the slope tan ω will be -ve.
• We do not have to worry about the sign of the slope of OA. All we need to write is that: Slope of OA is tan ω. The ω will take care of the sign. 
6. Our next task is to find the slope of L₄
• In step (5), we have calculated the slope of OA.
• OA is perpendicular to L₄. So the slope of L₄ will be the -ve reciprocal of the slope of OA.
• Thus we get:  
Slope of L₄ = $\frac{-1}{\tan \omega}$
7. Next, we have to find the coordinates of point A
• For that, we drop a perpendicular from A to the x-axis. M is the foot of the perpendicular.
• From the triangle OAM, we have:
    ♦ OM = p cos (360 - ω) = pcos ω
        ✰ See identity 9.g in the list of trigonometric identities.
    ♦ AM = p sin (360 - ω) = -p sin ω
        ✰ See identity 9.h in the list of trigonometric identities.
• So the coordinates of A will be (p cos ω , -p sin ω)
• Here also, we do not have to worry about the sign of the x and y coordinates. Since ω is in the fourth quadrant, sin ω will be -ve and cos ω will be +ve.
• We can write the coordinates of A as: (p cos ω , p sin ω)
8. So now we have the slope of L₄. We also have a point on L₄.
• Both are same as in case I. So we will get the same equation as in case I:
The equation of line L₄ is: $x \cos \omega~+~y \sin \omega = p$

---

• Based on the four cases, we can write:
Whatever be the orientation of the line, we can use the equation:
$x \cos \omega~+~y \sin \omega = p$
    ♦ The required signs will be taken care of by the angle ω.

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Let us see some solved examples:

Solved example 10.12
Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with +ve direction of x-axis is 15^o
Solution:
1. Given that:
p = 4 units and ω = 15^o
2. We can use the equation: $x \cos \omega~+~y \sin \omega = p$
• So we have to find cos ω and sin ω
3. We have calculated the cosine and sine of 15^o in chapter 3.
    ♦ cos ω = cos 15 = $\frac{\sqrt{3} + 1}{2 \sqrt{2}}$
    ♦ sin ω = sin 15 = $\frac{\sqrt{3} - 1}{2 \sqrt{2}}$
4. Substituting the known values in the equation in (2), we get:

$\begin{array}{ll}
{}&{x \cos \omega~+~y \sin \omega}
&{}={}& {p}
&{} \\

{\Rightarrow}&{\frac{(\sqrt{3} + 1)x}{2 \sqrt{2}}~+~\frac{(\sqrt{3} - 1)y}{2 \sqrt{2}}}
&{}={}& {4}
&{} \\

{\Rightarrow}&{(\sqrt{3} + 1)x~+~(\sqrt{3} - 1)y}
&{}={}& {8\sqrt{2}}
&{} \\

\end{array}$

5. The actual plot is shown in fig.10.31 below:

Fig.10.31

(i) The red line is the line for which we determined the equation.
(ii) Note the magenta dashed arc. It is drawn with center as O.
• This arc passes through:
    ♦ The point of intersection of the cyan and red lines.
    ♦ The point (0,4)
• So it is clear that, the length of the cyan line is four units.

Solved example 10.13
The Fahrenheit temperature F and absolute temperature K satisfy a linear relation. Given that K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F and find the value of F, when K = 0
Solution:
1. Given that: Fahrenheit temperature F and absolute temperature K satisfy a linear relation.
That means, if we plot the relation between F and K on a graph paper, the plot will be a straight line.
2. Also given that:
    ♦ K = 273 when F = 32
        ✰ This is just like: when x = 32, y is 273
    ♦ K = 373 when F = 212.
        ✰ This is just like: when x = 212, y is 373
• So we have two points on the straight line: (32, 273) and (212, 373).
3. It is important to keep in mind that, the points are in the form (F, K).
• That means,
    ♦ F is plotted along the x-axis.
    ♦ K is plotted along the y-axis.
4. We have two points on the line. So we can use the two point form:
$y-y_1~=~\frac{y_2 - y_1}{x_2 - x_1}(x-x_1)$
• For our present problem, we can modify this as:
$K-K_1~=~\frac{K_2 - K_1}{F_2 - F_1}(F-F_1)$
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{K-K_1}
&{}={}& {\frac{K_2 - K_1}{F_2 - F_1}(F-F_1)}
&{} \\

