Chapter 10.3 - Collinearity of Three Points

In the previous section, we saw solved example 2 and solved example 3. In those examples, we were given the slopes of two lines. We were asked to find the angle between the lines. In this section, we will see a solved example in which the angle between the two lines are given. Also the slope of one of the lines is given. We are asked to find the slope of the other line. Later in this section, we will see condition for collinearity also.

Solved example 4
If the angle between two lines is $\frac{\pi}{4}$ and slope of one of the lines is $\frac{1}{2}$, find the slope of the other line.
Solution:
1. Given that, $\theta~=~\frac{\pi}{4}~\text{and}~m_1~=~\frac{1}{2}$.
• We are asked to find m₂
2. We have: $\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
• Substituting the known values, we get:

$\begin{array}{ll}
{}&{\tan \frac{\pi}{4}}
&{}={}& {\frac{m_2~-~\frac{1}{2}}{1~+~\frac{1}{2}  × m_2}}
&{} \\

{\Rightarrow}&{1}
&{}={}& {\frac{2 m_2 ~-~1}{2~+~m_2}}
&{} \\

{\Rightarrow}&{2~+~m_2}
&{}={}& {2 m_2 ~-~1}
&{} \\

{\Rightarrow}&{2}
&{}={}& {m_2 ~-~1}
&{} \\

{\Rightarrow}&{m_2}
&{}={}& {3}
&{} \\

\end{array}$

3. In the above steps, we took $m_1 = \frac{1}{2}$ and calculated m₂.
• Let us now interchange the values.
   ♦ That is., take $m_2 = \frac{1}{2}$ and calculate m₁:

$\begin{array}{ll}
{}&{\tan \frac{\pi}{4}}
&{}={}& {\frac{\frac{1}{2}~-~m_1}{1~+~m_1  × \frac{1}{2}}}
&{} \\

{\Rightarrow}&{1}
&{}={}& {\frac{1~-~2 m_1}{2~+~m_1}}
&{} \\

{\Rightarrow}&{2~+~m_1}
&{}={}& {1~-~2 m_1}
&{} \\

{\Rightarrow}&{2~+~3 m_1}
&{}={}& {1}
&{} \\

{\Rightarrow}&{3 m_1}
&{}={}& {-1}
&{} \\

{\Rightarrow}&{m_1}
&{}={}& {-\frac{1}{3}}
&{} \\

\end{array}$

4. From the results in (2) and (3), we can write three points:
(i) A line L₁ has a slope of $\frac{1}{2}$ 
(ii) Another line L₂ (having a slope of 3) will make an angle $\frac{\pi}{4}$ with L₁.
(iii) Yet another line L₃ (having a slope of $-\frac{1}{3}$) will also make the same angle $\frac{\pi}{4}$ with L₁
5. The three lines are shown in fig.10.13 below:

Fig.10.13

• The red line is the given line with slope $\frac{1}{2}$
• We were asked to find the slope of another line in such a way that, the new line makes an angle $\frac{\pi}{4}$ with the given red line.
• We found out that, two lines make the same angle $\frac{\pi}{4}$ with the given red line:
   ♦ The magenta line with a slope of 3
   ♦ The green line with a slope of -1/3

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Based on solved examples 2, 3 and 4, we can write about two cases: Case I and Case II. They can be described in 2 steps:
1. Difference between the two cases:
This can be written in two steps:
(i) In case I, we will be given the slopes m₁ and m₂ of two lines. We will be asked to find the angle 𝜃 between those two lines.
(ii) In case II, we will be given the slope m₁ of one line. We will be asked to find the slope m₂ of another line in such a way that, the new line makes a certain angle 𝜃 with the original line.
2. How to solve each case:
This can be written in two steps:
(i) In case I, we use the equation: $\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
   ♦ If tan 𝜃 obtained is +ve, it indicates the acute angle between the lines.  
   ♦ If tan 𝜃 obtained is -ve, it indicates the obtuse angle between the lines.
• Whether it is acute or obtuse, does not matter. With just one angle, we can calculate all the other three angles around the point of intersection.
• There is no need to interchange m₁ and m₂ to explore different possibilities. Even if we interchange m₁ and m₂, we will get the same four angles around the point of intersection.
• Also there is no need to find the second principal solution.
(ii) In case II also, we use the equation: $\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
• First we find the unknown slope.
• Then we interchange m₁ and m₂ to find another slope.
• There will be two lines making the same angle.
• In this case it is necessary to interchange m₁ and m₂. Steps must be repeated after the interchange. This will give the third line.

