27.5 - Two Lines Parallel or Perpendicular

In the previous section, we saw the angle between lines in 3D. In this section, we will see the conditions for two lines to be parallel or perpendicular to each other.

First we will see angle in terms of direction cosines. It can be explained in 7 steps:
1. For $\small{L_1}$:
    ♦ Let the direction ratios be $\small{a_1,~b_1,~c_1}$
    ♦ Let the direction cosines be $\small{l_1,~m_1,~n_1}$
• Then we can write:
$\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}~=~\lambda_1}$

2. For $\small{L_2}$:
    ♦ Let the direction ratios be $\small{a_2,~b_2,~c_2}$
    ♦ Let the direction cosines be $\small{l_2,~m_2,~n_2}$
• Then we can write:
$\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}~=~\lambda_2}$

3. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2 ~+~ b_1 b_2~+~c_1 c_2}    & {~=~}    &{\left(\frac{l_1}{\lambda_1} \right)\left(\frac{l_2}{\lambda_2} \right)+\left(\frac{l_1}{\lambda_1} \right)\left(\frac{l_2}{\lambda_2} \right)+\left(\frac{l_1}{\lambda_1} \right)\left(\frac{l_2}{\lambda_2} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\lambda_1\lambda_2}}
\\ \end{array}}$

4. Also we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1^2 ~+~ b_1^2~+~c_1^2}    & {~=~}    &{\left(\frac{l_1}{\lambda_1} \right)^2~+~\left(\frac{m_1}{\lambda_1} \right)^2~+~\left(\frac{n_1}{\lambda_1} \right)^2}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_1^2~+~m_1^2~+~n_1^2}{\lambda_1^2}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\sqrt{a_1^2 ~+~ b_1^2~+~c_1^2}}    & {~=~}    &{\frac{\sqrt{l_1^2~+~m_1^2~+~n_1^2}}{\lambda_1}}
\\ \end{array}}$

5. Also we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2^2 ~+~ b_2^2~+~c_2^2}    & {~=~}    &{\left(\frac{l_2}{\lambda_2} \right)^2~+~\left(\frac{m_2}{\lambda_2} \right)^2~+~\left(\frac{n_2}{\lambda_2} \right)^2}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_2^2~+~m_2^2~+~n_2^2}{\lambda_2^2}}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\sqrt{a_2^2 ~+~ b_2^2~+~c_2^2}}    & {~=~}    &{\frac{\sqrt{l_2^2~+~m_2^2~+~n_2^2}}{\lambda_2}}
\\ \end{array}}$

6. In the previous section, we derived the formula:
$\small{\cos\theta~=~\frac{a_1 a_2 ~+~ b_1 b_2~+~c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2}\,\sqrt{a_2^2 + b_2^2 + c_2^2}}}$

7. Substituting from (3), (4) and (5), we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\theta}    & {~=~}    &{\frac{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\lambda_1\lambda_2}}{\left(\frac{\sqrt{l_1^2~+~m_1^2~+~n_1^2}}{\lambda_1} \right)\left(\frac{\sqrt{l_2^2~+~m_2^2~+~n_2^2}}{\lambda_2} \right)}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\left(\sqrt{l_1^2~+~m_1^2~+~n_1^2} \right)\left(\sqrt{l_2^2~+~m_2^2~+~n_2^2} \right)}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{l_1 l_2~+~m_1 m_2~+~n_1 n_2}{\left(1 \right)\left(1 \right)}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}
\\ \end{array}}$

◼ Remarks:
3 (magenta color): Here we use the fact that:
$\small{l^2 + m^2 +n^2 = 1}$

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Now we will see the condition for two lines to be perpendicular. It can be explained in 4 steps:
1. We have seen that $\small{\cos \theta = l_1 l_2~+~m_1 m_2~+~n_1 n_2}$ 

2. When the two lines are perpendicular, we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos\left(\frac{\pi}{2} \right)}    & {~=~}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{0}    & {~=~}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}
\\ \end{array}}$

3. So the condition for two lines to be perpendicular is:
$\small{l_1 l_2~+~m_1 m_2~+~n_1 n_2~=~0}$

4. The above condition is in terms of direction cosines. We can write it in terms of direction ratios also:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left(\lambda_1 a_1 \right)\left(\lambda_2 a_2 \right)~+~\left(\lambda_1 b_1 \right)\left(\lambda_2 b_2 \right)~+~\left(\lambda_1 c_1 \right)\left(\lambda_2 c_2 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\lambda_1 \lambda_2\left(a_1 a_2~+~b_1 b_2~+~c_1 c_2 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{0}
\\ \end{array}}$