{\Rightarrow}&{K-273}
&{}={}& {\frac{373 - 273}{212 - 32}(F-32)}
&{} \\

{\Rightarrow}&{K-273}
&{}={}& {\frac{100}{180}(F-32)}
&{} \\

{\Rightarrow}&{K-273}
&{}={}& {\frac{5}{9}(F-32)}
&{} \\

{\Rightarrow}&{9K-2457}
&{}={}& {5F-160}
&{} \\

{\Rightarrow}&{9K}
&{}={}& {5F+2297}
&{} \\

{\Rightarrow}&{K}
&{}={}& {\frac{5F}{9}+\frac{2297}{9}}
&{} \\

\end{array}$

• Thus we get a relation between K and F.
5. We want the value of F when K = 0
For that, we put K = 0 in the above relation. We get:

$\begin{array}{ll}
{}&{K}
&{}={}& {\frac{5F}{9}+\frac{2297}{9}}
&{} \\

{\Rightarrow}&{0}
&{}={}& {\frac{5F}{9}+\frac{2297}{9}}
&{} \\

{\Rightarrow}&{\frac{5F}{9}}
&{}={}& {-\frac{2297}{9}}
&{} \\

{\Rightarrow}&{5F}
&{}={}& {-2297}
&{} \\

{\Rightarrow}&{F}
&{}={}& {-459.4}
&{} \\

\end{array}$

6. The actual plot is shown in fig.10.32 below:

Fig.10.32

• Points A and B represent the given data.
• Point C represent the value of F when K is zero.

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The link to a few more solved examples is given below:

Exercise 10.2

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In the next section, we will see General equation of a line.

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Chapter 10.5 - Intercept Form

In the previous section, we saw some of the different forms of the equation of a line. In this section, we will see intercept form.

Intercept form

• Suppose that, we are given the following two items:
   ♦ x-intercept of a line.
   ♦ y-intercept of that line.
• Then we can quickly write the equation of that line. Let us see how it is done.

Case I
This can be written in 4 steps:
1. Let the x-intercept be 'a'.
• Then the point of intersection of the line with the x-axis will be (a,0).
    ♦ This is shown in fig.10.26(a) below:

Fig.10.26

2. Let the y-intercept be 'b'.
• Then the point of intersection of the line with the y-axis will be (0,b)
3. Now we have two points on the line. They are: (a,0) and (0,b).
• So we can use the two-point form: $y-y_1~=~\frac{y_2-y_1}{x_2-x_1} (x-x_1)$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{y-0}
&{}={}& {\frac{b-0}{0-a} (x-a)}
&{} \\

{\Rightarrow}&{y-0}
&{}={}& {\frac{-b}{a} (x-a)}
&{} \\

{\Rightarrow}&{ya}
&{}={}& {-bx+ab}
&{} \\

{\Rightarrow}&{\frac{y}{b}}
&{}={}& {\frac{-x}{a}~+~1}
&{\color{green}{\text{- - - (a)}}} \\

{\Rightarrow}&{\frac{x}{a} + \frac{y}{b}}
&{}={}& {1}
&{} \\

\end{array}$

◼ Remarks:
• Line marked as (a):
In this line, we divide both sides by ab
4. So we get an equation: $\frac{x}{a} + \frac{y}{b} = 1$
 

