27.1 - Direction Ratios of Two Parallel Lines

In the previous section, we saw the basic details about direction cosines and direction ratios. We saw the direction cosines of the line through two points P and Q. In this section, we will see more details about direction ratios.

Direction ratios of a line passing through two points

This can be explained in 6 steps:
1. Let $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ be two points in space.
2. We obtained the direction cosines of the line passing through P and Q:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l~=~\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{m~=~\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{}    &{n~=~\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
3. If we can find any three numbers a, b and c which satisfies the condition
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}}$, then we can say that a, b and c are the direction cosines of the line through P and Q
4. Let:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a}    & {~=~}    &{x_2 - x_1}
\\ {~\color{magenta}    2    }    &{}    &{b}    & {~=~}    &{y_2 - y_1}
\\ {~\color{magenta}    3    }    &{}    &{c}    & {~=~}    &{z_2 - z_1}
\\ \end{array}}$
5. We will check whether this a, b and c, satisfy the condition written in (3):
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{l}{a}}& {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|\left(x_2 - x_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{m}{b}}& {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|\left(y2 - y_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{n}{c}}& {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|\left(z_2 - z_1 \right)}}    & {~=~}    &{\frac{1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$
• The condition is satisfied.
• So $\small{\left(x_2 - x_1 \right),~\left(y_2 - y_1 \right),~\left(z_2 - z_1 \right)}$ are indeed direction ratios of the line through $\small{(x_1,x_2,x_3)~\text{and}~(x_1,x_2,x_3)}$
6. Note that, $\small{\left(x_1 - x_2 \right),~\left(y_1 - y_2 \right),~\left(z_1 - z_2 \right)}$ are also direction ratios of the line through $\small{(x_1,x_2,x_3)~\text{and}~(x_1,x_2,x_3)}$.
• This is shown below:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{l}{a}}& {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|\left(x_1 - x_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{m}{b}}& {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|\left(y1 - y_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{n}{c}}& {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|\left(z_1 - z_2 \right)}}    & {~=~}    &{\frac{-1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

Lines having proportional direction ratios

This can be explained in 10 steps:
1. Consider two directed lines L₁ and L₂
• L₁ has:
    ♦ direction cosines $\small{l_1,~m_1,~n_1}$
    ♦ direction ratios $\small{a_1,~b_1,~c_1}$
• L₂ has:
    ♦ direction cosines $\small{l_2,~m_2,~n_2}$
    ♦ direction ratios $\small{a_2,~b_2,~c_2}$

2. For L₁, we can write: $\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}}$

• Since the three fractions are equal, we can write:
$\small{\frac{l_1}{a_1}~=~\frac{m_1}{b_1}~=~\frac{n_1}{c_1}~=~\lambda_1}$

3. For L₂, we can write: $\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}}$

• Since the three fractions are equal, we can write:
$\small{\frac{l_2}{a_2}~=~\frac{m_2}{b_2}~=~\frac{n_2}{c_2}~=~\lambda_2}$

4. Suppose that, the direction ratios of the two lines satisfy the condition:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$

• Since the three fractions are equal, we can write:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}~=~\lambda_3}$

5. Substituting from (4) into (2), we get:
$\small{\frac{l_1}{a_2 \lambda_3}~=~\frac{m_1}{b_2 \lambda_3}~=~\frac{n_1}{c_2 \lambda_3}~=~\lambda_1}$

6. Substituting from (3) into (5), we get:
$\small{\frac{l_1}{\left(\frac{l_2}{\lambda_2} \right) \lambda_3}~=~\frac{m_1}{\left(\frac{m_2}{\lambda_2} \right) \lambda_3}~=~\frac{n_1}{\left(\frac{n_2}{\lambda_2} \right) \lambda_3}~=~\lambda_1}$

$\small{\Rightarrow \frac{l_1 \lambda_2}{l_2 \lambda_3}~=~\frac{m_1 \lambda_2}{m_2 \lambda_3}~=~\frac{n_1 \lambda_2}{n_2 \lambda_3}~=~\lambda_1}$

• Multiplying throughout by $\small{\left(\frac{\lambda_3}{\lambda_2} \right)}$, we get:

$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}~=~\lambda_1\left(\frac{\lambda_3}{\lambda_2} \right)~=~\lambda_4}$

• Note that $\small{\lambda_1,~\lambda_2}$ . . . etc., are just some real numbers. Multiplying them will give new real numbers.

7. So we get an important result:
When direction ratios of two lines satisfy the condition $\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}~=~\lambda_3}$,
their direction raios will satisfy the condition
$\small{\frac{l_1}{l_2}~=~\frac{m_1}{m_2}~=~\frac{n_1}{n_2}~=~\lambda_4}$  

We can write:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{b_1}{b_2}}    & {~=~}    &{\frac{c_1}{c_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{l_1}{l_2}}& {~=~}    &{\frac{m_1}{m_2}}    & {~=~}    &{\frac{n_1}{n_2}}
\\ \end{array}}$  

8. We have the direction cosines of L₁ and L₂. So we can write unit vectors parallel to them. We get:

$\small{\hat{L_1}= l_1\hat{i}+m_1\hat{j}+n_1\hat{k}}$

$\small{\hat{L_2}= l_2\hat{i}+m_2\hat{j}+n_2\hat{k}}$

9. Substituting from (7) into (8), we get:
$\small{\hat{L_1}= \left(l_2 \lambda_4 \right)\hat{i}+\left(m_2 \lambda_4 \right)\hat{j}+\left(n_2 \lambda_4 \right)\hat{k}}$

$\small{\Rightarrow \hat{L_1}= \lambda_4\left(l_2\hat{i}+m_2\hat{j}+n_2\hat{k} \right)}$

$\small{\Rightarrow \text{Unit vectors}~\hat{L_1}~\text{and}~\hat{L_2}~\text{are parallel}}$

$\small{\Rightarrow \text{Lines}~L_1~\text{and}~L_2~\text{are parallel}}$

10. So we can complete the result in (7). We get:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{b_1}{b_2}}    & {~=~}    &{\frac{c_1}{c_2}}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\frac{l_1}{l_2}}& {~=~}    &{\frac{m_1}{m_2}}    & {~=~}    &{\frac{n_1}{n_2}}
\\ {~\color{magenta}    3    }    &{{\Rightarrow}}    &{L_1}& {~~\parallel~}    &{L_2}    & {{}}    &{{}}
\\ \end{array}}$  

◼ Remarks:
• 1 (magenta color): This line gives the condition. It indicates that, the direction ratios of the two lines are proportional
• 2 (magenta color): If the condition is satisfied, the direction cosines of the two lines will be proportional.
• 3 (magenta color): If the condition is satisfied, the two lines will be parallel to each other.

   

Solved example 27.6
Show that the points A(2,3,−4), B(1,−2,3) and C(3,8,−11) are collinear.
Solution:
1. First we find the direction ratios of AB:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(1-2) = -1}
\\ {~\color{magenta}    2    }    &{{}}    &{b_1}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(-2-3) = -5}
\\ {~\color{magenta}    3    }    &{{}}    &{c_1}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(3-(-4)) = 7}
\\ \end{array}}$  

2. Next we find the direction ratios of BC:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(3-1) = 2}
\\ {~\color{magenta}    2    }    &{{}}    &{b_2}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(8-(-2)) = 10}
\\ {~\color{magenta}    3    }    &{{}}    &{c_2}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(-11-3) = -14}
\\ \end{array}}$ 

3. Next we calculate the three ratios:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{-1}{2}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{b_1}{b_2}}& {~=~}    &{\frac{-5}{10}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{c_1}{c_2}}& {~=~}    &{\frac{7}{-14}}    & {~=~}    &{-\frac{1}{2}}
\\ \end{array}}$  

4. We see that:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$
• That means, the direction ratios of the two lines AB and BC, are proportional.
• That means, AB and BC are parallel.

5. The two parallel lines AB and BC have one point B in common. So the points A, B and C are collinear.

Solved example 27.7
Show that the points (2,3,4), (−1,−2,1) and (5,8,7) are collinear.
Solution:
1. Let the three points be: A(2,3,4), B(−1,−2,1) and C(5,8,7)
• First we find the direction ratios of AB:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_1}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(-1-2) = -3}
\\ {~\color{magenta}    2    }    &{{}}    &{b_1}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(-2-3) = -5}
\\ {~\color{magenta}    3    }    &{{}}    &{c_1}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(1-4) = -3}
\\ \end{array}}$  

2. Next we find the direction ratios of BC:

$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{a_2}& {~=~}    &{x_2 - x_1}    & {~=~}    &{(5-(-1)) = 6}
\\ {~\color{magenta}    2    }    &{{}}    &{b_2}& {~=~}    &{y_2 - y_1}    & {~=~}    &{(8-(-2)) = 10}
\\ {~\color{magenta}    3    }    &{{}}    &{c_2}& {~=~}    &{z_2 - z_1}    & {~=~}    &{(7-1) = 6}
\\ \end{array}}$ 

3. Next we calculate the three ratios:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{a_1}{a_2}}& {~=~}    &{\frac{-3}{6}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    2    }    &{{}}    &{\frac{b_1}{b_2}}& {~=~}    &{\frac{-5}{10}}    & {~=~}    &{-\frac{1}{2}}
\\ {~\color{magenta}    3    }    &{{}}    &{\frac{c_1}{c_2}}& {~=~}    &{\frac{-3}{6}}    & {~=~}    &{-\frac{1}{2}}
\\ \end{array}}$  

4. We see that:
$\small{\frac{a_1}{a_2}~=~\frac{b_1}{b_2}~=~\frac{c_1}{c_2}}$
• That means, the direction ratios of the two lines AB and BC, are proportional.
• That means, AB and BC are parallel.

