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Geometry

Trigonometry & Coordinate

ALL MODULES

CAT 2025 Lesson : Trigonometry & Coordinate - Equation of a Line

3.3 Slope of a line

The slope of a line determines the flatness or steepness of the line with respect to the axes. Positive and negative slopes indicate the direction that the line takes with respect to the

y

-axis.

Where the angle formed by the line with the

x

-axis is θ, Slope (

m

) = tan θ

Where (

x_1, y_1

) and (

x_2, y_2

) are two points that the line passes through (and

x_1 ≠ x_{2}

), Slope (

m

) =

\dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}

The slope gives us the rate of change of

y

with respect to the change of

x

. For instance, if the slope of a line is 3, then 1 unit increase in the

x

-coordinate will result in a 3 units increase in the

y

-coordinate along the line. Similarly, if the slope is -0.5, then 1 unit increase in

x

-coordinate will result in 0.5 decrease in the

y

-coordinate (decrease in this case as the slope is negative).

Example 10

What is the slope of the line connecting the points

(i) (3, -2) and (5, 1) (ii) (-8, 12) and (3, -4) (iii) (2, 3) and (4, 7)

Solution

(i) Where (3, -2) and (5, 1) are two points that the line passes through
Slope (

m

) =

\dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{1 - (-2)}{5 - 3} = \dfrac{1 + 2}{5 - 3} = \dfrac{3}{2}

(ii) Where (

-8, 12

) and (

3, -4

) are two points that the line passes through
Slope

(m) = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-4 -12}{3 - (-8)} = \dfrac{-16}{11}

(iii) Where (

2, 3

) and (

4, 7

) are two points that the line passes through
Slope

(m) = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{7 - 3}{4 - 2} = \dfrac{4}{2}= 2

Answer: (i)

\dfrac{3}{2}

; (ii)

\dfrac{-16}{11}

; (iii)

2

3.4 Lines parallel to x-axis & y-axis

As shown below, few of the points that a line that is parallel to

x

-axis passes through are

(-2, k), (-1, k), (0, k), (1, k), (2, k)

. The

y

-coordinate of the points never changes. Therefore, the equation of such a line is

y = k

A vertical line parallel to the

y

-axis, as shown below, passes through are

(k, -2), (k, -1), (k, 0), (k, 1), (k, 2)

. The

x

-coordinate of the points never changes. Therefore, the equation of such a line is

x = k

3.5 Equation of a Line

Equations of a line can be formed or written in the following ways:

1) Slope & y-intercept form:

y = mx + c

This is one of the easiest ways to form an equation.

m

is the slope of the line and

c

is the

y

-intercept (note that when

x = 0, y = c

and therefore,

c

is the

y

-intercept).

m

indicates the direction and steepness of the line while

c

indicates the point where the line cuts the

y

-axis. An equation represented this way would help us to quickly draw or visualise the line.

2)

\dfrac{x}{a} + \dfrac{y}{b} = 1

where

a

and

b

are the

x

-intercept and

y

-intercept. This can be used when the two intercepts are known.

3)

ax + by + c = 0

This is a general form in which an equation of the line is represented.
(a) If

x = 0, y = - \dfrac{c}{b}

which is the

y

-intercept

(b) If

y = 0, x = - \dfrac{c}{a}

which is the

x

-intercept

(c) Equation can be re-written as

y = - \dfrac{ax}{b} - \dfrac{c}{b}

where the Slope is

\dfrac{-a}{b}

(a), (b) and (c) need not be memorised as you should be able to quickly deduce these.

Example 11

What is the equation of the line, if
(i) the slope and the

y

-intercept are

\dfrac{-3}{2}

and

5

respectively?
(ii)

x

-intercept and

y

-intercept are

2

and

-6

respectively?

Solution

(i) We can directly apply this in

y = mx + c

where

m

and

c

are the slope and y-intercept respectively.

y = \dfrac{-3}{2} x + 5

⇒

3x + 2y = 10

(ii) Applying the

y

-intercept, we get

y = mx

- 6 ----- (1)

An

x

-intercept of

2

means the line cuts the

x

-axis at

x = 2

, i.e., when

y = 0

. Substituting

x = 2

and

y = 0

, we get

0 = 2m - 6 ⇒ m = 3

Substituting

m = 3

in (1), we get

y = 3x - 6

⇒

3x - y = 6

Answer: (i)

3x + 2y = 10

; (ii)

3x - y = 6

Example 12

What is the equation of the line passing through the point (

4, -5

) with the slope

\dfrac{3}{2}

Solution

Working with the slope-intercept form of the equations

y = mx + c ⇒ y = \dfrac{3}{2} x + c

----- (1)

As the line passes through (

4, -5

), substituting

x = 4

and

y = -5

we get

-5 = \dfrac{3}{2} \times 4 + c

⇒

-10 = 12 + 2c

⇒

2c = -22

⇒

c = -11

Substituting

c = -11

in (1),

⇒

y = \dfrac{3}{2}x - 11 ⇒ 2y = 3x - 22

⇒

3x - 2y = +22

Answer:

3x - 2y = 22

Example 13

Find the area of the region (in units), where

2x + 3y -12 < 0, x > 0

and

y > 0

Solution


The graph of $x > 0$
The graph of $y > 0$
The graph of $2x + 3y -12 < 0$ The line $2x + 3y -12 = 0$ cuts the $x$ axis at ( $6, 0$ ) To find this coordinate, we need to substitute $y = 0$ in $2x + 3y -12 = 0$ ⇒ $2x + 3(0) - 12 = 0$ ⇒ $2x = 12 ⇒ x = 6$ . The line $2x + 3y -12 = 0$ cuts the $y$ axis at ( $4, 0$ ) To find this coordinate, we need to substitute $x = 0$ in $2x + 3y -12 = 0$ ⇒ $2(0) + 3y - 12 = 0$ ⇒ $3y = 12 ⇒ y = 4$ . The closed region will form a triangle with coordinates $(0, 0), (6, 0)$ and $(0, 4)$ Therefore, the area of the closed region = $\dfrac{1}{2} \times 4 \times 6 = 12$ units. Answer: $12$

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