Case II
This can be written in 4 steps:
1. In case I, we derived an equation based on fig.10.26(a) above.
• In this fig.,
    ♦ the line intersects the x-axis at the +ve side.
    ♦ the line intersects the y-axis also at the +ve side.
2. What if
    ♦ the line intersect the x-axis at the -ve side.
    ♦ the line intersect the y-axis at the +ve side.
• Such a situation is shown in fig.10.26(b) above.
3. But this situation should not discourage us. Because, we have the two points where the line cuts the x-axis and y-axis. The points are (-a,0) and (0,b).
• Using the two-point form as before, we get:
$\begin{array}{ll}
{}&{y-0}
&{}={}& {\frac{b-0}{0-~-a} (x-~-a)}
&{} \\

{\Rightarrow}&{y-0}
&{}={}& {\frac{b}{a} (x+a)}
&{} \\

{\Rightarrow}&{ya}
&{}={}& {bx+ab}
&{} \\

{\Rightarrow}&{\frac{y}{b}}
&{}={}& {\frac{x}{a}~+~1}
&{} \\

{\Rightarrow}&{\frac{-x}{a} + \frac{y}{b}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{x}{-a} + \frac{y}{b}}
&{}={}& {1}
&{} \\

\end{array}$

4. This is similar to the equation that we derived in case I. The 'a' has become '-a' because, the x-intercept in this case is -ve.

Case III
This can be written in 2 steps:
1. A third possible situation is shown in fig.10.27(a) below:

Fig.10.27

• Here the points are: (a,0) and (0,-b). Proceeding as before, we get:

$\begin{array}{ll}
{}&{y-0}
&{}={}& {\frac{-b-0}{0-a} (x-a)}
&{} \\

{\Rightarrow}&{y-0}
&{}={}& {\frac{-b}{-a} (x-a)}
&{} \\

{\Rightarrow}&{y-0}
&{}={}& {\frac{b}{a} (x-a)}
&{} \\

{\Rightarrow}&{ya}
&{}={}& {bx-ab}
&{} \\

{\Rightarrow}&{\frac{y}{b}}
&{}={}& {\frac{x}{a}~-~1}
&{} \\

{\Rightarrow}&{\frac{x}{a} - \frac{y}{b}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{x}{a} + \frac{y}{-b}}
&{}={}& {1}
&{} \\

\end{array}$

2. This is similar to the equation that we derived in case I. The 'b' has become '-b' because, the y-intercept in this case is -ve.  

Case IV
This can be written in 2 steps:
1. The fourth and last possible situation is shown in fig.10.27(b) above.

• Here the points are: (-a,0) and (0,-b). Proceeding as before, we get:

$\begin{array}{ll}
{}&{y-0}
&{}={}& {\frac{-b-0}{0-~-a} (x-~-a)}
&{} \\

{\Rightarrow}&{y-0}
&{}={}& {\frac{-b}{a} (x+a)}
&{} \\

{\Rightarrow}&{ya}
&{}={}& {-bx-ab}
&{} \\

{\Rightarrow}&{\frac{y}{b}}
&{}={}& {\frac{-x}{a}~-~1}
&{} \\

{\Rightarrow}&{\frac{x}{a} + \frac{y}{b}}
&{}={}& {-1}
&{} \\

{\Rightarrow}&{\frac{x}{-a} + \frac{y}{-b}}
&{}={}& {1}
&{} \\

\end{array}$

2. This is similar to the equation that we derived in case I.
    ♦ The 'a' has become '-a' because, the x-intercept in this case is -ve. 
    ♦ The 'b' has become '-b' because, the y-intercept in this case is -ve. 

---

Based on the four cases, we can write:
• In the intercept form, we can use the general equation: $\frac{x}{a} + \frac{y}{b} = 1$
• But care must be taken regarding the signs of 'a' and 'b'.
(This equation is easy to remember because, x is divided by the x-intercept. Similarly, y is divided by the y-intercept)

---

• Fig.10.28 below shows the actual plot of four lines.