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Solved example 10.5
Line through the points (-2,6) and (4,8) is perpendicular to the line through the points (8,12) and (x,24). Find the value of x.
Solution:
1. Let m₁ be the slope of the line through the points (-2,6) and (4,8)
Then we can write:
$\begin{array}{ll}
{m_1}&{}={}
&{\frac{y_2 - y_1}{x_2 - x_1}}& {} \\

{}&{}={}
&{\frac{8 - 6}{4 - (-2)}}& {} \\

{}&{}={}
&{\frac{2}{6}}& {} \\

{}&{}={}
&{\frac{1}{3}}& {} \\

\end{array}$

2. Let m₂ be the slope of the line through the points (8,12) and (x,24)
Then we can write:
$\begin{array}{ll}
{m_2}&{}={}
&{\frac{y_2 - y_1}{x_2 - x_1}}& {} \\

{}&{}={}
&{\frac{24 - 12}{x - 8}}& {} \\

{}&{}={}
&{\frac{12}{x-8}}& {} \\

\end{array}$

3. Given that, the two lines are perpendicular to each other.
So we can write:
$\begin{array}{ll}
{}&{m_1 m_2}
&{}={}& {-1} \\

{\Rightarrow}&{\frac{1}{3}~ × ~\frac{12}{x-8}}
&{}={}& {-1} \\

{\Rightarrow}&{\frac{4}{x-8}}
&{}={}& {-1} \\

{\Rightarrow}&{4}
&{}={}& {8-x} \\

{\Rightarrow}&{x}
&{}={}& {4} \\

\end{array}$

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Collinearity of three points

• Suppose that we are given the coordinates of three points on the xy-plane. We can check whether those three points lie on the same line. It can be explained in 5 steps:
1. In fig.10.14(a) below, we have three points P, Q and R on the xy-plane.

Fig.10.14

2. In fig.b, the red line connects P and Q. Similarly, the green line connects Q and R
3. So we have two lines: PQ and QR
The two lines have one common point, which is Q
4. If PQ and QR have the same slope, then PQ will be parallel to QR
• But Q is a common point. So, if PQ and QR are parallel, the only possibility is that, P, Q and R lie on the same line.
5. Based on the above steps, we can write the procedure:
(i) Find the slope of the line PQ
(ii) Find the slope of the line QR
(iii) If the two slopes are the same, then P, Q and R are collinear.

Solved example 10.6
Three points P (h,k), Q (x₁,y₁), and R (x₂,y₂) lie on a line. Show that:
(h-x₁) (y₂-y₁) = (k-y₁)(x₂-x₁)
Solution:
1. Slope of the line PQ = $\frac{y_1 - k}{x_1 - h}$
2. Slope of the line QR = $\frac{y_2 - y_1}{x_2 - x_1}$
3. Since P, Q and R lie on a line, the two slopes are equal. So we can write:
$\begin{array}{ll}
{}&{\frac{y_1 - k}{x_1 - h}}
&{}={}& {\frac{y_2 - y_1}{x_2 - x_1}} \\

{\Rightarrow}&{\frac{k - y_1}{h - x_1}}
&{}={}& {\frac{y_2 - y_1}{x_2 - x_1}}&{\color {green}{\text{- - - (a) }}} \\

{\Rightarrow}&{(h-x_1)(y_2 - y_1)}
&{}={}& {(k-y_1)(x_2 - x_1)} \\

\end{array}$

◼ Remarks:
• Line marked as (a):
Both numerator and denominator of the L.H.S are multiplied by -1.

Solved example 10.7
Fig.13.15 below, shows the Time-Distance graph of a car. When the stop-watch showed a time of zero, the total distance traveled by the car was 2 units. When the stop-watch showed a time of 3, the total distance traveled by the car was 8 units. Assuming the car to be in uniform motion, derive a relation between time (T) and distance traveled (D) by the car.

