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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