5. The two parallel lines AB and BC have one point B in common. So the points A, B and C are collinear.

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The link below gives a few more miscellaneous examples:

Exercise 27.1

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

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Chapter 27 - Three Dimensional Geometry

In the previous section, we completed a discussion on vector algebra. In this chapter, we will see Three dimensional geometry.

In class 11, we saw the basic details about three dimensional geometry. See chapter 12. In the present chapter, we will see some advanced topics related to three dimension. In the previous chapter, we saw vectors in three dimensional space. We can use those vectors for our present discussion. This will make the discussion more interesting.

Direction cosines of a line

This can be explained in 5 steps:
1. Consider a directed line L passing through the origin. It makes the following angles with the axes:
    ♦ $\small{\alpha}$ with the +ve direction of the x-axis
    ♦ $\small{\beta}$ with the +ve direction of the y-axis
    ♦ $\small{\gamma}$ with the +ve direction of the z-axis
• Then $\small{\cos \alpha,~\cos \beta~\text{and}~\cos \gamma}$ are the direction cosines of the line L

2. Suppose that, we reverse the direction of L. We will call the new line as L'.
• L' will make the following angles with the axes:
    ♦ $\small{\left( \pi - \alpha \right)}$ with the +ve direction of the x-axis
    ♦ $\small{\left( \pi - \beta \right)}$ with the +ve direction of the y-axis
    ♦ $\small{\left( \pi - \gamma \right)}$ with the +ve direction of the z-axis
• This can be easily proved if L lies in a two dimensional plane. In fig.27.1(a) below, $\small{\alpha~\text{and}~\beta}$ and are shown in white and green colors.

Fig.27.1

In fig.27.1(b), $\small{\alpha~\text{and}~\beta}$ are shown twice, by considering opposite angles. Based on those opposite angles, we can write:
• L' makes the following angles with the axes:
    ♦ $\small{\left( \pi - \alpha \right)}$ with the +ve direction of the x-axis
    ♦ $\small{\left( \pi - \beta \right)}$with the +ve direction of the y-axis
• So in the three dimensional space, $\small{\cos \left( \pi - \alpha \right),~\cos \left( \pi - \beta \right)~\text{and}~\cos \left( \pi - \gamma \right) }$ are the direction cosines of L'.
• We know that:
    ♦ $\small{\cos\left( \pi - \alpha \right)~=~-\cos \alpha}$
    ♦ $\small{\left(\cos \pi - \beta \right)~=~-\cos \beta}$
    ♦ $\small{\left(\cos \pi - \gamma \right)~=~-\cos \gamma}$
• That means, when a directed line is reversed, the direction cosines are also reversed.

3. If we are given a simple line in space, it can be extended in two opposite directions. So a simple line will have two sets of direction cosines.
• While doing problems, we must know which set to use. So we must specify the direction of the line.
• In other words, while doing problems in three dimensional geometry, we deal with directed lines only.

4. For a directed line, there will be a unique set of direction cosines. Those direction cosines are denoted by $\small{l,~m~\text{and}~n}$

5. Suppose that, the given line does not pass through the origin. Then we draw a line through the origin and parallel to the given line. The direction cosines of this new line will be same as those of the given line.

Relation between direction cosines of a line

This can be explained in 9 steps:
1. In fig.27.2 below, a line L is shown. This line does not pass through the origin. So we draw a line L₁ passing through the origin and parallel to L

Fig.27.2

2. Mark any convenient point P(x,y,z) on L₁
• Imagine a vector $\small{\vec{OP}}$ with initial point O and terminal point P. Since the coordinates of P are (x,y,z), we can write:
$\small{\vec{OP}~=~x\hat{i}+y\hat{j}+z\hat{k}}$

3. Based on the above component form, we can write:
    ♦ Projection of $\small{\vec{OP}}$ on the x-axis = x
    ♦ Projection of $\small{\vec{OP}}$ on the y-axis = y
    ♦ Projection of $\small{\vec{OP}}$ on the z-axis = z

4. The line L₁ makes an angle $\small{\alpha}$ with the +ve side of the x-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\alpha}$ with the +ve side of the x-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \alpha = x}$. Recall a similar fig., based on which, we discussed about vectors. (See fig.26.4 of the first section of the previous chapter 26)
• From this, we get:
$\small{\cos \alpha = \frac{x}{\left|\vec{OP} \right|} \Rightarrow l = \frac{x}{\left|\vec{OP} \right|}}$

5. The line L₁ makes an angle $\small{\beta}$ with the +ve side of the y-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\beta}$ with the +ve side of the y-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \beta = y}$.
• From this, we get:
$\small{\cos \beta= \frac{y}{\left|\vec{OP} \right|} \Rightarrow m = \frac{y}{\left|\vec{OP} \right|}}$
(See fig.26.5 of the first section of the previous chapter 26)

6. The line L₁ makes an angle $\small{\gamma}$ with the +ve side of the z-axis. So $\small{\vec{OP}}$ also makes the same angle $\small{\gamma}$ with the +ve side of the z-axis. Then we can write: $\small{\left|\vec{OP} \right| \cos \gamma = z}$.
• From this, we get:
$\small{\cos \gamma = \frac{z}{\left|\vec{OP} \right|} \Rightarrow n = \frac{z}{\left|\vec{OP} \right|}}$
(See fig.26.6 of the first section of the previous chapter 26)

7. In the above steps (4), (5) and (6), we made use of our knowledge about vectors. Thereby, we obtained the direction cosines of L₁.
• We can write:
To obtain the direction cosines of L₁, we require three items:
    ♦ Any convenient point P on L₁
    ♦ The coordinates of P
    ♦ The length OP

8. Our next task is to find $\small{\left|\vec{OP} \right|}$
• This is the magnitude of the vector $\small{\vec{OP}}$.
• So it is simply, the distance between O and P.
• We have the coordinates of both O and P. So the distance between them is $\small{\sqrt{x^2 + y^2 + z^2}}$
That is., $\small{\left|\vec{OP} \right| = \sqrt{x^2 + y^2 + z^2}}$

9. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l^2 + m^2 + n^2}    & {~=~}    &{\frac{x^2}{\left|\vec{OP} \right|^2}~+~\frac{y^2}{\left|\vec{OP} \right|^2}~+~\frac{z^2}{\left|\vec{OP} \right|^2}}
\\ {~\color{magenta}    2    }    &{}    &{}    & {~=~}    &{\frac{x^2 ~+~y^2~+~z^2}{\left|\vec{OP} \right|^2}}
\\ {~\color{magenta}    3    }    &{}    &{}    & {~=~}    &{\frac{x^2 ~+~y^2~+~z^2}{x^2 ~+~y^2~+~z^2}}
\\ {~\color{magenta}    4    }    &{}    &{}    & {~=~}    &{1}
\\ \end{array}}$

Direction Ratios

This can be explained in 6 steps:
1. Consider a line with direction cosines l, m and n
2. Suppose that, there exist three numbers a, b and c which satisfy the following condition:
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}}$
3. If the above condition is satisfied, then we say that:
a, b and c are direction ratios of the line.