Fig.10.15

Solution:
1. From the given data,
   ♦ we have the first point P (0,2) 
   ♦ we have the second point Q (3,8)
2. Let us mark a third point R on the graph.
   ♦ The x-coordinate of R will be T 
   ♦ The y-coordinate of R will be D
3. We can write:
The three points P (0,2), Q (3,8) and R (T,D) lie on the same line.
4. Let us write the slopes:
   ♦ Slope of the line segment PQ = $\frac{8-2}{3-0}~=~\frac{6}{3}~=~2$  
   ♦ Slope of the line segment QR = $\frac{D-8}{T-3}$
5. Since the three points are collinear, we can equate the slopes. So we get:
$\begin{array}{ll}
{}&{2}
&{}={}& {\frac{D - 8}{T - 3}} \\

{\Rightarrow}&{2T - 6}
&{}={}& {D-8} \\

{\Rightarrow}&{2T + 2}
&{}={}& {D} \\

{\Rightarrow}&{D}
&{}={}& {2(T+1)} \\

\end{array}$

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Link to a few more solved examples is given below:

Exercise 10.1

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

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Chapter 10.2 - Angle Between Two Lines

In the previous section, we saw slope and inclination of lines. In this section, we will see angle between two lines.

• Consider two lines lying in a plane. One of the following two possibilities will always occur:
(i) The two lines are parallel.
   ♦ They will never intersect.
(ii) The two lines are non-parallel
   ♦ Two non-parallel lines will surely intersect at a point.
• Two non-parallel lines may be lying far away from each other. But if we extend them suitably, they will surely intersect at a point.
• When two lines intersect, an angle will be formed between them.
• It is possible to find that angle using the slopes of the lines. Let us see how it is done. It can be written in 5 steps:

1. Fig.10.8(a) below shows two non-vertical lines L₁ and L₂.
    ♦ L₁ has an inclination of 𝛼
    ♦ L₂ has an inclination of β
    ♦ The slope of L₁ is m₁
    ♦ The slope of L₂ is m₂

Fig.10.8

2. Formation of triangle PQR:
    ♦ L₁ and L₂ intersect at R.
    ♦ L₁ intersect the x-axis at P
    ♦ L₂ intersect the x-axis at Q
• Thus we get a right angled triangle PQR
3. Formation of two angles:
• L₁ and L₂ intersect at R. The angle formed at R is marked as 𝜃.
• When 𝜃 is formed, another angle ɸ is also formed at the point of intersection.
• We see that: 𝜃 + ɸ = 180^o
4. Relation between 𝛼₁, 𝛼₂ and 𝜃:
• 𝛼₂ is an exterior angle of the triangle PQR
• For this exterior angle 𝛼₂, the angles 𝛼₁ and ∠PRQ are remote interior angles.
    ♦ ∠PRQ and 𝜃 are opposite angles
    ♦ So ∠PRQ = 𝜃
• Then we get: 𝛼₂ = 𝛼₁ + 𝜃
Which is same as: 𝜃 = 𝛼₂ - 𝛼₁
• This result can be used to find a relation between 𝜃, m₁ and m₂:

$\begin{array}{ll}
{}&{\theta}
&{}={}& {\alpha_2 - \alpha_1}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\tan (\alpha_2 - \alpha_1)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{\tan \alpha_2~-~\tan \alpha_1}{1~+~\tan \alpha_1 \tan \alpha_2}}
&{\color {green}{\text{- - - (a) }}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{m_2~-~m_1}{1~+~m_1 m_2}}
&{\color {green}{\text{- - - (b) }}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
(i) Line marked as (a):
See identity 11 in the list of identities.
(ii) Line marked as (b):
   ♦ tan 𝛼₁ is m₁
   ♦ tan 𝛼₂ is m₂


5. Using this relation, we can find 𝜃

---

Let us see an example:
Solved example 2
Two lines lying in the XY-plane have slopes $\frac{1}{2}$ and 3. Find the angle between them.
Solution:
1. Let m₁ = $\frac{1}{2}$ and m₂ = 3
• We are asked to find 𝜃
2. We have the equation: $\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\tan \theta}
&{}={}& {\frac{3~-~\frac{1}{2}}{1~+~\frac{1}{2}  × 3}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{6 ~-~1}{2~+~3}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{5}{5}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {1}
&{} \\

