Chapter 11.1 - General Equation of A Circle

In the previous section, we saw the basic details about conic sections. In this section, we will see some advanced details about circles.

A circle can be defined in 5 steps:
1. In fig.11.11(a) below, a curve is drawn in red color.
• A point C is marked inside the curve.

Fig.11.11

2. Mark a few convenient points P₁, P₂, P₃, P₄, P₅, . . .  on the curve.
• In the fig.(a), three points P₁, P₂ and P₃ are marked.
3. Measure the distances CP₁, CP₂, CP₃, CP₄, CP₅, . . .  
• If all those distances are the same, then the curve is a circle.
4. The point C is called the centre of the circle.
5. The distance from C to any point on the circle is called the radius of the circle.
• It is denoted using the letter r.

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Our next task is to derive the equation of a circle. It can be done in 4 steps:
1. In fig.11.12(b) above, the coordinates of C are (h,k)
2. Mark any convenient point P on the circle. Let the coordinates of P be (x,y)
3. Then the distance CP can be calculated using the distance formula. We get:
CP = $\sqrt{(x-h)^2 + (y-k)^2}$ 
4. But CP = r. So we can write:
$r = \sqrt{(x-h)^2 + (y-k)^2}$
• Squaring both sides, we get:
$r^2 = (x-h)^2 + (y-k)^2$
• This can be rearranged as:
$(x-h)^2 + (y-k)^2 = r^2$
• This is the equation of a circle.

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The equation of a circle can be written in expanded form also. It's details can be written in 5 steps:
1. Opening the brackets in the above equation, we get:
r² = x² - 2hx + h² + y² - 2ky + k² 
2. This can be rearranged as:
x² - 2hx + h² + y² - 2ky + k² - r² = 0
3. Writing similar terms together, we get:
x² + y² - 2hx - 2ky + h² + k² - r² = 0
4. We can describe the terms as follows:
• First term is a term in x².
• Second term is a term in y².
• Third term is a term in x.
• Fourth term is a term in y.
• The remaining terms are constant terms.
5. So we can write the equation as:
[x²] + [y²] + [-2hx] + [-2ky] + [h² + k² - r²] = 0
• Based on this we can write:
(i) The coefficient of x² term is equal to 1.
(ii) The coefficient of y² term is equal to 1.
(iii) The coefficient of x term is equal to -2h.
   ♦ So by dividing this coefficient by -2, we will get h.
   ♦ Thus we get the x-coordinate of the centre C.
(iv) The coefficient of y term is equal to -2k.
   ♦ So by dividing this coefficient by -2, we will get k.
   ♦ Thus we get the y-coordinate of the centre C.
(v) The constant term is h² + k² - r².
   ♦ Once we know k, h and the constant term, we can easily calculate the radius r.

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Now we will see some solved examples:
Solved example 11.1
Derive the equation of a circle whose center is at O
Solution:
1. In fig.11.11(c) above, the center of the circle is at O
2. Let P(x,y) be any point on the circle.
3. Using the distance formula, we can write:
OP = $\sqrt{(x-0)^2 + (y-0)^2}~=~\sqrt{x^2 + y^2}$ 
4. But OP = r. So we can write:
$r = \sqrt{x^2 + y^2}$
• Squaring both sides, we get:
$r^2 = x^2 + y^2$
• This can be rearranged as:
$x^2 + y^2 = r^2$
• This is the equation of a circle with radius r and center O.