4. In step (2), there are three fractions. Since they are equal, we can write:
$\small{\frac{l}{a}~=~\frac{m}{b}~=~\frac{n}{c}~=~k}$, where k is a real number.
• So we get:
$\small{l=ak,~m=bk~\text{and}~n=ck}$

5. Now we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l^2 + m^2 + n^2}    & {~=~}    &{a^2 k^2~+~b^2 k^2~+~c^2 k^2}
\\ {~\color{magenta}    2    }    &{\Rightarrow}    &{1}    & {~=~}    &{k^2(a^2~+~b^2~+~c^2)}
\\ {~\color{magenta}    3    }    &{\Rightarrow}    &{k^2}    & {~=~}    &{\frac{1}{a^2 ~+~b^2~+~c^2}}
\\ {~\color{magenta}    4    }    &{\Rightarrow}    &{k}    & {~=~}    &{\pm\frac{1}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ \end{array}}$

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

6. So based on step (4), we get the direction cosines:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l}    & {~=~}    &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ {~\color{magenta}    2    }    &{{}}    &{m}    & {~=~}    &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ {~\color{magenta}    3    }    &{{}}    &{n}    & {~=~}    &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}}
\\ \end{array}}$

• We must select the sign carefully. It depends on the direction of the line. This will become more clear when we do some solved examples

Solved example 27.1
If a line makes angles 90°, 60° and 30° with the positive direction of x, y and z axis respectively, find its direction cosines.
Solution:
1. The line makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$  

2. The line makes angle 60° with the positive direction of the y axis. So $\small{\beta = 60^{\circ}}$
• Then the direction cosine $\small{m = \cos 60 = \frac{1}{2}}$ 

3. The line makes angle 30° with the positive direction of the z axis. So $\small{\gamma = 30^{\circ}}$
• Then the direction cosine $\small{n = \cos 30 = \frac{\sqrt{3}}{2}}$

Solved example 27.2
If a line has direction ratios 2, −1, −2, determine its direction cosines.
Solution:
1. We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l}    & {~=~}    &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{2}{\sqrt{2^2 ~+~(-1)^2~+~(-2)^2}}~=~\pm\frac{2}{3}}
\\ {~\color{magenta}    2    }    &{{}}    &{m}    & {~=~}    &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-1}{3}}
\\ {~\color{magenta}    3    }    &{{}}    &{n}    & {~=~}    &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-2}{3}}
\\ \end{array}}$

2. So the two sets of direction cosines are:
$\small{\frac{2}{3},~\frac{-1}{3},~\frac{-2}{3}~\text{and}~\frac{-2}{3},~\frac{1}{3},~\frac{2}{3}}$

3. Depending on the direction of the directed line, we must choose the appropriate set.
• Let there be two points A and B on the line.
• If a person traveling from A to B has the direction cosines $\small{\frac{2}{3},~\frac{-1}{3},~\frac{-2}{3}}$, then a person traveling from B to A will have the direction cosines $\small{\frac{-2}{3},~\frac{1}{3},~\frac{2}{3}}$.  

Solved example 27.3
If a line has direction ratios −18, 12, −4, then what are its direction cosines.
Solution:
1. We have:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{l}    & {~=~}    &{\pm\frac{a}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-18}{\sqrt{(-18)^2 ~+~12^2~+~(-4)^2}}~=~\pm\frac{-18}{22}~=~\pm\frac{-9}{11}}
\\ {~\color{magenta}    2    }    &{{}}    &{m}    & {~=~}    &{\pm\frac{b}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{12}{22}~=~\pm\frac{6}{11}}
\\ {~\color{magenta}    3    }    &{{}}    &{n}    & {~=~}    &{\pm\frac{c}{\sqrt{a^2 ~+~b^2~+~c^2}}~=~\pm\frac{-4}{22}~=~\pm\frac{-2}{11}}
\\ \end{array}}$

2. So the two sets of direction cosines are:
$\small{\frac{-9}{11},~\frac{6}{11},~\frac{-2}{11}~\text{and}~\frac{9}{11},~\frac{-6}{11},~\frac{2}{11}}$

3. Depending on the direction of the directed line, we must choose the appropriate set.
• Let there be two points A and B on the line.
• If a person traveling from A to B has the direction cosines $\small{\frac{-9}{11},~\frac{6}{11},~\frac{-2}{11}}$, then a person traveling from B to A will have the direction cosines $\small{\frac{9}{11},~\frac{-6}{11},~\frac{2}{11}}$.

Direction cosines of a line passing through two points

This can be explained in 9 steps:
1. In fig.27.3 below, $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ are two points in space.

Fig.27.3

• There can be one and only one line through both P and Q. That line is shown in magenta color. We want the direction cosines of this magenta line.
2. Imagine that, a vector $\small{\vec{PQ}}$ is present between P and Q. Based on our knowledge in vector algebra, we can write the component form of $\small{\vec{PQ}}$:
$\small{\vec{PQ}~=~\left(x_2 - x_1 \right)\hat{i}+\left(y_2 - y_1 \right)\hat{j}+\left(z_2 - z_1 \right)\hat{k}}$
3. Let $\small{\vec{PQ}}$ make angles $\small{\alpha,~\beta~\text{and}~\gamma}$ with the positive side of the x, y and z-axis respectively. Then the magenta line will also be making the same angles with the three axes.
4. $\small{\left(x_2 - x_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the x-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{x_2 - x_1}    & {~=~}    &{\left|\vec{PQ} \right|\cos \alpha}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

5. $\small{\left(y_2 - y_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the y-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y_2 - y_1}    & {~=~}    &{\left|\vec{PQ} \right|\cos \beta}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

6. $\small{\left(z_2 - z_1 \right)}$ is the projection of $\small{\vec{PQ}}$ on the z-axis. So we can write:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{z_2 - z_1}    & {~=~}    &{\left|\vec{PQ} \right|\cos \gamma}
\\ {~\color{magenta}    2    }    &{{\Rightarrow}}    &{\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\left|\vec{PQ} \right|}}
\\ \end{array}}$

7. So we obtained the three direction cosines of $\small{\vec{PQ}}$. Note that, $\small{\left|\vec{PQ} \right|}$ can be easily determined because, we have the coordinates of both P and Q. We get:
$\small{\left|\vec{PQ} \right| = \sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}$

8. The magenta line has the same direction cosines as that of $\small{\vec{PQ}}$. So we can compile the result.
• The direction cosines of a line through two points $\small{P(x_1,x_2,x_3)~\text{and}~Q(x_1,x_2,x_3)}$ are:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    3    }    &{}    &{\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ \end{array}}$

9. The direction cosines written in (8) above, are the ones for a person traveling from P to Q. If he travels from Q to P, he must use the reversed direction cosines:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \alpha}    & {~=~}    &{\frac{x_1 - x_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    2    }    &{}    &{\cos \beta}    & {~=~}    &{\frac{y_1 - y_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ {~\color{magenta}    3    }    &{}    &{\cos \gamma}    & {~=~}    &{\frac{z_1 - z_2}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}}
\\ \end{array}}$
   

Solved example 27.4
Find the direction cosines of the line passing through the two points (−2,4,−5) and (1,2,3).
Solution:
1. Let the two points be: P(−2,4,−5) and Q(1,2,3).
2. So for a person traveling in the direction P to Q, the direction cosines are:
$\small{\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\cos \alpha}    & {~=~}    &{\frac{x_2 - x_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{1-(-2)}{\sqrt{\left(1-(-2) \right)^2 + \left(2 - 4 \right)^2 + \left(3 - (-5) \right)^2}}~=~\frac{3}{\sqrt{\left(3 \right)^2 + \left(-2 \right)^2 + \left(8 \right)^2}}~=~\frac{3}{\sqrt{77}}}
\\ {~\color{magenta}    2    }    &{}    &{\cos \beta}    & {~=~}    &{\frac{y_2 - y_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{2-4}{\sqrt{77}}~=~\frac{-2}{\sqrt{77}}}
\\ {~\color{magenta}    3    }    &{}    &{\cos \gamma}    & {~=~}    &{\frac{z_2 - z_1}{\sqrt{\left(x_2 - x_1 \right)^2 + \left(y_2 - y_1 \right)^2 + \left( z_2 - z_1 \right)^2}}~=~\frac{3-(-5)}{\sqrt{77}}~=~\frac{8}{\sqrt{77}}}
\\ \end{array}}$

3. Also, for a person traveling in the direction Q to P, the direction cosines are:
$\small{\frac{-3}{\sqrt{77}},~\frac{2}{\sqrt{77}},~\frac{-8}{\sqrt{77}}}$

Solved example 27.5
Find the direction cosines of x, y and z-axis.
Solution:
Part (a): Direction cosines of x-axis
1. The x-axis makes angle 0° with the positive direction of the x axis. So $\small{\alpha = 0^{\circ}}$
• Then the direction cosine $\small{l = \cos 0 = 1}$ 
2. The x-axis makes angle 90° with the positive direction of the y axis. So $\small{\beta = 90^{\circ}}$
• Then the direction cosine $\small{m = \cos 90 = 0}$
3. The x-axis makes angle 90° with the positive direction of the z axis. So $\small{\gamma = 90^{\circ}}$
• Then the direction cosine $\small{n = \cos 90 = 0}$
4. So the direction cosines of the x-axis are: 1,0,0

Part (b): Direction cosines of y-axis
1. The y-axis makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$ 
2. The y-axis makes angle 0° with the positive direction of the y axis. So $\small{\beta = 0^{\circ}}$
• Then the direction cosine $\small{m = \cos 0 = 1}$
3. The y-axis makes angle 90° with the positive direction of the z axis. So $\small{\gamma = 90^{\circ}}$
• Then the direction cosine $\small{n = \cos 90 = 0}$
4. So the direction cosines of the y-axis are: 0,1,0

Part (b): Direction cosines of z-axis
1. The z-axis makes angle 90° with the positive direction of the x axis. So $\small{\alpha = 90^{\circ}}$
• Then the direction cosine $\small{l = \cos 90 = 0}$ 
2. The z-axis makes angle 90° with the positive direction of the y axis. So $\small{\beta = 90^{\circ}}$
• Then the direction cosine $\small{m = \cos 90 = 0}$
3. The y-axis makes angle 0° with the positive direction of the z axis. So $\small{\gamma = 0^{\circ}}$
• Then the direction cosine $\small{n = \cos 0 = 1}$
4. So the direction cosines of the z-axis are: 0,0,1

---

In the next section, we will see more details about direction ratios.