1. Equation of the red line L₁ is: $\frac{x}{-3} + \frac{y}{4} = 1$
• We see that:
    ♦ L₁ cuts the x-axis at (-3,0) 
    ♦ L₁ cuts the y-axis at (0,4)

Fig.10.28

 

2. Equation of the green line L₂ is: $\frac{x}{-6} + \frac{y}{-4} = 1$
• We see that:
    ♦ L₂ cuts the x-axis at (-6,0) 
    ♦ L₂ cuts the y-axis at (0,-4)

3. Equation of the magenta line L₃ is: $\frac{x}{3} + \frac{y}{5} = 1$
• We see that:
    ♦ L₃ cuts the x-axis at (3,0) 
    ♦ L₃ cuts the y-axis at (0,5)      

4. Equation of the pink line L₄ is: $\frac{x}{6} + \frac{y}{-3} = 1$
• We see that:
    ♦ L₄ cuts the x-axis at (6,0) 
    ♦ L₄ cuts the y-axis at (0,-3)

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In the next section, we will see normal form.

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Chapter 10.4 - Various Forms of Equations of A Line

In the previous section, we saw collinearity of three points. In this section, we will see the various forms of the equation of a line.

Some basics can be written in 3 steps:
1. Consider a line L lying in the xy-plane.
• We know that in any line, there will be infinite number of points.
• But there are other infinite points which do not lie on the line.
2. Then what is the difference between the following two types of points:
(i) Points which lie on L
(ii) Points which do not lie on L  
• The answer is:
   ♦ All points lying on L, will satisfy a particular condition.
   ♦ The points which do not lie on L, will not satisfy that condition.
3. We can write this is another way also:
Take any point on the xy-plane. If that point satisfies a particular condition, then that point will lie on L.

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• So our next aim is to find the conditions for different types of lines.
• First we will see horizontal lines.

Equation of a horizontal line

This can be written in 9 steps.
1. The red line L₁ in fig.10.16(a) is parallel to the x-axis.
   ♦ It is a horizontal line.
   ♦ It is at a distance of ‘a’ from the x-axis.

Fig.10.16

2. Consider any point on L₁. The ordinate (y-coordinate) of that point will be 'a'.
• Conversely, any point whose ordinate is a, will lie on L₁
3. So we can write:
The condition for a point to lie on L₁ is that, the ordinate of that point must be ‘a’
4. In algebraic form, we can write: y = a
• This algebraic form is the equation of line L₁
• We can write:
Equation of L₁ is: y = a
5. Now consider the red line L₁' in fig.10.16(b) above.
   ♦ It is parallel to the x-axis
   ♦ It is a horizontal line.
   ♦ It is at a distance of ‘a’ from the x-axis.
6. Consider any point on L₁'. The ordinate (y-coordinate) of that point will be '-a'.
• Conversely, any point whose ordinate is -a, will lie on L₁'
7. So we can write:
The condition for a point to lie on L₁' is that, the ordinate of that point must be ‘-a’
8. In algebraic form, we can write: y = -a
• This algebraic form is the equation of line L₁'
• We can write:
Equation of L₁' is: y = -a
9. In general, we can write:
Equation of any horizontal line will be in the form: $y=\pm a$
   ♦ Where ‘a’ is a constant.
         ✰ We use the ‘+’ sign when the horizontal line is above the x-axis.
         ✰ We use the ‘-’ sign when the horizontal line is below the x-axis.

Equation of a vertical line

This can be written in 9 steps.
1. The red line L₂ in fig.10.17(a) is parallel to the y-axis.
   ♦ It is a vertical line.
   ♦ It is at a distance of ‘b’ from the y-axis.

Fig.10.17

2. Consider any point on L₂. The abscissa (x-coordinate) of that point will be 'b'.
• Conversely, any point whose abscissa is 'b', will lie on L₂
3. So we can write:
The condition for a point to lie on L₂ is that, the abscissa of that point must be ‘b’
4. In algebraic form, we can write: x = b
• This algebraic form is the equation of line L₂
• We can write:
Equation of L₂ is: x = b
5. Now consider the red line L₂' in fig.10.17(b) above.
   ♦ It is parallel to the y-axis
   ♦ It is a vertical line.
   ♦ It is at a distance of ‘b’ from the y-axis.
6. Consider any point on L₂'. The abscissa (x-coordinate) of that point will be '-b'.
• Conversely, any point whose abscissa is -b, will lie on L₂'
7. So we can write:
The condition for a point to lie on L₂' is that, the abscissa of that point must be ‘-b’
8. In algebraic form, we can write: x = -b
• This algebraic form is the equation of line L₂'
• We can write:
Equation of L₂' is: x = -b
9. In general, we can write:
Equation of any vertical line will be in the form: $x=\pm b$
   ♦ Where ‘b’ is a constant.
         ✰ We use the ‘+’ sign when the vertical line is on the right side of the y-axis.
         ✰ We use the ‘-’ sign when the vertical line is on the left side of the y-axis.