Easier method:
1. The general equation of a circle with radius r and center at (h,k) is:
$(x-h)^2 + (y-k)^2 = r^2$
2. If the center is at O, both h and k are zero.
• Substituting these in the general equation, we get:
$(x-0)^2 + (y-0)^2 = r^2$   
• This is same as: $x^2 + y^2 = r^2$

Solved example 11.2
Find the equation of the circle with center at (3/2, 7/5) and radius 5 units.
Solution:
1. The general equation of a circle with radius r and center at (h,k) is:
$(x-h)^2 + (y-k)^2 = r^2$
2. The center is given as (3/2, 7/5).
• Substituting these in the general equation, we get:

$\begin{array}{ll}
{}&{\left(x-\frac{3}{2}\right)^2 + \left(y-\frac{7}{5}\right)^2}
& {~=~}& {6^2}
&{}&{}&{} \\

{\Rightarrow}&{x^2 - 3x + \frac{9}{4}~+~y^2 - \frac{14y}{5} + \frac{49}{25}}
& {~=~}& {36}
&{}&{}&{} \\

{\Rightarrow}&{100 \times \left[x^2 - 3x + \frac{9}{4}~+~y^2 - \frac{14y}{5} + \frac{49}{25} \right]}
& {~=~36 \times 100}& {}
&{\color{green}{\text{LCM of 4, 5, 25 is 100}}}&{}&{} \\

{\Rightarrow}&{100 x^2 - 300 x + 25 × 9 + 100 y^2 - 20 × 14 × y + 4 × 49}
& {~=~3600}& {}
&{}&{}&{} \\

{\Rightarrow}&{100 x^2 + 100 y^2 - 300 x - 280 y + 225 + 196}
& {~=~3600}& {}
&{}&{}&{} \\

{\Rightarrow}&{100 x^2 + 100 y^2 - 300 x - 280 y - 3179}
& {~=~0}& {}
&{}&{}&{} \\

\end{array}$

3. The actual plot is shown in fig.11.12 below:

Fig.11.12

Solved example 11.3
Find the center and radius of the circle x² + y² + 8x +10y - 8 = 0
Solution:
1. The general equation of a circle is:
[x²] + [y²] + [-2hx] + [-2ky] + [h² + k² - r²] = 0
• The given equation can be written in the general form:
[x²] + [y²] + [8x] + [10y] + [-8] = 0
2. So we can write:
(i) coefficient of x = -2h = 8. So h = -4
(ii) coefficient of y = -2k = 10. So k = -5
(iii) constant term = h² + k² - r² = -8
• Substituting the values of h and k, we get:

$\begin{array}{ll}
{}&{(-4)^2 + (-5)^2 - r^2}
& {~=~}& {-8}
&{}&{}&{} \\

{\Rightarrow}&{16 + 25 - r^2}
& {~=~}& {-8}
&{}&{}&{} \\

{\Rightarrow}&{41 - r^2}
& {~=~}& {-8}
&{}&{}&{} \\

{\Rightarrow}&{r^2}
& {~=~}& {49}
&{}&{}&{} \\

{\Rightarrow}&{r}
& {~=~}& {\sqrt{49}}
&{}&{}&{} \\

{\Rightarrow}&{r}
& {~=~}& {\pm 7}
&{}&{}&{} \\

\end{array}$

• Radius cannot be -ve. We must take the +ve value.
3. We can write:
• Center (h,k) of the circle is (-4, -5)
• Radius r of the circle is 7 units.
• Once we get the center and radius, we can use the simple form also:

$\begin{array}{ll}
{}&{(x-h)^2 + (y-k)^2}
& {~=~}& {r^2}
&{}&{}&{} \\

{\Rightarrow}&{(x~-~-4)^2 + (y~-~-5)^2}
& {~=~}& {7^2}
&{}&{}&{} \\

{\Rightarrow}&{(x+4)^2 + (y+5)^2}
& {~=~}& {49}
&{}&{}&{} \\

\end{array}$

4. The actual plot is shown below:

Fig.11.13

Solved example 11.4
Find the equation of the circle which passes through the points (2,-2) and (3,4) and whose center lies on the line x+y=2.
Solution:
1.The general equation of a circle is: (x-h)² + (y-k)² = r²
• Where (h,k) is the center and r is the radius.
2. In our present case, the circle passes through (2,-2). Substituting these values in the general equation, we get:
(2-h)² + (-2-k)² = r²
⇒ (2-h)² + [-1(2+k)]² = r²
⇒ (2-h)² + (2+k)² = r²
• Expanding this and rearranging, we get:
8 – 4h + 4k +h² + k² - r² = 0
3. In our present case, the circle passes through (3,4) also. Substituting these values in the general equation, we get:
(3-h)² + (4-k)² = r²
• Expanding this and rearranging, we get:
25 – 6h - 8k + h² + k² - r² = 0
4. So now we have two equations:
(i) From (2), we have: 8 – 4h + 4k +h² + k² - r² = 0
(ii) From (3), we have: 25 – 6h - 8k + h² + k² - r² = 0
5. Subtracting 4(ii) from 4(i), we get: -17 + 2h +12k = 0
6. Given that, the center lies on the line x+y=2.
• We assumed that, the center is (h,k). Substituting these in the equation of the line, we get: h+k=2
• So h = 2-k
7. Substituting this value of h in (5), we get:
-17 + 2(2-k) + 12k = 0
⇒ -17 + 4 – 2k  + 12k = 0
⇒ -13 + 10k = 0
⇒ k = 1.3
8. Substituting this value of k in (6), we get: h = 2-1.3 = 0.7
9. Now we have the center of the circle (h,k) = (0.7, 1.3)
• (2,-2) is a point on the circle.
• So using the distance formula, we can write:
r² = [(2 - 0.7)² + (-2 - 1.3)²] = [1.3² + 3.3²] = 12.58
10. Now we can write the equation of the circle:
(x - 0.7)² + (y - 1.3)² = 12.58
• The actual plot is shown below:

Fig.11.14

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The link below gives a few more solved examples:

Exercise 11.1

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In the next section, we will see parabola.

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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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Normal Form Example

It is necessary to practice how to give appropriate signs for sine and cosine terms in the normal form. We have seen an example in section 10.8. A similar example is given below:

Equation of a line is 3x - 5y - 15 = 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 = -5 and C = -15
2. Calculation of p

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

{}&{}
&{}={}& {\pm \frac{-15}{\sqrt{3^2~+~(-5)^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{15}{\sqrt{34}}}
&{} \\

\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~+~(-5)^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{3}{\sqrt{34}}}
&{} \\

\end{array}$

3. Calculation of sin 𝜔

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

{}&{}
&{}={}& {\pm \frac{-5}{\sqrt{3^2~+~(-5)^2}}}
&{} \\

{}&{}
&{}={}& {\pm \frac{5}{\sqrt{34}}}
&{} \\

\end{array}$

4. Now we determine the signs:
(i) Slope of the line = $m=-\frac{A}{B}~=~-\frac{3}{-5}~=~\frac{3}{5}$
• This is +ve. So we move along the left arrow of the chart in fig.13.36.
• The chart is shown again below:

(ii) x-intercept of the line = $a=-\frac{C}{A}~=~-\frac{-15}{3}~=~5$
• This is +ve. So sin 𝜔 is -ve and cos 𝜔 is +ve.

5. So we can write:

$\begin{array}{ll}
{}&{\sin \omega }
&{}={}& {- \frac{5}{\sqrt{35}}}
&{} \\

{}&{\cos \omega}
&{}={}& {\frac{3}{\sqrt{35}}}
&{} \\

\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{35}}~+~y × - \frac{5}{\sqrt{35}}}
&{}={}& {\frac{15}{\sqrt{35}}}
&{} \\

{\Rightarrow}&{ \frac{3x}{\sqrt{35}}~+~ - \frac{5y}{\sqrt{35}}}
&{}={}& {\frac{15}{\sqrt{35}}}
&{} \\

{\Rightarrow}&{\frac{3x}{\sqrt{35}}~-~ \frac{5y}{\sqrt{35}}}
&{}={}& {\frac{15}{\sqrt{35}}}
&{} \\