\end{array}$

3. This is a trigonometrical equation. We have learned how to solve them in a previous chapter (see solved example 3.65 in section 3.17).
2. Given that: tan 𝜃 = 1
• We know that 45^o is a principal solution.
• That means, we can put 45^o in place of 𝜃.
3. We will now find another principal solution.
• We have seen that, 𝜃 = 45^o is a principal solution.
• When the input 𝜃 is 45^o, the left side becomes tan 45^o
• This gives us: tan 𝜃 = tan 45^o = 1
• We want another angle such that, it's tangent is also 1
• Such an angle can be calculated using identities 9.f and 9.e [See the list of identities]: tan (π+x) = tan x
• We get:
tan 45^o = tan (180^o + 45^o) = tan 225^o
• So we can write:
tan 𝜃 = tan 45^o = tan 225^o
• Picking the first and last items, we get:
𝜃 = 225^o
   ♦ 225^o is greater than 0^o
   ♦ 225^o is less than 360^o  also.
   ♦ So it can be accepted as a principal solution.
• Thus we have two principal solutions for 𝜃:
𝜃 = 45^o and 𝜃 = 225^o
4. Fig.10.9 below shows how the two angles 45^o and 225^o are related to the given lines.

Fig.10.9

• The red line has a slope of $\frac{1}{2}$
Two convenient points points A and B are marked on this line.
• The green line has a slope of 3
Two convenient points points C and D are marked on this line.
• The two lines intersect at E
• ∠AEC (measured in the anti-clockwise direction) is the angle between the two lines. It's value is 45^o
• ∠BEC (measured in the anti-clockwise direction) is also an angle between the two lines. It's value is (180 + 45) = 225^o
5. In step (1), we took m₁ = $\frac{1}{2}$ and m₂ = 3.
• Since the order is not specified, we can take the reverse order also.
   ♦ That is., we interchange m₁ and m₂.
• Let us see if there is any change in the angles when we do the interchange. The following steps from (6) to (11) will give us the angles.
6. Let m₁ = 3 and m₂ = $\frac{1}{2}$
• We are asked to find 𝜃
7. We have the equation: $\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\tan \theta}
&{}={}& {\frac{\frac{1}{2}~-~3}{1~+~ 3 × \frac{1}{2}}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{1~-~6}{2~+~3}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {-\frac{5}{5}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {-1}
&{} \\

\end{array}$

8. This is a trigonometrical equation. We have learned how to solve them in a previous chapter (see solved example 3.65 in section 3.17).
9. Given that: tan 𝜃 = -1
• We know that 135^o is a principal solution.
• That means, we can put 135^o in place of 𝜃.
10. We will now find another principal solution.
• We have seen that, 𝜃 = 135^o is a principal solution.
• When the input 𝜃 is 135^o, the left side becomes tan 135^o
• This gives us: tan 𝜃 = tan 135^o = -1
• We want another angle such that, it's tangent is also -1
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
tan 135^o = tan (180^o + 135^o) = tan 315^o
• So we can write:
tan 𝜃 = tan 135^o = tan 315^o
• Picking the first and last items, we get:
𝜃 = 315^o
   ♦ 315^o is greater than 0^o
   ♦ 315^o is less than 360^o  also.
   ♦ So it can be accepted as a principal solution.
• Thus we have two principal solutions for 𝜃:
𝜃 = 135^o and 𝜃 = 315^o
11. Fig.10.10 below shows how the two angles 135^o and 315^o are related to the given lines.

Fig.10.10

• The red line has a slope of $\frac{1}{2}$
Two convenient points points A and B are marked on this line.
• The green line has a slope of 3
Two convenient points points C and D are marked on this line.
• The two lines intersect at E
• ∠DEA (measured in the anti-clockwise direction) is the angle between the two lines. It's value is 135^o
• ∠DEB (measured in the anti-clockwise direction) is also an angle between the two lines. It's value is (∠DEA + ∠AEC + ∠CEB) = (135 + 45 +135) = 315^o
12. The figs.10.9 and 10.10 are exactly the same. They appear different because, in the first fig, we marked the acute angle between the given lines. In the second fig., we marked the obtuse angle. 
• Based on this, we can write a conclusion. It can be written in 3 steps:
(i) First we find tan 𝜃 using the equation:
$\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
(ii) If tan 𝜃 obtained is +ve, the 𝜃 so obtained is the acute angle between the given lines.
(iii) If tan 𝜃 obtained is -ve, the 𝜃 so obtained is the obtuse angle between the given lines.