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We have already seen the condition for two lines to be parallel in a previous section 27.1.
1. The condition in terms of direction cosines is:
$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}}$

2. The condition in terms of direction ratios is:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

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Now we will see some solved examples

Solved example 27.19
Show that the three lines with direction cosines
$\small{\frac{12}{13},~\frac{-3}{13},~\frac{-4}{13};~~~~~~\frac{4}{13},~\frac{12}{13},~\frac{3}{13};~~~~~~\frac{3}{13},~\frac{-4}{13},~\frac{12}{13}}$
are mutually perpendicular.
Solution:
1. Let us name the three lines and classify the direction cosines:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{L_1}    & {~:~}    &{\frac{12}{13},~\frac{-3}{13},~\frac{-4}{13}}
\\ {~\color{magenta}    2    }    &{}    &{L_2}    & {~:~}    &{\frac{4}{13},~\frac{12}{13},~\frac{3}{13}}
\\ {~\color{magenta}    3    }    &{}    &{L_3}    & {~:~}    &{\frac{3}{13},~\frac{-4}{13},~\frac{12}{13}}
\\ \end{array}}$

2. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{l_1 l_2~+~m_1 m_2~+~n_1 n_2~=~0}$

3. Let us check the given $\small{L_1~\text{and}~L_2}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_1 l_2~+~m_1 m_2~+~n_1 n_2}    & {~=~}    &{\left(\frac{12}{13} \right)\left(\frac{4}{13} \right)+\left(\frac{-3}{13} \right)\left(\frac{12}{13} \right)+\left(\frac{-4}{13} \right)\left(\frac{3}{13} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{48 -36 - 12}{169}~=~\frac{0}{169}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$
• So $\small{L_1~\text{and}~L_2}$ are mutually perpendicular.

4. Let us check the given $\small{L_2~\text{and}~L_3}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_2 l_3~+~m_2 m_3~+~n_2 n_3}    & {~=~}    &{\left(\frac{4}{13} \right)\left(\frac{3}{13} \right)
+\left(\frac{12}{13} \right)\left(\frac{-4}{13} \right)
+\left(\frac{3}{13} \right)\left(\frac{12}{13} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{12 -48 + 36}{169}~=~\frac{0}{169}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$
• So $\small{L_2~\text{and}~L_3}$ are mutually perpendicular.

5. Let us check the given $\small{L_3~\text{and}~L_1}$:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l_3 l_1~+~m_3 m_1~+~n_3 n_1}    & {~=~}    &{\left(\frac{3}{13} \right)\left(\frac{12}{13} \right)
+\left(\frac{-4}{13} \right)\left(\frac{-3}{13} \right)
+\left(\frac{12}{13} \right)\left(\frac{-4}{13} \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{36 +12 - 48}{169}~=~\frac{0}{169}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$
• So $\small{L_3~\text{and}~L_1}$ are mutually perpendicular.

6. Based on (3), (4) and (5), we can write:
The three lines are mutually perpendicular.

Solved example 27.20
Show that the line through the points (1,−1,2) and (3,4,−2) is perpendicular to the line through the points (0,3,2) and (3,5,6)
Solution:
1. For a line through two points $\small{U\left(x_1,y_1,z_1 \right)~\text{and}~V\left(x_2,y_2,z_2 \right)}$, we can consider $\small{\left(x_2 - x_1 \right),~\left(y_2 - y_1 \right)~\text{and}~\left(z_2 - z_1 \right)}$ as a set of direction ratios.

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
2, 5 and −4

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
3, 2 and 4

4. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{a_1 a_2~+~b_1 b_2~+~c_1 c_2~=~0}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{\left(2 \right)\left(3 \right)+\left(5 \right)\left(2 \right)+\left(-4 \right)\left(4 \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{6+10-16}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$

6. So the two lines are perpendicular to each other.

Solved example 27.21
Show that the line through the points (4,7,8) and (2,3,4) is  parallel to the line through the points (−1,−2,1) and (1,2,5)
Solution:
1. For a line through two points $\small{U\left(x_1,y_1,z_1 \right)~\text{and}~V\left(x_2,y_2,z_2 \right)}$, we can consider $\small{\left(x_2 - x_1 \right),~\left(y_2 - y_1 \right)~\text{and}~\left(z_2 - z_1 \right)}$ as a set of direction ratios.

