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 = 33.69o
(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 213.69o
9. So the normal form can be written as:
x cos 213.69o + y sin 213.69o = $\frac{6}{\sqrt{13}}$
• 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 30o is the correct value of 𝜔
9. So the normal form can be written as:
x cos cos 30o + y sin 30o = 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 = 150o
5. We need only one principal solution. It indicates that, one of the angls between the two given lines is 150o.
• If one of the angles is 150o, then obviously, the other angle will be (180 - 150) = 30o
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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