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Solved example 3
Two lines lying in the XY-plane have slopes $\frac{1}{2}$ and $-\frac{1}{3}$. Find the angle between them.
Solution:
1. Let m₁ = $\frac{1}{2}$ and m₂ = $-\frac{1}{3}$
• We are asked to find 𝜃
2. We have the equation: $\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\tan \theta}
&{}={}& {\frac{-\frac{1}{3}~-~\frac{1}{2}}{1~+~\frac{1}{2}  × -\frac{1}{3}}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{-2 ~-~3}{6~-~1}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{-5}{5}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {-1}
&{} \\

\end{array}$

3. This is a trigonometrical equation. We have learned how to solve them in a previous chapter (see solved example 3.65 in section 3.17).
2. Given that: tan 𝜃 = -1
• We know that 135^o is a principal solution.
• That means, we can put 135^o in place of 𝜃.
3. We will now find another principal solution.
• We have seen that, 𝜃 = 135^o is a principal solution.
• When the input 𝜃 is 135^o, the left side becomes tan 135^o
• This gives us: tan 𝜃 = tan 135^o = -1
• We want another angle such that, it's tangent is also -1
• Such an angle can be calculated using identities 9.f and 9.e [See the list of identities]: tan (π+x) = tan x
• We get:
tan 135^o = tan (180^o + 135^o) = tan 315^o
• So we can write:
tan 𝜃 = tan 135^o = tan 315^o
• Picking the first and last items, we get:
𝜃 = 315^o
   ♦ 315^o is greater than 0^o
   ♦ 315^o is less than 360^o  also.
   ♦ So it can be accepted as a principal solution.
• Thus we have two principal solutions for 𝜃:
𝜃 = 135^o and 𝜃 = 315^o
4. Fig.10.11 below shows how the two angles 135^o and 315^o are related to the given lines.

Fig.10.11

• The red line has a slope of $\frac{1}{2}$
Two convenient points points A and B are marked on this line.
• The green line has a slope of $-\frac{1}{3}$
Two convenient points points C and D are marked on this line.
• The two lines intersect at E
• ∠BEC (measured in the anti-clockwise direction) is the angle between the two lines. It's value is 135^o
• ∠BED (measured in the anti-clockwise direction) is also an angle between the two lines. It's value is (∠BEA + ∠AED) = (180 + 135) = 315^o
5. In step (1), we took m₁ = $\frac{1}{2}$ and m₂ = $-\frac{1}{3}$.
• Since the order is not specified, we can take the reverse order also.
• Since the order is not specified, we can take the reverse order also.
   ♦ That is., we interchange m₁ and m₂.
• Let us see if there is any change in the angles when order is reversed. The following steps from (6) to (11) will give us the angles.
6. Let m₁ = $-\frac{1}{3}$ and m₂ = $\frac{1}{2}$
• We are asked to find 𝜃
7. We have the equation: $\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
• Substituting the known values, we get:
$\begin{array}{ll}
{}&{\tan \theta}
&{}={}& {\frac{\frac{1}{2}~-~-\frac{1}{3}}{1~+~ -\frac{1}{3} × \frac{1}{2}}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{3~+~2}{6~-~1}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{5}{5}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {1}
&{} \\

\end{array}$

8. This is a trigonometrical equation. We have learned how to solve them in a previous chapter (see solved example 3.65 in section 3.17).
9. Given that: tan 𝜃 = 1
• We know that 45^o is a principal solution.
• That means, we can put 45^o in place of 𝜃.
10. We will now find another principal solution.
• We have seen that, 𝜃 = 45^o is a principal solution.
• When the input 𝜃 is 45^o, the left side becomes tan 45^o
• This gives us: tan 𝜃 = tan 45^o = 1
• We want another angle such that, it's tangent is also 1
• Such an angle can be calculated using identities 9.f and 9.e: tan (π+x) = tan x
• We get:
tan 45^o = tan (180^o + 45^o) = tan 225^o
• So we can write:
tan 𝜃 = tan 45^o = tan 225^o
• Picking the first and last items, we get:
𝜃 = 225^o
   ♦ 225^o is greater than 0^o
   ♦ 225^o is less than 360^o  also.
   ♦ So it can be accepted as a principal solution.
• Thus we have two principal solutions for 𝜃:
𝜃 = 45^o and 𝜃 = 225^o
11. Fig.10.12 below shows how the two angles 45^o and 225^o are related to the given lines.