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
−2, −4 and −4

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
2, 4 and 4

4. If two lines $\small{L_1~\text{and}~L_2}$ are parallel, we have:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{-2}{2}}& {~=~}    &{\frac{-4}{4}}    & {~=~}    &{\frac{-4}{4}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{-1}& {~=~}    &{-1}    & {~=~}    &{-1}
\\ \end{array}}$ 

6. So the two lines are parallel.

Solved example 27.22
Show that the lines
$\small{\frac{x-5}{7}~=~\frac{y+2}{-5}~=~\frac{z}{1}}$ and
$\small{\frac{x}{1}~=~\frac{y}{2}~=~\frac{z}{3}}$
are perpendicular to each other
Solution:
1. When the equation of the line is in Cartesian form, the denominators can be considered as a set of direction ratios.

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
7, −5 and 1

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
1, 2 and 3

4. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{a_1 a_2~+~b_1 b_2~+~c_1 c_2~=~0}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{\left(7 \right)\left(1\right)+\left(-5 \right)\left(2 \right)+\left(1 \right)\left(3 \right)}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{7-10+3}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{0}
\\ \end{array}}$

6. So the two lines are perpendicular to each other.

Solved example 27.23
Find the value of $\small{p}$ so that the lines
$\small{\frac{1-x}{3}~=~\frac{7y-14}{2p}~=~\frac{z-3}{2}}$ and
$\small{\frac{7-7x}{3p}~=~\frac{y-5}{1}~=~\frac{6-z}{5}}$
are perpendicular to each other
Solution:
1. When the equation of the line is in Cartesian form, the denominators can be considered as a set of direction ratios.
• But we need to convert the equations into standard form.
• For the first line, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{1-x}{3}}& {~=~}    &{\frac{7y-14}{2p}}    & {~=~}    &{\frac{z-3}{2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{(-1)(1-x)}{(-1)(3)}}& {~=~}    &{\frac{(7y-14)/7}{(2p)/7}}    & {~=~}    &{\frac{z-3}{2}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x-1}{-3}}& {~=~}    &{\frac{y-2}{(2/7)p}}    & {~=~}    &{\frac{z-3}{2}}
\\ \end{array}}$
• For the second line, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{7-7x}{3p}}& {~=~}    &{\frac{y-5}{1}}    & {~=~}    &{\frac{6-z}{5}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{(-1)(1/7)(7-7x)}{(-1)(1/7)(3p)}}& {~=~}    &{\frac{y-5}{1}}    & {~=~}    &{\frac{(-1)(6-z)}{(-1)(5)}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{\frac{x-1}{(-3/7)p}}& {~=~}    &{\frac{y-5}{1}}    & {~=~}    &{\frac{z-6}{-5}}
\\ \end{array}}$ 

2. So for the first line, the direction ratios $\small{a_1,b_1,c_1}$ are:
−3, (2/7)p and 2

3. Similarly, for the second line, the direction ratios $\small{a_2,b_2,c_2}$ are:
(−3/7)p, 1 and −5

4. If two lines $\small{L_1~\text{and}~L_2}$ are perpendicular, we have:
$\small{a_1 a_2~+~b_1 b_2~+~c_1 c_2~=~0}$

5. Substituting the values, we get:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1 a_2~+~b_1 b_2~+~c_1 c_2}    & {~=~}    &{0}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{\left(-3 \right)\left((-3/7)p\right)+\left((2/7)p \right)\left(1 \right)+\left(2 \right)\left(-5 \right)}    & {~=~}    &{0}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{\frac{9p}{7} + \frac{2p}{7} - 10}    & {~=~}    &{0}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{\frac{11p - 70}{7}}    & {~=~}    &{0}
\\ {~\color{magenta}    5    }    &{\Rightarrow}    &{11p - 70}    & {~=~}    &{0}
\\ {~\color{magenta}    6    }    &{\Rightarrow}    &{p}    & {~=~}    &{\frac{70}{11}}
\\ \end{array}}$

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

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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.

---

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.

---

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.

---

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.

---

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.

---

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.

---

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.

---

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 = $\sqrt{3}$

---

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

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