Solved Example 10.8
Write the equation of the line parallel to x-axis and passing through (-2,3). Also write the equation of the line parallel to y-axis and passing through the same point.
Solution:
1. First we will consider the line parallel to the x-axis.
• It's equation will be of the form $y = \pm a$
• This line passes through (-2,3). So it's distance from the x-axis is 3 units.
   ♦ That means, a = 3
• Also, (-2,3) is above the x-axis.
   ♦ So we use the +ve sign
• So the required equation is: $y = 3$
2. Next we will consider the line parallel to the y-axis.
• It's equation will be of the form $x = \pm b$
• This line passes through (-2,3). So it's distance from the y-axis is 2 units.
   ♦ That means, a = 2
• Also, (-2,3) is on the left side of the y-axis.
   ♦ So we use the -ve sign
• So the required equation is: $x = -2$
3. The actual lines are shown in the fig.10.18 below:

Fig.10.18

• In the above fig,
   ♦ The red line is the required line parallel to the x-axis.
   ♦ The green line is the required line parallel to the y-axis.

---

Point-slope form

• Suppose that, we are given the following two items:
   ♦ Slope of a line.
   ♦ A point through which the line passes.
• Then we can quickly write the equation of that line. Let us see how it is done. It can be written in 7 steps:
1. Let the slope of the line be m
2. Let the line pass through a known point P₀ (x₀,y₀)
• By saying "Known point", we mean that, the values of x₀ and y₀ are known.
3. Let P(x,y) be any point on the line.
4. Now we have two points on the line. They are: P₀(x₀,y₀) and P(x,y). This is shown in fig.10.19 below:

Fig.10.19

5. If we have two points on a line, we can write the slope of the line.
• So in our present case, the slope will be $\frac{y-y_0}{x-x_0}$
6. But we are already given the slope. It is m. So we can write:

$\begin{array}{ll}
{}&{m}
&{}={}& {\frac{y-y_0}{x-x_0}}
&{} \\

{\Rightarrow}&{y-y_0}
&{}={}& {m(x-x_0)}
&{} \\
\end{array}$

7. So we get an equation: y-y₀ = m(x-x₀)
• We derived this equation by assuming that, the point (x,y) lies on the line (see fig.10.19).
• That means,
   ♦ if we take any point outside the line, this equation will not be satisfied.
   ♦ if we take any point on the line, this equation will be satisfied.
• So we can say that, y-y₀ = m(x-x₀) is the equation of the line.
   ♦ This is the equation of a line written in slope-point form.
         ✰ The slope is m.
         ✰ The point is (x₀,y₀).

---

Let us see a solved example:
Solved example 10.9
Three lines L₁, L₂ and L₃ pass through the point (-3,4). Write the equation of each of those lines, if their slopes are -3, 2 and 4 respectively. 
Solution:
• Given that,
   ♦ (x₀,y₀) = (-3,4)
   ♦ m₁ = -3
   ♦ m₂ = 2
   ♦ m₃ = 4
• We have the slope-point form: y-y₀ = m(x-x₀)
1. So for L₁, the equation will be:
$\begin{array}{ll}
{}&{y-4}
&{}={}& {-3(x-~-3)}
&{} \\