Fig.10.12

• The red line has a slope of $\frac{1}{2}$
Two convenient points points A and B are marked on this line.
• The green line has a slope of $-\frac{1}{3}$
Two convenient points points C and D are marked on this line.
• The two lines intersect at E
• ∠CEA (measured in the anti-clockwise direction) is the angle between the two lines. It's value is 45^o
• ∠CEB (measured in the anti-clockwise direction) is also an angle between the two lines. It's value is (CEA + 180) = (45 +180) = 225^o
12. The figs.10.11 and 10.12 are exactly the same. They appear different because, in the first fig, we marked the obtuse angle between the given lines. In the second fig., we marked the acute angle. 
• Based on this, we can write the same conclusion that we wrote in the previous solved example. It can be written in 3 steps:
(i) First we find tan 𝜃 using the equation:
$\tan \theta~=~\frac{m_2~-~m_1}{1~+~m_1 m_2}$
(ii) If tan 𝜃 obtained is +ve, the 𝜃 so obtained is the acute angle between the given lines.
(iii) If tan 𝜃 obtained is -ve, the 𝜃 so obtained is the obtuse angle between the given lines.

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In the next section, we will see a few more solved examples.

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Chapter 10.1 - Slope of a Line

In the previous section, we saw some basic concepts and formulae in coordinate geometry. In this section, we will see slope of lines.

Inclination of a line and Slope of a line

First we will see inclination of a line. It can be written in 4 steps:
1. In fig.10.5(a) below,
(i) The angle between the line l₁ and the +ve side of the x-axis is 𝜃₁.
(ii) The angle 𝜃₁ is measured from the +ve side of the x-axis in the anti-clockwise direction.
• If the above two conditions are satisfied, then 𝜃₁ is called the inclination of the line l₁.

Fig.10.5

2. Consider the line l₂ in fig.10.5(b) above.
• Based on what we saw in fig.(a), we can write:
Inclination of the line l₂ is  𝜃₂
• This is because, the two conditions are satisfied:
(i) 𝜃₂ is measured from the +ve side of the x-axis.
(ii) 𝜃₂ is measured in the anti-clockwise direction.
3. Based on the two figs. (a) and (b), we can write:
• Inclination of a line parallel to the x-axis will be 0^o
• Inclination of a line parallel to the y-axis will be 90^o
4. So it is obvious that:
    ♦ Inclination of a line can never be less than 0^o.
    ♦ Inclination of a line can never be greater than 180^o.
• We can write: 0^o≤ 𝜃 ≤ 180^o.

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Now we will see slope of a line. It can be written in steps:
1. We have seen how to write the inclination 𝜃 of any line l
• Now, tan 𝜃 is called the slope of that line l
2. Another name for slope is gradient.
• So we can write in either of the two ways:
    ♦ tan 𝜃 is the slope of the line l
    ♦ tan 𝜃 is the gradient of the line l
3. Slope of a line whose inclination is zero (a horizontal line) = tan 0 = 0
4. Slope of a line whose inclination is 90^o (a vertical line) = tan 90
    ♦ But tan 90 gives a number which does not exist.
    ♦ So slope of such a line is not defined
5. Based on this, we can write:
    ♦ Slope of x-axis = 0
    ♦ Slope of y-axis is not defined.

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Slope from coordinates

• We know that, a line can be completely determined if we know the coordinates of any two points on it.
• If we know those coordinates, we can find the slope also. Let us see how it is done. It can be written in 7 steps:
1. In fig.10.6(a) below, the non-vertical line l has an inclination of 𝜃
• P (x₁, y₁) and Q (x₂,y₂) are two points on the line.