\end{array}$

8. Calculation of 𝜔:
(i) From (5), we have: tan 𝜔 = -5/3
• Using a scientific calculator, we get: 𝜔 = -59.04^o
(ii) In our present case, 𝜔 is in the fourth quadrant (sin is -ve and cos is +ve).
• -59.04^o indeed lies in the fourth quadrant. But we want the +ve angle. The +ve angle corresponding to -59.04^o is (360 - 59.04) = 300.96^o
• We get: tan 𝜔 = tan -59.04 = tan 300.96
(iii) So the value of 𝜔 is 300.96^o

9. So the normal form can be written as:
x cos 300.96^o + y sin 300.96^o = $\frac{15}{\sqrt{35}}$

• The actual plot is shown below:

 

 

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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Chapter 10.7 - General Equation of A Line

In the previous section, we saw the normal form of the equation of a line. In this section, we will see the general equation of a line.

Some basics about the general equation can be written in 4 steps:
1. Consider the equation: Ax + By + C = 0
• It is a first degree equation in two variables.
2. Let us put some random values for A, B and C:
A = 2, B = 3 and C = 5. Now the equation becomes: 2x + 3y + 5 = 0
• In this equation, if we put some random values for x, we will get corresponding values for y.
• For example:
    ♦ If x = 2, then y will be -3
    ♦ If x = 3, then y will be -3.67
    ♦ If x = 6, then y will be -5.67
    ♦ If x = 7, then y will be -6.33
3. So we get some coordinates: (2,-3), (3,-3.67), (6,-5.67), (7,-6.33)
• If we plot these coordinates, they will lie on a straight line. This is shown in fig.10.33 below:

Fig.10.33

4. So we can write:
• Any equation of the form Ax + By + C = 0 is called general linear equation.
• It is also called general equation of a line.
◼ But there is one condition:
A and B should not be zero simultaneously.
• This can be explained in 3 steps:
(i) If A = 0, the general equation will become: By + C = 0
    ♦ This is OK because, By + C = 0 will give a straight line.
(ii) If B = 0, the general equation will become: Ax + C = 0
    ♦ This is OK because, Ax + C = 0 will give a straight line.
(iii) If A = 0 and B = 0, the general equation will become: C = 0
    ♦ This is not OK because, C = 0 will not give a straight line.

---

• Now we will see the different forms of Ax + By + C = 0
• We have seen:
    ♦ Slope-intercept form
    ♦ Intercept form
    ♦ Normal form.
• The general equation can be written in each of these three forms.

First we will see the slope-intercept form. It can be written in 11 steps:
1. The slope-intercept form is: y = mx + c
2. The general form is: Ax + By + C = 0
3. Let the equations in both (1) and (2) represent the same line.
4. The point at which (1) intersects the x-axis can be obtained by putting y = 0.
So we get:

$\begin{array}{ll}
{}&{y}
&{}={}& {mx+c}
&{} \\

{\Rightarrow}&{0}
&{}={}& {mx+c}
&{} \\

{\Rightarrow}&{x}
&{}={}& {-\frac{c}{m}}
&{} \\

\end{array}$

5. So the line in (1) intersects the x-axis at $\left(-\frac{c}{m},~0 \right)$
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.

• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{A × -\frac{c}{m} ~+~B × 0~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{-\frac{Ac}{m}~+~C}
&{}={}& {0}
&{} \\

\end{array}$

6. The point at which (1) intersects the y-axis can be obtained by putting x = 0.
So we get:

$\begin{array}{ll}
{}&{y}
&{}={}& {mx+c}
&{} \\

{\Rightarrow}&{y}
&{}={}& {m × 0+c}
&{} \\

{\Rightarrow}&{y}
&{}={}& {c}
&{} \\

\end{array}$

7. So the line in (1) intersects the y-axis at (0,c)
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.

• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{A × 0 ~+~B × c~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{B × c~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{c}
&{}={}& {-\frac{C}{B}}
&{} \\

\end{array}$

8. Substituting this value of c in (5), we get:

$\begin{array}{ll}
{}&{-\frac{Ac}{m}~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{-\frac{A}{m} × -\frac{C}{B}~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{-\frac{A}{m} × -\frac{1}{B}~+~1}
&{}={}& {0}
&{} \\