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Chapter 10.10 - Miscellaneous Examples

In the previous section, we saw the distance of a point from a line. In this section, we will see some miscellaneous examples.

Solved example 10.20
If the lines 2x+y-3=0, 5x+ky-3=0 and 3x-y-2=0 are concurrent, find the value of k.
Solution:
1. Three lines are said to be concurrent if all three of them pass through a common point.
• In such a situation,
    ♦ We can consider the point of intersection of any two of those lines.
    ♦ Let this point of intersection be P.
    ♦ Then the third line will also pass through P.
2. In our present case,
    ♦ The first and third lines do not have any unknown quantities.
    ♦ The second line has an unknown quantity 'k'.
• So we will solve the first and third equations. The solution will give the point of intersection of the first and third lines.
3. We first multiply (I) by 3. We get: 6x+3y-9=0 - - - (IV)
• Then we multiply (III) by 2. We get: 6x-2y-4=0 - - - (V)
• Subtracting (V) from (IV), we get: y = 1
• Substituting y = 1 in (I), we get: x = 1
4. So the point of intersection of (I) and (III) is (1,1)
5. Line (II) also passes through (1,1).
• So we can write: 5 × 1 + k × 1 - 3 = 0
• From this we get: k = -2
6. The actual plot is shown in fig.10.42 below:

Fig.10.42

 

Solved example 10.21
Find the distance of the line 4x-y=0 from the point P(4,1) measured along the line making an angle of 135^o with the +ve x-axis.
Solution:
1. Consider a line which makes 135^o with the +ve side of the x-axis.
• We can draw infinite number of such lines. But only one of them will pass through the point P(4,1). This line is shown in red color in the rough sketch below:

Fig.10.43 Rough sketch

2. The given line 4x-y=0 is shown in yellow color.
• Consider this line 4x-y=0
    ♦ Here, A = 4, B = -1 and C = 0
    ♦ So slope = -A/B = 4
        ✰ Since slope is +ve, the line has an upward slope
        ✰ Since C = 0, the line passes through the origin
• Even when we draw rough sketches, it is better to take care of such details.
3. Any two non-parallel lines will intersect at a point.
• In our case, the red and yellow lines intersect at Q.
4. We are required to find the length PQ.
5. For that, first we find the equation of the red line.
• The slope of the red line will be tan 135^o = -1
• So we have the slope of the red line. We also have a point on the red line. We can use the point-slope form:

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

{\Rightarrow}&{y-1}
&{}={}& {-1(x-4)}
&{} \\

{\Rightarrow}&{y-1}
&{}={}& {-x+4}
&{} \\

{\Rightarrow}&{x+y-5}
&{}={}& {0}
&{} \\

\end{array}$

6. Point Q can be determined by solving the equations of the two lines:
4x-y=0 - - - (I)
x+y-5=0 - - - (II)
7. Solving the two equations, we get: x = 1 and y = 4
• So the coordinates of Q are (1,4)
8, Now we have the coordinates of both P and Q.
    ♦ Coordinates of P are (4,1)
    ♦ Coordinates of Q are (1,4)
• We can use the distance formula:

$\begin{array}{ll}
{}&{d}
&{}={}& {\sqrt{(x_2 - x_1)^2~+~(y_2 - y_1)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{(1 - 4)^2~+~(4 - 1)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{(- 3)^2~+~(3)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{9~+~9}}
&{} \\

{}&{}
&{}={}& {\sqrt{18}}
&{} \\

{}&{}
&{}={}& {3\sqrt{2}~=~4.24~\text{units}}
&{} \\

\end{array}$

9. The actual plot is shown below:

Fig.10.44

 

Solved example 10.22
Assuming that straight lines work as the plane mirror for a point, find the image of the point (1,2) in the line x-3y+4=0
Solution:
1. In the rough sketch below, the line x-3y+4=0 is shown in red color. This line is to be considered as the mirror line.
• We want the image of the point P(1,2)

Fig.10.45 Rough sketch

2. Let Q(h,k) be the image of P.
• Then PQ will be perpendicular to the mirror line. Using this property, we can find the equation of PQ:
(i) For the mirror line, A = 1, B = -3 and C = 4
• So slope = -A/B = 1/3
• Then slope of the perpendicular line PQ = -3
(ii) We have the point P(1,2) and the slope of PQ. So we can use the point-slope form:

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

{\Rightarrow}&{y-2}
&{}={}& {-3(x-1)}
&{} \\

{\Rightarrow}&{y-2}
&{}={}& {-3x+3}
&{} \\

{\Rightarrow}&{3x+y-5}
&{}={}& {0}
&{} \\

\end{array}$

3. Let PQ intersect the mirror line at R.
• We can find the coordinates of R by solving the two equations:
x-3y+4=0 - - - (I)
    ♦ This is the equation of the mirror line.
3x+y-5=0 - - - (II)
    ♦ This is the equation of PQ
• Solving the two equations, we get: x = 11/10 and y = 17/10
• So the coordinates of R are (11/10, 17/10)
4. Since Q(h,k) is the image of P(1,2), the point R will be the midpoint of PQ.
We can find the coordinates of the midpoint as follows:

$\begin{array}{ll}
{}&{\text{Midpoint}}
&{}={}& {\left( \frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2} \right)}
&{} \\

{}&{}
&{}={}& {\left( \frac{1 + h}{2},\frac{2 + k}{2} \right)}
&{} \\

\end{array}$

5. Using the results in (3) and (4), we can write:
(i) (1+h)/2 = 11/10
    ♦ So h = 6/5
(ii) (2+k)/2 = 17/10
    ♦ So k = 7/5
6. So the image of P(1,2) is Q(6/5, 7/5) = Q(1.2, 1.4)
• The actual plot is shown below:

Fig.10.46

 

Solved example 10.23
Show that the area of the triangle formed by the lines y= m₁x+c₁, y=m₂x+c₂ and x = 0 is $\frac{\left(c_1 - c_2 \right)^2}{2|m_1 - m_2|}$
Solution:
1. The equation x=0 is the equation of the y-axis.
• So the triangle is formed between the two lines and the y-axis. This is shown in the rough sketch below:

Fig.10.47 Rough sketch

• The line y= m₁x+c₁ is shown in green color. It intersects the y-axis at P.
• The line y= m₂x+c₂ is shown in red color. It intersects the y-axis at Q.
• The two lines intersect at R.
• We need to find the area of the triangle PQR.
2. The given lines are in slope-intercept form. So we can easily write the intercepts that the lines make with the y-axis.
• We can write:
    ♦ The y-intercept of the line y= m₁x+c₁ is c₁. So the coordinates of P are (0,c₁) 
    ♦ The y-intercept of the line y= m₂x+c₂ is c₂. So the coordinates of Q are (0,c₂) 
3. Our next task is to find the coordinates of R. This can be done by solving the two equations. It is shown below:

$\begin{array}{ll}
{}&{m_1 x}
& {~-~y}& {~=~}
&{~-~}&{c_1}&{\color{green}{\text{ - - - (I)}}} \\

{}&{m_2 x}
& {~-~y}& {~=~}
&{~-~}&{c_2}&{\color{green}{\text{ - - - (II)}}} \\

{}&{m_1 x - m_2 x}
& {~-~y + y}& {~=~}
&{~-~}&{c_1 + c_2}&{\color{green}{\text{ - - - I - II}}} \\

{\Rightarrow}&{(m_1 - m_2) x}
& {~-~0}& {~=~}
&{~-~}&{c_1 + c_2}&{} \\

{\Rightarrow}&{}
& {x}& {~=~}
&{}&{\frac{c_2 - c_1}{m_1 - m_2}}&{} \\

{}&{m_1 \times \frac{c_2 - c_1}{m_1 - m_2}}
& {~-~y}& {~=~}
&{~-~}&{c_1}&{\color{green}{\text{substituting for x in I}}} \\

{\Rightarrow}&{}
& {~-~y}& {~=~}
&{~-~}&{c_1~-~m_1 \times \frac{c_2 - c_1}{m_1 - m_2}}&{} \\

{\Rightarrow}&{}
& {~+~y}& {~=~}
&{~+~}&{C_1~+~m_1 \times \frac{c_2 - c_1}{m_1 - m_2}}&{} \\

{\Rightarrow}&{}
& {y}& {~=~}
&{}&{\frac{c_1(m_1 - m_2)~+~m_1(c_2 - c_1)}{m_1 - m_2}}&{} \\

{\Rightarrow}&{}
& {y}& {~=~}
&{}&{\frac{c_1 m_1 - c_1 m_2~+~m_1 c_2 - m_1 c_1}{m_1 - m_2}}&{} \\

{\Rightarrow}&{}
& {y}& {~=~}
&{}&{\frac{m_1 c_2~-~c_1 m_2}{m_1 - m_2}}&{} \\

{\Rightarrow}&{}
& {y}& {~=~}
&{}&{\frac{m_1 c_2~-~m_2 c_1}{m_1 - m_2}}&{} \\

\end{array}$

• Thus we get the coordinates of R: $\left(\frac{c_2 - c_1}{m_1 - m_2},~\frac{m_1 c_2~-~m_2 c_1}{m_1 - m_2} \right)$