Fig.10.6

• Note:
In fig.10.6(a), x₁ cannot be equal to x₂. If they are equal, line l will become vertical. The slope of a vertical line is not defined.
2. A line QR is drawn perpendicular to the x-axis.
• Another line PM is drawn perpendicular to QR.
3. In the triangle PMQ, angle QPM will be equal to 𝜃.
So we can write:
$\text{Slope of}~l~=~\tan \theta ~=~\frac{QM}{PM}$
4. Now, the coordinates of M will be (x₂,y₁)
So the distance QM wil be (y₂ – y₁)
5. Since the coordinates of M are (x₂,y₁), the distance PM will be (x₂ – x₁)
6. Based on steps (3), (4) and (5), we get:
$\text{Slope of}~l~=~\tan \theta ~=~\frac{QM}{PM}~=~\frac{y_2 - y_1}{x_2 - x_1}$
7. Thus we get a simple method to find the slope using coordinates.

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In the fig.10.6(a) above, the inclination 𝜃 is an acute angle. Let us see the case when 𝜃 is an obtuse angle. It can be written in steps:
1. In fig.10.6(b) above, the non-vertical line l has an inclination of 𝜃
• P (x₁, y₁) and Q (x₂,y₂) are two points on the line.
• Note:
In fig.10.6(b), x₁ cannot be equal to x₂. If they are equal, line l will become vertical. The slope of a vertical line is not defined.
2. A line QR is drawn perpendicular to the x-axis.
• Another line PM is drawn perpendicular to QR.
3. In the triangle PMQ, angle QPM will be equal to (180 - 𝜃).
So we can write:
$\text{Slope of}~l~=~\tan (180~-~\theta) ~=~-\tan \theta~=~\frac{QM}{PM}$
[See identities 9(c) and 9(d) in the list of identities]
4. Now, the coordinates of M will be (x₂,y₁)
So the distance QM wil be (y₂ – y₁)
5. Since the coordinates of M are (x₂,y₁), the distance PM will be (x₁ – x₂)
• Note that, in this case, we subtract x₂ from x₁ because, x₁ is larger than x₂.
6. Based on steps (3), (4) and (5), we get:
$\text{Slope of}~l~=~\tan (180~-~\theta) ~=~-\tan \theta~=~\frac{QM}{PM}~=~\frac{y_2 - y_1}{x_1 - x_2}$
• The -ve sign can be avoided by rearranging the denominator. We get:
$\text{Slope of}~l~=~\tan \theta~=~\frac{y_2 - y_1}{x_2 - x_1}$
7. Thus here also, we get the same simple method to find the slope using coordinates.

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• If a line is horizontal, all points on that line will have the same y coordinates.
    ♦ That is., y₂ will be equal to y₁.
• Then the numerator will become zero.
• Consequently, tan 𝜃 will become zero.
• So the above result is applicable to all non-vertical lines, even if the line is horizontal.
• We can write:
If P (x₁, y₁) and Q (x₂,y₂) are two points on a non-vertical line, then the slope of that line will be equal to $\frac{y_2 - y_1}{x_2 - x_1}$
• Note that, in the numerator, we have a distance. Distance can be measured in meter or centimeter. In the denominator also, we have distance. So the units will cancel each other. That means: Slope has no units. It is just a number.

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Condition for parallelism

This can be written in steps:
1. Fig.10.7(a) below shows two non-vertical lines l₁ and l₂.
    ♦ l₁ has an inclination of 𝛼
    ♦ l₂ has an inclination of β
    ♦ The slope of l₁ is m₁
    ♦ The slope of l₂ is m₂

Fig.10.7

2. From our earlier geometry classes, we know this:
    ♦ If l₁ and l₂ are parallel, then 𝛼 = β
3. But if 𝛼 = β, then tan 𝛼 = tan β
    ♦ If tan 𝛼 = tan β, then m₁ = m₂
• So the condition for two lines l₁ and l₂ to be parallel is:
Their slopes m₁ and m₂ must be equal.

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Let us prove the converse.
• We need to prove that, if slopes m₁ and m₂ of the two lines l₁ and l₂ are equal, then the two lines are parallel.
It can be proved in 5 steps:
1. Given that, m₁ = m₂
Then we can write: tan 𝛼 = tan β
2. This is a trigonometrical equation. We have learned how to solve them in a previous chapter (see solved example 3.65 in section 3.17).
3. So, if tan 𝛼 = tan β, we can write:
𝛼 = n𝞹 + β
• Let us put the smallest possible value of 'n', which is 1
We get: 𝛼 = 𝞹 + β
4. But (𝞹 + β) is greater than 180^o
• Inclination cannot be greater than 180^o
• So the only possibility is that 𝛼 = β.
If 𝛼 = β, then the two lines are parallel
5. Thus we arrive at the conclusion:
If slopes m₁ and m₂ of two lines are equal, then the lines will be parallel.