{\Rightarrow}&{\frac{A}{B m}~+~1}
&{}={}& {0}
&{} \\

{\Rightarrow}&{\frac{A}{Bm}}
&{}={}& {-1}
&{} \\

{\Rightarrow}&{m}
&{}={}& {-\frac{A}{B}}
&{} \\

\end{array}$

9. So we get the following results:
    ♦ From (7), we get: $c~=~-\frac{C}{B}$
    ♦ From (8), we get: $m~=~-\frac{A}{B}$
10. Now we can write:
• If we are given the equation of a line in the general form Ax + By + C = 0, then:
    ♦ Slope of that line can be obtained as: $m~=~-\frac{A}{B}$
    ♦ y-intercept of that line can be obtained as: $c~=~-\frac{C}{B}$
11. If B = 0, then the general equation becomes: $x~=~-\frac{C}{A}$
    ♦ This is an equation of vertical line.
    ♦ It’s slope is undefined.
        ✰ Indeed "$m~=~-\frac{A}{B}$" is undefined if B = 0
    ♦ It intersects the x axis at $-\frac{C}{A}$.

---

Next we will see the intercept form. It can be written in 10 steps:
1. The intercept form is: $\frac{x}{a}~+~\frac{y}{b}~=~1$
2. The general form is: Ax + By + C = 0
3. Let the equations in both (1) and (2) represent the same line.
4. The point at which (1) intersects the x-axis can be obtained by putting y = 0.
So we get:

$\begin{array}{ll}
{}&{\frac{x}{a}~+~\frac{y}{b}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{x}{a}~+~\frac{0}{b}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{x}{a}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{x}
&{}={}& {a}
&{} \\

\end{array}$

5. So the line in (1) intersects the x-axis at (a,0)
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.

• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{A × a ~+~B × 0~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{Aa~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{a}
&{}={}& {-\frac{C}{A}}
&{} \\

\end{array}$

6. The point at which (1) intersects the y-axis can be obtained by putting x = 0.
So we get:

$\begin{array}{ll}
{}&{\frac{x}{a}~+~\frac{y}{b}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{0}{a}~+~\frac{y}{b}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{y}{b}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{y}
&{}={}& {b}
&{} \\

\end{array}$

7. So the line in (1) intersects the x-axis at (0,b)
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.

• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{A × 0 ~+~B × b~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{B × b~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{b}
&{}={}& {-\frac{C}{B}}
&{} \\

\end{array}$

8. So we get the following results:
    ♦ From (5), we get: $a~=~-\frac{C}{A}$
    ♦ From (7), we get: $b~=~-\frac{C}{B}$
9. Now we can write:
• If we are given the equation of a line in the general form Ax + By + C = 0, then:
    ♦ x-intercept of that line can be obtained as: $a~=~-\frac{C}{A}$
    ♦ y-intercept of that line can be obtained as: $b~=~-\frac{C}{B}$
        ✰ Note that, we obtained the same y-intercept in slope-intercept form also.
10.If C = 0, then:
    ♦ x-intercept = $-\frac{0}{A}$ = 0
    ♦ y-intercept = $-\frac{0}{B}$ = 0
• That means, the line will pass through the origin.
◼ We can write:
In the general equation of a line, if "term with no variable" is absent, then that line will pass through the origin.

---

Next we will see the normal form. It can be written in 12 steps:
1. The normal form is: $x \cos \omega~+~y \sin \omega~=~p$
2. The general form is: Ax + By + C = 0
3. Let the equations in both (1) and (2) represent the same line.
4. The point at which (1) intersects the x-axis can be obtained by putting y = 0.
So we get:

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

{\Rightarrow}&{x \cos \omega~+~0 × \sin \omega}
&{}={}& {p}
&{} \\

{\Rightarrow}&{x \cos \omega}
&{}={}& {p}
&{} \\

{\Rightarrow}&{x}
&{}={}& {\frac{p}{\cos \omega}}
&{} \\

\end{array}$

5. So the line in (1) intersects the x-axis at $\left(\frac{p}{\cos \omega},~0 \right)$
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.

• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{A × \frac{p}{\cos \omega} ~+~B × 0~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{\frac{A p}{\cos \omega}~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{\cos \omega}
&{}={}& {-\frac{Ap}{C}}
&{} \\

\end{array}$

6. The point at which (1) intersects the y-axis can be obtained by putting x = 0.
So we get:

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

{\Rightarrow}&{0 × \cos \omega~+~ y \sin \omega}
&{}={}& {p}
&{} \\

{\Rightarrow}&{y \sin \omega}
&{}={}& {p}
&{} \\

{\Rightarrow}&{y}
&{}={}& {\frac{p}{\sin \omega}}
&{} \\

\end{array}$

7. So the line in (1) intersects the x-axis at $\left(0,~\frac{p}{\sin \omega} \right)$
• But the equations in both (1) and (2) represent the same line. So this point must satisfy (2) also.

• We can write:
$\begin{array}{ll}
{}&{Ax + By + C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{A × 0 ~+~B × \frac{p}{\sin \omega}~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{\frac{B p}{\sin \omega}~+~C}
&{}={}& {0}
&{} \\

{\Rightarrow}&{\sin \omega}
&{}={}& {-\frac{Bp}{C}}
&{} \\

\end{array}$

8. So we get the following results:
    ♦ From (5), we get: $\cos \omega~=~-\frac{Ap}{C}$
    ♦ From (7), we get: $\sin \omega~=~-\frac{Bp}{C}$
9. But $\sin^2 \omega~+~\cos^2 \omega~=~1$
So we get:
$\begin{array}{ll}
{}&{\left(-\frac{Bp}{C} \right)^2~+~\left(-\frac{Ap}{C} \right)^2}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{B^2 p^2}{C^2}~+~\frac{A^2 p^2}{C^2}}
&{}={}& {1}
&{} \\

{\Rightarrow}&{\frac{ p^2(A^2~+~B^2)}{C^2}}
&{}={}& {1}
&{} \\

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

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

\end{array}$

10. We can substitute this value of p in (5). We get:

$\begin{array}{ll}
{}&{\cos \omega}
&{}={}& {-\frac{Ap}{C}}
&{} \\

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

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

\end{array}$

11. Similarly, we can substitute this value of p in (7). We get:

$\begin{array}{ll}
{}&{\sin \omega}
&{}={}& {-\frac{Bp}{C}}
&{} \\

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

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

\end{array}$

12. Thus we get three results:
    ♦ From (9), we get: $p~=~\pm \frac{C}{\sqrt{A^2~+~B^2}}$ 
    ♦ From (10), we get: $\cos \omega~=~\pm \frac{A}{\sqrt{A^2~+~B^2}}$ 
    ♦ From (11), we get: $\sin \omega~=~\pm \frac{B}{\sqrt{A^2~+~B^2}}$

---

Let us compile the above results. It can be done in 5 steps:
1. Slope of a given line can be calculated as: $m~=~-\frac{A}{B}$
• This result is available from slope-intercept form.
2. x-intercept of a given line can be calculated as: $c~=~-\frac{C}{A}$
• This result is available from intercept form.
3. y-intercept of a given line can be calculated as: $c~=~-\frac{C}{B}$
• This result is available from both slope-intercept form and intercept form.
4. For writing the normal form, we can use three results:
    ♦ $p~=~\pm \frac{C}{\sqrt{A^2~+~B^2}}$ 
    ♦ $\cos \omega~=~\pm \frac{A}{\sqrt{A^2~+~B^2}}$ 
    ♦ $\sin \omega~=~\pm \frac{B}{\sqrt{A^2~+~B^2}}$
5. These results are easy to remember because, they follow a pattern.

---

• In the above results, we have $\pm$ sign for p, $\cos \omega$ and $\sin \omega$.
• So two questions arise:
    ♦ When do we use the '+' sign ?  
    ♦ When do we use the '-' sign ?
• We will see the answers in the next section.

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