4. So the vertices of the triangle are:
• $P (0,~c_1)$
• $Q (0,~c_2)$
• $R \left(\frac{c_2 - c_1}{m_1 - m_2},~\frac{m_1 c_2~-~m_2 c_1}{m_1 - m_2} \right)$

5. Now we can find the area of the triangle:

$\begin{array}{ll}
{}&{\text{Area}}
&{}={}& {\frac{1}{2} \left|x_1(y_3 - y_2)~+~x_2(y_1 - y_3)~+~x_3(y_2 - y_1) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2} \left|0 \times (y_3 - y_2)~+~0 \times (y_1 - y_3)~+~x_3(y_2 - y_1) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2} \left|x_3(y_2 - y_1) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2} \left|\frac{c_2 - c_1}{m_1 - m_2} \times (c_2 - c_1) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2} \left|\frac{(c_2 - c_1)^2}{m_1 - m_2} \right|}
&{} \\

{}&{}
&{}={}& {\frac{(c_2 - c_1)^2}{\left|2(m_1 - m_2) \right|}}
&{} \\

\end{array}$  

Solved example 10.24
A line is such that it's segment between the lines 5x-y+4=0 and 3x+4y-4=0 is bisected at the point (1,5). Obtain it's equation.
Solution:
1. Consider the rough sketch below:

Fig.10.48 Rough sketch

• The given lines are shown in red and green colors.
    ♦ The red line 5x-y+4=0 intersects the yellow line at P
    ♦ The green line 3x+4y-4=0 intersects the yellow line at Q
• Midpoint of PQ is R(1,5)
• We are asked to find the equation of the yellow line.
2. Let 'm' be the slope of the yellow line.
• The yellow line passes through (1,5)
• So the equation of the yellow line will be y-5 = m(x-1)
• This is same as y-5 = mx - m
• This is same as mx - y = m - 5
3. The coordinates of P can be calculated by solving the equations of the red line and the yellow line.

$\begin{array}{ll}
{}&{m x}
&{~-~}& {y}& {~=~}
&{m-5}&{}&{}&{\color{green}{\text{ - - - (I)}}} \\

{}&{5 x}
&{~-~}& {y}& {~=~}
&{-4}&{}&{}&{\color{green}{\text{ - - - (II)}}} \\

{}&{mx - 5 x}
&{~-~}& {y + y}& {~=~}
&{m - 5 + 4}&{}&{}&{\color{green}{\text{ - - - (I) - (II)}}} \\

{\Rightarrow}&{mx - 5 x}
&{~-~}& {0}& {~=~}
&{m - 1}&{}&{}&{} \\

{\Rightarrow}&{(m - 5) x}
&{}& {}& {~=~}
&{m - 1}&{}&{}&{} \\

{\Rightarrow}&{x}
&{}& {}& {~=~}
&{\frac{m-1}{m-5}}&{}&{}&{} \\

\end{array}$

• We can write: x coordinate of P is: $\frac{m-1}{m-5}$
• In this problem, we do not need to find the y coordinate of P.

4. The coordinates of Q can be calculated by solving the equations of the green line and the yellow line.

$\begin{array}{ll}
{}&{m x}
&{~-~}& {y}& {~=~}
&{m-5}&{}&{}&{\color{green}{\text{ - - - (I)}}} \\

{}&{3x}
&{~+~}& {4y}& {~=~}
&{4}&{}&{}&{\color{green}{\text{ - - - (II)}}} \\

{}&{4mx}
&{~-~}& {4y}& {~=~}
&{4m - 20}&{}&{}&{\color{green}{\text{ - - - [(I) × 4] - - -(III)}}} \\

{}&{4mx + 3x}
&{~-~}& {4y + 4y}& {~=~}
&{4m - 20 + 4}&{}&{}&{\color{green}{\text{ - - - (III) + (II)}}} \\

{\Rightarrow}&{4mx + 3x}
&{~-~}& {0}& {~=~}
&{4m - 16}&{}&{}&{} \\

{\Rightarrow}&{(4m + 3) x}
&{}& {}& {~=~}
&{4m - 16}&{}&{}&{} \\

{\Rightarrow}&{x}
&{}& {}& {~=~}
&{\frac{4m-16}{4m+3}}&{}&{}&{} \\

\end{array}$

• We can write: x coordinate of Q is: $\frac{4m-16}{4m+3}$
• In this problem, we do not need to find the y coordinate of Q.

5. Now we have the required x coordinates:
    ♦ From (3) we  have the x coordinate of P.
    ♦ From (4) we  have the x coordinate of Q.
• The average of these two quantities will be the x coordinate of R (Recall the mid-point formula)
• So we can write:

$\begin{array}{ll}
{}&{\text{x coordinate of R}}
&{}={}& {\frac{\frac{m-1}{m-5}~+~\frac{4m-16}{4m+3}}{2}}
&{} \\

{}&{}
&{}={}& {\frac{\frac{(m-1)(4m+3)~+~(4m-16)(m-5)}{(m-5)(4m+3)}}{2}}
&{} \\

{}&{}
&{}={}& {\frac{(m-1)(4m+3)~+~(4m-16)(m-5)}{2(m-5)(4m+3)}}
&{} \\

{}&{}
&{}={}& {\frac{(4m^2 + 3m - 4m - 3)~+~(4m^2 - 20m - 16m + 80)}{2(4m^2 + 3m - 20m - 15)}}
&{} \\

{}&{}
&{}={}& {\frac{(4m^2  - m - 3)~+~(4m^2  - 36m + 80)}{2(4m^2  - 17m - 15)}}
&{} \\

{}&{}
&{}={}& {\frac{8m^2 - 37m + 77}{8m^2  - 34m - 30}}
&{} \\

\end{array}$

6. Thus the x coordinate of R is $\frac{8m^2 - 37m + 77}{8m^2  - 34m - 30}$.
• But x coordinate of R is given as 1. So we can equate them:

$\begin{array}{ll}
{}&{\frac{8m^2 - 37m + 77}{8m^2  - 34m - 30}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{8m^2 - 37m + 77}
&{}={}& {8m^2  - 34m - 30}
&{} \\

{\Rightarrow}&{- 37m + 77}
&{}={}& {- 34m - 30}
&{} \\

{\Rightarrow}&{77 + 30}
&{}={}& {37m - 34m}
&{} \\

{\Rightarrow}&{107}
&{}={}& {3m}
&{} \\

{\Rightarrow}&{m}
&{}={}& {\frac{107}{3}}
&{} \\

\end{array}$

7. We have obtained the value of 'm'. So based on the result in (2), we can obtain the required equation:

$\begin{array}{ll}
{}&{mx - y}
&{}={}& {m-5}
&{} \\

{\Rightarrow}&{\left( \frac{107}{3} \times x \right) - y}
&{}={}& {\frac{107}{3} - 5}
&{} \\

{\Rightarrow}&{107x - 3y}
&{}={}& {107 - 15}
&{} \\

{\Rightarrow}&{107x - 3y -92}
&{}={}& {0}
&{} \\

\end{array}$

8. The actual plot is shown below:

Fig.10.49

Solved example 10.25
Show that the path of a moving point such that it's distances from two lines 3x-2y=5 and 3x+2y=5 are equal is a straight line.
Solution:
1. In the rough sketch below,
• The yellow line is the path of a moving point.
• The line 3x-2y-5=0 is shown in red color.
• The line 3x+2y-5=0 is shown in green color.

Fig.10.50 Rough sketch

2. Let us consider the distances:
• When the point is at A(x₁,y₁),
    ♦ it's distance from the red line is AA'.
    ♦ it's distance from the green line is AA''.
        ✰ We are give that, AA' = AA''.
• When the point is at B(x₂,y₂),
    ♦ it's distance from the red line is BB'.
    ♦ it's distance from the green line is BB''.
        ✰ We are give that, BB' = BB''
3. Which ever be the point on the yellow line, the distances from the red and green lines will be equal.
• So let us consider a general point P(x,y)
◼ Distance of P from the red line is PP'. It can be calculated as follows:

$\begin{array}{ll}
{}&{PP'}
&{}={}& {\frac{\left|Ax_1 + B x_1 + C \right|}{\sqrt{A^2 + B^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|3x - 2y - 5 \right|}{\sqrt{3^2 + 2^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|3x - 2y - 5 \right|}{\sqrt{13}}}
&{} \\

\end{array}$

◼ Distance of P from the green line is PP''. It can be calculated as follows:

$\begin{array}{ll}
{}&{PP''}
&{}={}& {\frac{\left|Ax_1 + B x_1 + C \right|}{\sqrt{A^2 + B^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|3x + 2y - 5 \right|}{\sqrt{3^2 + 2^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|3x + 2y - 5 \right|}{\sqrt{13}}}
&{} \\