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Condition for perpendicularity

This can be written in 3 steps:
1. Fig.10.7(b) above shows two non-vertical lines l₁ and l₂.
    ♦ l₁ has an inclination of 𝛼
    ♦ l₂ has an inclination of β
    ♦ The slope of l₁ is m₁
    ♦ The slope of l₂ is m₂
2. Formation of triangle PQR:
    ♦ l₁ and l₂ intersect at R.
    ♦ l₁ intersect the x-axis at P
    ♦ l₂ intersect the x-axis at q
• Thus we get a right angled triangle PQR
3. Relation between 𝛼 and β:
• β is an exterior angle of the triangle PQR
• For this exterior angle β, the angles 𝛼 and 90^o are remote interior angles.
• So we get: β = 𝛼 + 90
• This result can be used to find a relation between the two slopes:

$\begin{array}{ll}
{}&{\beta}
&{}={}& {\alpha + 90^o}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \beta}
&{}={}& {\tan (\alpha + 90^o)}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \beta}
&{}={}& {- \cot \alpha}
&{\color {green}{\text{- - - (a) }}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \beta}
&{}={}& {- \frac{1}{\tan \alpha}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{m_2}
&{}={}& {- \frac{1}{m_1}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{m_1 m_2}
&{}={}& {- 1}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
• Line marked as (a):
See identities 9(a) and 9(b) in the list of identities.
tan (90+𝜃) = -cot 𝜃
• So the condition for two lines l₁ and l₂ to be perpendicular is:
Product of their slopes (m₁ m₂) must be -1.

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Let us prove the converse.
• We need to prove that, if product of the slopes (m₁ m₂) of the two lines l₁ and l₂ is -1, then the two lines are perpendicular to each other.
It can be proved in 3 steps:

1. Relation between 𝛼 and β:

$\begin{array}{ll}
{}&{m_1 m_2}
&{}={}& {-1}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \alpha × \tan \beta}
&{}={}& {-1}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \alpha}
&{}={}& {- \frac{1}{ \tan \beta}}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \alpha}
&{}={}& {- \cot \beta}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\tan \alpha}
&{}={}& {\tan (90^o + \beta)}
&{\color {green}{\text{- - - (a) }}}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

{\Rightarrow}&{\alpha}
&{}={}& {90^o + \beta}
&{}& {}
&{}& {}
&{}& {}&{}& {} &{} &{} &{} \\

\end{array}$

◼ Remarks:
• Line marked as (a):
See identities 9(a) and 9(b) in the list of identities.
tan (90+𝜃) = -cot 𝜃

2. So 𝛼 and β differ by 90^o
• Then based on fig.10.7(b), the two lines will be perpendicular to each other.
3. So it is clear that, if (m₁m₂) = -1, then the two lines are perpendicular to each other.

Let us see a solved example

Solved example 10.1
Find the slope of the following lines:
(a) Line passing through the points (3,-2) and (-1,4)
(b) Line passing through the points (3,-2) and (7,-2)
(c) Line passing through the points (3,-2) and (3,4)
(d) Line making an inclination of 60^o with the positive direction of the x-axis.
Solution:
If a line pass through two points (x₁,y₁) and (x₂,y₂), then the slope of that line will be $\frac{y_2 - y_1}{x_2 - x_1}$ 
Part (a): Slope = $\frac{4 - -2}{-1 - 3}~=~\frac{6}{-4}~=~-\frac{3}{2}$
Part (b): Slope = $\frac{-2 - -2}{7 - 3}~=~\frac{0}{4}~=~0$
Part (c): Slope = $\frac{4 - -2}{3 - 3}~=~\frac{6}{0},~\text{which is not defined}$
Part (d): Slope is the tangent of inclination. So we get:
Slope = tan 60^o = $\sqrt{3}$

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In the next section, we will see angle between two lines.

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