{\Rightarrow}&{y-4}
&{}={}& {-3x-9}
&{} \\

{\Rightarrow}&{3x + y +5}
&{}={}& {0}
&{} \\
\end{array}$

2. Similarly for L₂, the equation will be:
$\begin{array}{ll}
{}&{y-4}
&{}={}& {2(x-~-3)}
&{} \\

{\Rightarrow}&{y-4}
&{}={}& {2x+6}
&{} \\

{\Rightarrow}&{2x - y +10}
&{}={}& {0}
&{} \\
\end{array}$

3. Finally for L₃, the equation will be:
$\begin{array}{ll}
{}&{y-4}
&{}={}& {4(x-~-3)}
&{} \\

{\Rightarrow}&{y-4}
&{}={}& {4x+12}
&{} \\

{\Rightarrow}&{4x - y +16}
&{}={}& {0}
&{} \\
\end{array}$

4. The actual plot of the three lines are shown in fig.10.20 below:

Fig.10.20

5. We can infer some general information from the plot. They can be written in 3 steps:
(i) If the slope of a line is +ve, that line will have an upward slope. That means, when we move along the x-axis towards the +ve side, the line will be sloping upwards.
   ♦ Lines L₂ and L₃ are examples.
(ii) If the slope of a line is -ve, that line will have a downward slope. That means, when we move along the x-axis towards the +ve side, the line will be sloping downwards.
   ♦ Line L₁ is an example.
(iii) Compare the slopes of two lines.
   ♦ The line with the greater slope will be steeper.
   ♦ The line with the lesser slope will be flatter.
• This will become clear when we compare L₂ and L₃.

---

Two-point form

• Suppose that, we are given two known points on a line.
• Then we can quickly write the equation of that line. Let us see how it is done. It can be written in 6 steps:
1. Let the line pass through two known points P₁ (x₁,y₁) and P₂ (x₂,y₂)
• By saying "Known points", we mean that, the values of x₁, y₁, x₂ and y₂ are known.
2. Let P(x,y) be any point on the line.
3. Now we have three points on the line. They are: P₁(x₁,y₁), P₂(x₂,y₂) and P(x,y). This is shown in fig.10.21 below:

Fig.10.21

4. If there are three points, we can write slopes.
• So in our present case,
   ♦ the slope of line P₁P will be $\frac{y-y_1}{x-x_1}$
   ♦ the slope of line P₁P₂ will be $\frac{y_2-y_1}{x_2-x_1}$
5. But since the three points are collinear, the slopes will be equal. So we can write:

$\begin{array}{ll}
{}&{\frac{y-y_1}{x-x_1}}
&{}={}& {\frac{y_2-y_1}{x_2-x_1}}
&{} \\

{\Rightarrow}&{y-y_1}
&{}={}& {\frac{y_2-y_1}{x_2-x_1} (x-x_0)}
&{} \\
\end{array}$

6. So we get an equation: $y-y_1~=~\frac{y_2-y_1}{x_2-x_1} (x-x_1)$
• We derived this equation by assuming that, the point (x,y) lies on the line (see fig.10.21).
• That means,
   ♦ if we take any point outside the line, this equation will not be satisfied.
   ♦ if we take any point on the line, this equation will be satisfied.
• So we can say that, $y-y_1~=~\frac{y_2-y_1}{x_2-x_1} (x-x_1)$ is the equation of the line.
   ♦ This is the equation of a line written in two-point form.
         ✰ The two points are: (x₁,y₁) and (x₂,y₂).

---

Let us see a solved example:
Solved example 10.10
Write the equations of the following lines:
(i) Line L₁, passing through (-1, -3) and (2, 4). 
(ii) Line L₂, passing through (-3, 7) and (6, 7). 
Solution:
Part (i):
• We have the two-point form: $y-y_1~=~\frac{y_2-y_1}{x_2-x_1} (x-x_1)$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{y-~-3}
&{}={}& {\frac{4-~-3}{2-~-1} (x-~-1)}
&{} \\