\end{array}$

4. The two distances PP' and PP'' will be equal. So we can write:

$\begin{array}{ll}
{}&{PP'}
&{}={}& {PP''}
&{} \\

{\Rightarrow}&{\frac{\left|3x - 2y - 5 \right|}{\sqrt{13}}}
&{}={}& {\frac{\left|3x + 2y - 5 \right|}{\sqrt{13}}}
&{} \\

{\Rightarrow}&{\left|3x - 2y - 5 \right|}
&{}={}& {\left|3x + 2y - 5 \right|}
&{} \\

\end{array}$

5. The above result gives us two possibilities:
(i) 3x - 2y - 5 = 3x + 2y - 5
(ii) 3x - 2y - 5 = -(3x + 2y - 5)
• From (i) we get: 4y = 0, which gives y = 0
• From (ii) we get: 6x = 10, which gives x = 5/3
6. So there are two possibilities for the yellow line. We can write:
• If a point moves along the line y = 0, it's distances from two lines 3x-2y=5 and 3x+2y=5 will be equal.
    ♦ Note that, y = 0 is the x-axis.
• If a point moves along the line x = 5/3, it's distances from two lines 3x-2y=5 and 3x+2y=5 will be equal.
    ♦ Note that, x = 5/3 is a vertical line.
7. The actual plot is shown below:

Fig.10.51

---

Link to a few more solved examples is given below:

Miscellaneous Exercise on Chapter 10 - Part I

Miscellaneous Exercise on Chapter 10 - Part II

---

In the next chapter, we will see conic sections.

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Chapter 10.9 - Distance of a Point from a Line

In the previous section, we saw the details about normal form. In this section, we will see distance of a point from a line. Later in this section we will see distance between two parallel lines also.

• We can derive an expression for the distance in 9 steps:
1. In fig.10.38 below, equation of line L is Ax + By + C = 0

Fig.10.38

2. This line intersects the x-axis at Q.
• We know that, the x-intercept will be $-\frac{C}{A}$. So the coordinates of Q will be: $\left(-\frac{C}{A},0 \right)$
3. This line intersects the y-axis at R.
• We know that, the y-intercept will be $-\frac{C}{B}$. So the coordinates of R will be $\left(0,-\frac{C}{B} \right)$
4. P(x₁,y₁) is a point on the xy-plane.
• A perpendicular is dropped from P onto the line L
    ♦ Foot of the perpendicular is M.
    ♦ The length of the line segment PM is d.
    ♦ d is called the distance of the point P from line L.
5. It is possible to draw infinite lines from P onto line L.
• But it is possible to draw only one perpendicular.
• The length of that perpendicular line is called the distance of the point from L.
• The other lengths are not eligible to be considered as the distance of the point from L.
6. Our aim is to derive an expression for d
• Consider the triangle PQR. We have the coordinates of all three vertices:
$P(x_1,y_1),~Q\left(-\frac{C}{A},0 \right),~R\left(0,-\frac{C}{B} \right)$
• So we can find the area of that triangle. We get:

$\begin{array}{ll}
{}&{Area}
&{}={}& {\frac{1}{2}\left|x_1(y_2 - y_3)~+~x_2(y_3 - y_1)~+~x_3(y_1 - y_2) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2}\left|x_1 \left(0~-~-\frac{C}{B}\right)~+~-\frac{C}{A} × \left(-\frac{C}{B}~-~y_1 \right)~+~0 × \left(y_1~-~0\right) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2}\left|\frac{C x_1}{B}~+~\frac{C^2}{AB}~+~\frac{C y_1}{A} \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2}\left| \frac{C}{AB}\left(\frac{C x_1}{B} × \frac{AB}{C}~+~\frac{C^2}{AB} × \frac{AB}{C}~+~\frac{C y_1}{A} × \frac{AB}{C}\right) \right|}
&{\color{green}{\text{- - - (a)}}} \\

{}&{}
&{}={}& {\frac{1}{2}\left| \frac{C}{AB}\left(\frac{x_1}{1} × \frac{A}{1}~+~C~+~\frac{y_1}{1} × \frac{B}{1}\right) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2}\left| \frac{C}{AB}\left(A x_1~+~B y_1~+~C\right) \right|}
&{} \\

{}&{}
&{}={}& {\frac{1}{2} × \left| \frac{C}{AB}\right| × \left|\left(A x_1~+~B y_1~+~C\right) \right|}
&{} \\

\end{array}$

◼ Remarks:
Line marked as (a):
In this line, we multiply each term by $\frac{C}{AB}$ and $\frac{AB}{C}$.

7. We know that, area of any triangle = 1/2 × Base × Altitude
• In our present case,
    ♦ Base of triangle PQR is QR
    ♦ Altitude of triangle PQR is d
• We know the coordinates of both Q and R:
$Q\left(-\frac{C}{A},0 \right),~R\left(0,-\frac{C}{B} \right)$
• So the length QR can be obtained as:

$\begin{array}{ll}
{}&{QR}
&{}={}& {\sqrt{(x_2 - x_1)^2~+~(y_2 - y_1)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{\left(0~-~ -\frac{C}{A}\right)^2~+~\left(-\frac{C}{B} - 0 \right)^2}}
&{} \\

{}&{}
&{}={}& {\sqrt{\frac{C^2}{A^2}~+~\frac{C^2}{B^2}}}
&{} \\

{}&{}
&{}={}& {\sqrt{ \frac{C^2}{A^2 B^2} \left(\frac{C^2}{A^2} × \frac{A^2 B^2}{C^2}~+~\frac{C^2}{B^2} × \frac{A^2 B^2}{C^2}\right)}}
&{\color{green}{\text{- - - (a)}}} \\

{}&{}
&{}={}& {\sqrt{ \frac{C^2}{A^2 B^2} \left(B^2~+~A^2\right)}}
&{} \\

{}&{}
&{}={}& {\sqrt{ \frac{C^2}{A^2 B^2}} × \sqrt{\left(B^2~+~A^2\right)}}
&{} \\

{}&{}
&{}={}& {\left| \frac{C}{AB} \right| × \sqrt{\left(A^2~+~B^2\right)}}
&{} \\

\end{array}$

◼ Remarks:
Line marked as (a):
In this line, we multiply each term by $\frac{C^2}{A^2 B^2}$ and $\frac{A^2 B^2}{C^2}$.

• Now we can obtain the area of the triangle PQR:

$\begin{array}{ll}
{}&{Area}
&{}={}& {\frac{1}{2} × \text{Base} × \text{Altitude}}
&{} \\

{}&{}
&{}={}& {\frac{1}{2} × QR ×d}
&{} \\

{}&{}
&{}={}& {\frac{1}{2} × \left| \frac{C}{AB} \right| × \sqrt{\left(A^2~+~B^2\right)} × d}
&{} \\

\end{array}$

8. Equating the areas:
    ♦ In (6), we obtained the area of triangle PQR.
    ♦ In (7) also, we obtained the area of triangle PQR.
• We can equate the two areas:

$\begin{array}{ll}
{}&{\frac{1}{2} × \left| \frac{C}{AB}\right| × \left|\left(A x_1~+~B y_1~+~C\right) \right|}
&{}={}& {\frac{1}{2} × \left| \frac{C}{AB} \right| × \sqrt{\left(A^2~+~B^2\right)} × d}
&{} \\

{\Rightarrow}&{\left|\left(A x_1~+~B y_1~+~C\right) \right|}
&{}={}& {\sqrt{\left(A^2~+~B^2\right)} × d}
&{} \\

{\Rightarrow}&{d}
&{}={}& {\frac{\left|\left(A x_1~+~B y_1~+~C\right) \right|}{\sqrt{\left(A^2~+~B^2\right)}}}
&{} \\

\end{array}$

9. Thus we get a formula to find the distance.
$d~=~\frac{\left|A x_1~+~B y_1~+~C \right|}{\sqrt{A^2~+~B^2}}$

---

Distance between two parallel lines

• We have derived a formula for the distance of a point from a line. Using this formula, we can derive another formula for the distance between two parallel lines. It can be derived in 5 steps:
1. In fig.10.39 below,
    ♦ equation of line L₁ is A₁x + B₁y + C₁ = 0
    ♦ equation of line L₂ is A₂x + B₂y + C₂ = 0

Fig.10.39

2. Line L₁ intersects the x-axis at P
• Then the coordinates of P will be $\left(-\frac{C_1}{A_1},0 \right)$
3. A perpendicular line is dropped from P onto the line L₂
• Length of this perpendicular line is d
• Then we can write:
The distance between the two parallel lines is d.
4. We can use the formula derived earlier to find d.
• For applying the formula, we take P as the point and L₂ as the line. Thus we get:

$\begin{array}{ll}
{}&{d}
&{}={}& {\frac{\left|A x_1~+~B y_1~+~C \right|}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|A_2 × -\frac{C_1}{A_1}~+~B_2  × 0~+~C_2 \right|}{\sqrt{A_2^2~+~B_2^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|A_2 × -\frac{C_1}{A_1}~+~C_2 \right|}{\sqrt{A_2^2~+~B_2^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left| \frac{-A_2 C_1~+~A_1 C_2}{A_1}\right|}{\sqrt{A_2^2~+~B_2^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left| \frac{A_1 C_2~-~A_2 C_1}{A_1}\right|}{\sqrt{A_2^2~+~B_2^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\frac{\left|A_1 C_2~-~A_2 C_1 \right|}{\left|A_1\right|}}{\sqrt{A_2^2~+~B_2^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|A_1 C_2~-~A_2 C_1 \right|}{\left|A_1\right| × \sqrt{A_2^2~+~B_2^2}}}
&{} \\

\end{array}$

5. Thus we get a formula to find the distance between two parallel lines.
$d~=~\frac{\left|A_1 C_2~-~A_2 C_1 \right|}{\left|A_1\right| × \sqrt{A_2^2~+~B_2^2}}$

---

Now we will see two solved examples

Solved example 10.18
Find the distance of the point (3,-5) from the line 3x - 4y - 26 = 0
Solution:
1. The given line is: 3x - 4y - 26 = 0
• So we get: A = 3, B = -4 and C = -26
2. We have: $d~=~\frac{\left|A x_1~+~B y_1~+~C \right|}{\sqrt{A^2~+~B^2}}$

• Substituting the known values, we get:

$\begin{array}{ll}
{}&{d}
&{}={}& {\frac{\left|3 × 3~+~-4 × -5~+~-26 \right|}{\sqrt{3^2~+~(-4)^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|9~+~20~+~-26 \right|}{\sqrt{9~+~16}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|3 \right|}{\sqrt{25}}}
&{} \\

{}&{}
&{}={}& {\frac{3}{5}~=~0.6~\text{units}}
&{} \\

\end{array}$

3. The actual plot is shown below:

Fig.10.40

 

Solved example 10.19
Find the distance between the parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 5 = 0
Solution:
1. Let the line L₁ be: 3x - 4y + 7 = 0
• Then we get: A₁ = 3, B₁ = -4 and C₁ = 7
2. Let the line L₂ be: 3x - 4y + 5 = 0
• Then we get: A₂ = 3, B₂ = -4 and C₂ = 5
3. We have: $d~=~\frac{\left|A_1 C_2~-~A_2 C_1 \right|}{\left|A_1\right| × \sqrt{A_2^2~+~B_2^2}}$

• Substituting the known values, we get:

$\begin{array}{ll}
{}&{d}
&{}={}& {\frac{\left|3 × 5~-~3 × 7 \right|}{\left|3 \right| × \sqrt{3^2~+~(-4)^2}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|15~-~21 \right|}{\left|3 \right| × \sqrt{9~+~16}}}
&{} \\

{}&{}
&{}={}& {\frac{\left|-6 \right|}{\left|3 \right| × 5}}
&{} \\

{}&{}
&{}={}& {\frac{6}{3× 5}}
&{} \\

{}&{}
&{}={}& {\frac{2}{5}~=~0.4~\text{units}}
&{} \\

\end{array}$

4. The actual plot is shown below:

Fig.10.41

---

Link to a few more solved examples is given below:

Exercise 10.3

---

In the next section, we will see some miscellaneous examples.

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Chapter 10.8 - More Details about Normal Form

In the previous section, we saw how slope, intercepts, 𝜔 and p can be obtained from the general equation of a line. We saw that, $\pm$ sign is present for p, cos 𝜔 and sin 𝜔. In this section, we will see how to apply those signs.

• We know that, sign of sin 𝜔, cos 𝜔 etc., will depend upon the position of the angle 𝜔.
• For example:
    ♦ If 𝜔 is in the I quadrant, sin 𝜔 will be +ve, cos 𝜔 will be +ve.
    ♦ If 𝜔 is in the III quadrant, sin 𝜔 will be -ve, cos 𝜔 will be +ve.
• There is an easy method to find the position of 𝜔. It involves the use of the flow chart in fig.10.34 below. It can be written in 5 steps:

Fig.10.34

1. We are given the equation of a line in the form Ax +By +C = 0.
• We want to find the signs of sin 𝜔, cos 𝜔 and p.
• The first step is to calculate the slope using the equation: $m=-\frac{A}{B}$
• If the slope is +ve, we move along the left arrow of the chart. All our works will then be in the left side of the vertical cyan line. 
• If the slope is -ve, we move along the right arrow of the chart. All our works will then be in the right side of the vertical cyan line.
2. Suppose that, the slope is +ve. Then we are in the left side of the cyan line.
• The second step is to calculate the x-intercept ‘a’ using the equation: $a=-\frac{C}{A}$
3. If ‘a’ is +ve, then we move along the left arrow.
• The y-intercept ‘b’ will be -ve.
• This condition is shown in fig.10.35(d) below. Note that in the fig.d, the line has a +ve slope, and x-intercept ‘a’ is +ve. Then the y-intercept will be -ve.

Fig.10.35

4. From the fig.d, it is clear that, if ‘a’ is +ve and ‘b’ is -ve, then the normal p will lie in the IV quadrant.
• Then sin will be -ve and cos will be +ve
• Thus we get the required signs.
5. With practice, we will not need to go through all the above steps. We can simply move along the appropriate arrows and write:
If m is +ve and a is +ve, then: sin is -ve and cos is +ve.
6. The sign of p will be always +ve because, it is a distance.

---

Let us see another case from the chart. It can be written in 5 steps:
1. Suppose that, the slope is -ve. Then we are in the right side of the cyan line.
• The second step is to calculate the x-intercept ‘a’ using the equation: $a=-\frac{C}{A}$
2. If ‘a’ is -ve, then we move along the right arrow.
• The y-intercept ‘b’ will be -ve.
• This condition is shown in fig.10.35(c) above. Note that in the fig.c, the line have a -ve slope, and x-intercept ‘a’ is -ve. Then the y-intercept will be -ve.
3. From the fig.c, it is clear that, if ‘a’ is -ve and ‘b’ is -ve, then the normal p will lie in the III quadrant.
• Then sin will be -ve and cos will be -ve
• Thus we get the required signs.
4. With practice, we will not need to go through all the above steps. We can simply move along the appropriate arrows and write:
If m is -ve and a is -ve, then: sin is -ve and cos is -ve.
5. The sign of p will be always +ve because, it is a distance.

---

Once we obtain a through knowledge on fig.10.34 and fig.10.35, we can avoid some intermediate steps and use a simplified chart. It is shown in fig.10.36 below:

Fig.10.36

---

Let us see an example where we use the simplified chart:

Equation of a line is 3x + 2y + 6 = 0. Reduce this equation to normal form. Find the values of p and 𝜔.
Solution:
1. From the given equation, we get:
A = 3, B = 2 and C = 6
2. Calculation of p

$\begin{array}{ll}
{}&{p}
&{}={}& {\pm \frac{C}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{6}{\sqrt{3^2~+~2^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{6}{\sqrt{13}}}
&{} \\

\end{array}$

2. Calculation of cos 𝜔

$\begin{array}{ll}
{}&{\cos \omega }
&{}={}& {\pm \frac{A}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{3}{\sqrt{3^2~+~2^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{3}{\sqrt{13}}}
&{} \\

\end{array}$

3. Calculation of sin 𝜔

$\begin{array}{ll}
{}&{\sin \omega }
&{}={}& {\pm \frac{B}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{2}{\sqrt{3^2~+~2^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{2}{\sqrt{13}}}
&{} \\

\end{array}$

4. Now we determine the signs:
(i) Slope of the line = $m=-\frac{A}{B}~=~-\frac{3}{2}$
• This is -ve. So we move along the right arrow of the chart in fig.13.36
(ii) x-intercept of the line = $a=-\frac{C}{A}~=~-\frac{6}{3}~=~-2$
• This is -ve. So sin 𝜔 is -ve and cos 𝜔 is -ve.

5. So we can write:

$\begin{array}{ll}
{}&{\sin \omega }
&{}={}& {- \frac{2}{\sqrt{13}}}
&{} \\

{}&{\cos \omega}
&{}={}& {- \frac{3}{\sqrt{13}}}
&{} \\

\end{array}$

6.  The sign of p will be always +ve because, it is a distance.

7. Now we can write the normal form:

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

{\Rightarrow}&{x × - \frac{3}{\sqrt{13}}~+~y × - \frac{2}{\sqrt{13}}}
&{}={}& {\frac{6}{\sqrt{13}}}
&{} \\

{\Rightarrow}&{- \frac{3x}{\sqrt{13}}~+~ - \frac{2y}{\sqrt{13}}}
&{}={}& {\frac{6}{\sqrt{13}}}
&{} \\

{\Rightarrow}&{\frac{3x}{\sqrt{13}}~+~ \frac{2y}{\sqrt{13}}}
&{}={}& {-\frac{6}{\sqrt{13}}}
&{} \\

\end{array}$

8. Calculation of 𝜔:
(i) From (5), we have: tan 𝜔 = 2/3
• Using a scientific calculator, we get: 𝜔 = 33.69^o
(ii) But in our present case, 𝜔 is in the third quadrant (both sin and cos are -ve).
• So we find the other value of 𝜔 using the identity: tan x = tan (180 + x)
• We get: tan 𝜔 = tan 33.69 = tan (180 + 33.69) = tan 213.69
(iii) So the value of 𝜔 is 213.69^o

9. So the normal form can be written as:
x cos 213.69^o + y sin 213.69^o = $\frac{6}{\sqrt{13}}$

• The actual plot is shown below:

Fig.10.37

 

• Another example can be seen here.

---

Now we will see some solved examples.

Solved example 10.13
The equation of a line is 3x - 4y + 10 = 0. Find it's (i) slope (ii) x - and y-intercepts.
Solution:
1. From the given equation, we can write:
A = 3, B = -4 and C = 10
2. Slope can be calculated as:
$m=-\frac{A}{B}~=~-\frac{3}{-4}~=~\frac{3}{4}$
3. x-intercept can be calculated as:
$a=-\frac{C}{A}~=~-\frac{10}{3}$
4. y-intercept can be calculated as:
$b=-\frac{C}{B}~=~-\frac{10}{-4}~=~\frac{5}{2}$

Solved example 10.14
Reduce the equation (√3)x + y - 8 = 0 into normal form. Find the values of p and 𝜔.
Solution:
1. From the given equation, we get:
A = √3, B = 1 and C = -8
2. Calculation of p

$\begin{array}{ll}
{}&{p}
&{}={}& {\pm \frac{C}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{-8}{\sqrt{(\sqrt{3})^2~+~1^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{8}{\sqrt{4}}}
&{} \\

{}&{}
&{}={}& {\pm 4}
&{} \\

\end{array}$

2. Calculation of cos 𝜔

$\begin{array}{ll}
{}&{\cos \omega }
&{}={}& {\pm \frac{A}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{\sqrt{3}}{\sqrt{(\sqrt{3})^2~+~1^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{\sqrt{3}}{\sqrt{4}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{\sqrt{3}}{2}}
&{} \\

\end{array}$

3. Calculation of sin 𝜔

$\begin{array}{ll}
{}&{\sin \omega }
&{}={}& {\pm \frac{B}{\sqrt{A^2~+~B^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{1}{\sqrt{(\sqrt{3})^2~+~1^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{1}{\sqrt{4}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{1}{2}}
&{} \\

\end{array}$

4. Now we determine the signs:
(i) Slope of the line = $m=-\frac{A}{B}~=~-\frac{\sqrt{3}}{1}~=~-\sqrt{3}$
• This is -ve. So we move along the right arrow of the chart in fig.13.36
(ii) x-intercept of the line = $a=-\frac{C}{A}~=~-\frac{-8}{\sqrt{3}}~=~\frac{8}{\sqrt{3}}$
• This is +ve. So sin 𝜔 is +ve and cos 𝜔 is +ve.

5. So we can write:

$\begin{array}{ll}
{}&{\sin \omega }
&{}={}& {\frac{1}{2}}
&{} \\

{}&{\cos \omega}
&{}={}& {\frac{\sqrt{3}}{2}}
&{} \\

\end{array}$

6.  The sign of p will be always +ve because, it is a distance.

7. Now we can write the normal form:

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

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

{\Rightarrow}&{\frac{\sqrt{3}x}{2}~+~\frac{y}{2}}
&{}={}& {4}
&{} \\

\end{array}$

8. Calculation of 𝜔:
(i) From (5), we have: $\tan \omega ~=~\frac{1}{\sqrt{3}}$
• We know that $\tan 30 ~=~\frac{1}{\sqrt{3}}$
(ii) In our present case, 𝜔 is in the first quadrant (both sin and cos are +ve).
• So 30^o is the correct value of 𝜔

9. So the normal form can be written as:
x cos cos 30^o + y sin 30^o = 4

Solved example 10.15
Find the angle between the lines y - (√3)x - 5 = 0 and (√3)y - x + 6 = 0
Solution:
1. The first line given is: y - (√3)x - 5 = 0
• Rearranging this into the general form, we get: (√3)x - y + 5 = 0
• So we can write:
A = √3, B = -1 and C = 5
• Thus we get: $m_1=-\frac{A}{B}~=~-\frac{\sqrt{3}}{-1}~=~\sqrt{3}$
2. The second line given is: (√3)y - x + 6 = 0
• Rearranging this into the general form, we get: x - (√3)y - 6 = 0
• So we can write:
A = 1, B = -√3 and C = -6
• Thus we get: $m_2=-\frac{A}{B}~=~-\frac{1}{-\sqrt{3}}~=~\frac{1}{\sqrt{3}}$

3. We have two slopes and we are asked to find the angle between the lines
   ♦ So this problem belongs to case I.
• We have seen the details about case I and case II here.
• Since this problem belongs to case I, there is no need to interchange the slopes and explore the two possibilities.
• 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}{\sqrt{3}}~-~\sqrt{3}}{1~+~\sqrt{3}  × \frac{1}{\sqrt{3}}}}
&{} \\

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

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{-2}{2 \sqrt{3}}}
&{} \\

{\Rightarrow}&{\tan \theta}
&{}={}& {\frac{-1}{\sqrt{3}}}
&{} \\

\end{array}$

4. So we have to solve the equation: tan 𝜃 = $-\frac{1}{\sqrt{3}}$
It can be solved in 5 steps:
(i) Given that, tan θ = $-\frac{1}{\sqrt{3}}$
(ii) We know that, tan 30 = $\frac{1}{\sqrt{3}}$
• Using the identities 9.d and 9.c, we have: tan (180 – θ) = - tan θ
(See the list of identities here)
• So we can write: tan (180 – 30) = -tan 30
• That means: tan 150 = -tan 30
(iii) But tan 30 = $\frac{1}{\sqrt{3}}$
• So the result in (ii) becomes:
tan 150 = -tan 30 = $-\frac{1}{\sqrt{3}}$
(iv) We are given that, tan θ = $-\frac{1}{\sqrt{3}}$
• Including this in (iii), we get:
tan 150 = -tan 30 = $-\frac{1}{\sqrt{3}}$ = tan θ
• Picking the first and last items , we get:
tan 150 = tan θ
(v) So the first principal solution is: θ = 150^o
5. We need only one principal solution. It indicates that, one of the angls between the two given lines is 150^o.
• If one of the angles is 150^o, then obviously, the other angle will be (180 - 150) = 30^o

Solved example 10.16
Show that the two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, where b₁, b₂ ≠ 0 are: (i) Parallel if $\frac{a_1}{b_1}~=~\frac{a_2}{b_2}$ and (ii) Perpendicular if a₁a₂ + b₁b₂ = 0
Solution:
1. The first line given is: a₁x + b₁ y + c₁ = 0
• So we can write:
A = a₁, B = b₁ and C = c₁
• Thus we get: $m_1=-\frac{A}{B}~=~-\frac{a_1}{b_1}$
2. The second line given is: a₂x + b₂ y + c₂ = 0
• So we can write:
A = a₂, B = b₂ and C = c₂
• Thus we get: $m_2=-\frac{A}{B}~=~-\frac{a_2}{b_2}$
3. If the two lines are parallel, then the two slopes will be equal.
• In such a situation, we get:

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

{\Rightarrow}&{-\frac{a_1}{b_1}}
&{}={}& {-\frac{a_2}{b_2}}
&{} \\

{\Rightarrow}&{\frac{a_1}{b_1}}
&{}={}& {\frac{a_2}{b_2}}
&{} \\

\end{array}$

4. If the two lines are perpendicular, then m₁ will be the -ve reciprocal of m₂.
• In such a situation, we get:

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

{\Rightarrow}&{-\frac{a_1}{b_1}}
&{}={}& {\frac{-1}{-\frac{a_2}{b_2}}}
&{} \\

{\Rightarrow}&{-\frac{a_1}{b_1}}
&{}={}& {\frac{1}{\frac{a_2}{b_2}}}
&{} \\

{\Rightarrow}&{-\frac{a_1}{b_1}}
&{}={}& {\frac{b_2}{a_2}}
&{} \\

{\Rightarrow}&{-a_1 a_2}
&{}={}& {b_1 b_2}
&{} \\

{\Rightarrow}&{a_1 a_2~+~b_1 b_2}
&{}={}& {0}
&{} \\

\end{array}$

Solved example 10.17
Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1, -2)
Solution:
1. The line given is: x - 2y + 3 = 0
• So we can write:
A = 1, B = -2 and C = 5
• Thus we get: $m=-\frac{A}{B}~=~-\frac{1}{-2}~=~\frac{1}{2}$
2. So slope of the line perpendicular to the given line will be the -ve reciprocal, which is -2.
3. Now we have the slope of the required line. A point on the line is given.
We can use the point-slope form: y - y₀ = m(x-x₀)

$\begin{array}{ll}
{}&{y~-~-2}
&{}={}& {-2(x~-~1)}
&{} \\

{\Rightarrow}&{y~+~2}
&{}={}& {-2x~+~2}
&{} \\

{\Rightarrow}&{2x~+~y}
&{}={}& {0}
&{} \\

\end{array}$

---

In the next section, we will see distance of a point from a line.

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