{\Rightarrow}&{y+3}
&{}={}& {\frac{7}{3} (x+1)}
&{} \\

{\Rightarrow}&{3y+9}
&{}={}& {7x+7}
&{} \\

{\Rightarrow}&{7x-3y-2}
&{}={}& {0}
&{} \\

\end{array}$

Part (ii):
• We have the two-point form: $y-y_1~=~\frac{y_2-y_1}{x_2-x_1} (x-x_1)$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{y-7}
&{}={}& {\frac{7-7}{6-~-3} (x-~-3)}
&{} \\

{\Rightarrow}&{y-7}
&{}={}& {\frac{0}{9} (x+3)}
&{} \\

{\Rightarrow}&{y-7}
&{}={}& {0}
&{} \\

{\Rightarrow}&{y}
&{}={}& {7}
&{} \\

\end{array}$

• The actual plots are shown in fig.10.22 below:

Fig.10.22

• Note:
In the case of L₂, we see that, it is horizontal. Indeed we obtained the slope $\frac{y_2 - y_1}{x_2 - x_1}$ = 0 for L₂

---

Slope-intercept form

• Let us first see, what intercept is. It can be written in 6 steps:
1. In fig.10.23(a) below, the line L₁ intersects the x-axis at (a,0)
• The distance of the ‘point of intersection’ from the origin O is ‘a’.
• We say that, the x-intercept of the line is +a.

Fig.10.23

2. In fig.10.23(b) above, the line L₂ intersects the x-axis at (-b,0)
• The distance of the ‘point of intersection’ from the origin O is ‘b’
• We say that, the x-intercept of the line is -b.
3. In fig.10.24(a) below, the line L₃ intersects the y-axis at (c,0)
• The distance of the ‘point of intersection’ from the origin O is ‘c’.
• We say that, the y-intercept of the line is +c.

Fig.10.,24

4. In fig.10.24(b) above, the line L₄ intersects the x-axis at (-d,0)
• The distance of the ‘point of intersection’ from the origin O is ‘d’
• We say that, the x-intercept of the line is -d.
5. So intercept is the distance of the point of intersection from O with the proper sign.
6. While writing intercepts, it is better to remember that:
• Any non-parallel line will intersect the x-axis.
   ♦ In some cases, the point of intersection will be on the right side of y-axis.
         ✰ Then the point of intersection will be of the type (a,0)
   ♦ In the remaining cases, the point of intersection will be on the left side of y-axis.
         ✰ Then the point of intersection will be of the type (-b,0)
• Any non-parallel line will intersect the y-axis.
   ♦ In some cases, the point of intersection will be above the x-axis.
         ✰ Then the point of intersection will be of the type (0,c)
   ♦ In the remaining cases, the point of intersection will be below the x-axis.
         ✰ Then the point of intersection will be of the type (0,-d)

---

• Suppose that, we are given the following two items:
   ♦ Slope of a line.
   ♦ x-intercept or y-intercept of the line.
• Then we can quickly write the equation of that line. Let us see how it is done.

Case I: The x-intercept is +ve.
This can be written in 4 steps:
1. Let the slope of the line be m
2. Let the x-intercept be +a
• Then we can say that, the line passes through (a,0). See fig.10.23(a) above.
3. Now we have two items:
   ♦ Slope of the line.
   ♦ A point through which the line passes.
4. So we can use the point-slope form: y-y₀ = m(x-x₀)
• Here, x₀ is 'a' and y₀ is 0
• Thus the equation becomes:
$\begin{array}{ll}
{}&{y-0}
&{}={}& {m(x-a)}
&{} \\

{\Rightarrow}&{y}
&{}={}& {m(x-a)}
&{} \\
\end{array}$

Case II: The x-intercept is -ve.
This can be written in 4 steps:
1. Let the slope of the line be m
2. Let the x-intercept be -b
• Then we can say that, the line passes through (-b,0). See fig.10.23(b) above.
3. Now we have two items:
   ♦ Slope of the line.
   ♦ A point through which the line passes.
4. So we can use the point-slope form: y-y₀ = m(x-x₀)
• Here, x₀ is '-b' and y₀ is 0
• Thus the equation becomes:
$\begin{array}{ll}
{}&{y-0}
&{}={}& {m(x- -b)}
&{} \\

{\Rightarrow}&{y}
&{}={}& {m(x+b)}
&{} \\
\end{array}$

• Based on cases I and II, we can write:
When the slope and x-intercept is given, we can use the general form: y = m(x-a)
    ♦ When the intercept is +ve, we give a +ve value for 'a'.
    ♦ When the intercept is -ve, we give a -ve value for 'a'.

Case III: The y-intercept is +ve.
This can be written in 4 steps:
1. Let the slope of the line be m
2. Let the y-intercept be +c
• Then we can say that, the line passes through (0,c). See fig.10.24(a) above.
3. Now we have two items:
   ♦ Slope of the line.
   ♦ A point through which the line passes.
4. So we can use the point-slope form: y-y₀ = m(x-x₀)
• Here, x₀ is 0 and y₀ is 'c'
• Thus the equation becomes:
$\begin{array}{ll}
{}&{y-c}
&{}={}& {m(x-0)}
&{} \\

{\Rightarrow}&{y-c}
&{}={}& {mx}
&{} \\

{\Rightarrow}&{y}
&{}={}& {mx + c}
&{} \\
\end{array}$

Case IV: The y-intercept is -ve.
This can be written in 4 steps:
1. Let the slope of the line be m
2. Let the y-intercept be -d
• Then we can say that, the line passes through (0,-d). See fig.10.24(b) above.
3. Now we have two items:
   ♦ Slope of the line.
   ♦ A point through which the line passes.
4. So we can use the point-slope form: y-y₀ = m(x-x₀)
• Here, x₀ is 0 and y₀ is '-d'
• Thus the equation becomes:
$\begin{array}{ll}
{}&{y-~-d}
&{}={}& {m(x-0)}
&{} \\

{\Rightarrow}&{y+d}
&{}={}& {mx}
&{} \\

{\Rightarrow}&{y}
&{}={}& {mx - d}
&{} \\
\end{array}$

• Based on cases III and IV, we can write:
When the slope and y-intercept is given, we can use the general form: y = mx+c
    ♦ When the intercept is +ve, we give a +ve value for 'c'.
    ♦ When the intercept is -ve, we give a -ve value for 'c'.

---

Let us see a solved example:
Solved example 10.11
Write the equations of the following lines:
(i) Line L₁ for which tan 𝜃 = 1/3 and x-intercept is 4. 
(ii) Line L₂ for which tan 𝜃 = 1/2 and y-intercept is -(11/2).
𝜃 is the inclination of the line. 
Solution:
Part (i):
• 𝜃 is the inclination of the line.  So tan 𝜃 will be the slope.
   ♦ So we can write: m = 1/3
• x-intercept is 4.
   ♦ We use the general equation: y = m(x-a)
   ♦ Here 'a' is 4.
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{y}
&{}={}& {\frac{1}{3} (x-4)}
&{} \\

{\Rightarrow}&{3y}
&{}={}& {x-4}
&{} \\

{\Rightarrow}&{x-3y-4}
&{}={}& {0}
&{} \\

\end{array}$

Part (ii):
• 𝜃 is the inclination of the line.  So tan 𝜃 will be the slope.
   ♦ So we can write: m = 1/2
• y-intercept is -(11/2).
   ♦ We use the general equation: y = mx + c
   ♦ Here 'c' is -(11/2)
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{y}
&{}={}& {\frac{1 x}{2} - \frac{11}{2}}
&{} \\

{\Rightarrow}&{2y}
&{}={}& {x-11}
&{} \\

{\Rightarrow}&{x-2y-11}
&{}={}& {0}
&{} \\

\end{array}$

• The actual plots are shown in fig.10.25 below:

Fig.10.25

• We can see that:
   ♦ L₁ intersects the x-axis at (4,0)
   ♦ L₂ intersects the x-axis at (0,-11/2)

---

In the next section, we will see intercept form and normal form.

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