# FINDING THE EQUATION OF A LINE

## About "Finding the Equation of a Line"

Finding the Equation of a Line :

In this section, we are going to learn, how to find the equation of a line from the information given.

Base on the information, we can use different forms equation of line.

For any straight line, if we want to find the equation, we must have the following information.

(i)  Slope and y-intercept

(ii)  A point and slope

(iii) Two points

(iv) Two intercepts (x-intercept and y-intercept)

If we have any one of the four information given above, we will be able to find the equation of the line using the formulas given below.

Now, let us look at the formulas for different forms equation of a line.

## Different Forms of Equation of Line

1. Slope - Intercept form equation of a line

Here,

Slope of the line  =  m

y-intercept  =  b

2.Point - slope form equation of a line

Here,

Slope of the line  =  m

Point  =  (x₁ , y)

3. Two-points form equation of a line

Here, the two points are (x₁ , y) and (x₂ , y)

4.  Intercept form equation of a line

Here,

x- intercept  =  a

y- intercept  =  b

Apart from the above forms of equation of straight line, there are some other ways to get equation of a straight line.

1. If a straight line is passing through a point (0,k) on y-axis and parallel to x-axis, then  the equation of the straight line is   y  =  k

2. If a straight line is passing through a point (c,0) on x-axis and parallel to y-axis, then  the equation of the straight line is   x  =  c

3. Equation of x-axis is y  =  0.

(Because, the value of "y" in all the points  on x-axis is zero)

4. Equation of y-axis is  x  =  0.

(Because, the value of "x" in all the points  on y-axis is zero)

5. General equation of a straight line is

ax + by + c  =  0

## Finding the Equation of a Line - Examples

Example 1 :

Find the equations of the straight lines parallel to the coordinate axes and passing through the point (3, -4).

Solution :

Let L and L' be the straight lines passing through the point (3, - 4) and parallel to x-axis and y-axis respectively.

The y-coordinate of every point on the line L is – 4.

Hence, the equation of the line L is y = - 4

Similarly, the x-coordinate of every point on the straight line L' is 3

Hence, the equation of the line L' is x = 3.

Example 2 :

Find the general equation of the straight line passing through the point (-2, 3) with slope 1/3.

Solution :

Given : Point  =  (-2, 3)  and  slope  m  =  1/3

So, the equation of the straight line in point-slope form is

y - y1  =  m(x - x1)

Substitute (x1 , y1) = (-2 , 3) and m = 1/3.

y - 3  =  1/3 ⋅ (x + 2)

Multiply each side by 3.

3(y - 3)  =  x + 2

Simplify.

3y - 9  =  x + 2

Subtract 3y from each side.

-9  =  x - 3y + 2

0  =  x - 3y + 11

So, the general equation of straight line is x - 3y + 11 = 0.

Example 3 :

Find the general equation of the straight line passing through the point (2, 5) with slope -5.

Solution :

In this problem, instead of using point-slope form, we can use slope-intercept form also to find the equation of the line.

y  =  mx + b

Substitute m  =  -5.

y  =  -5x + b ----(1)

Substitute (x, y)  =  (2, 5)

5  =  -5(2) + b

Simplify.

5  =  -10 + b

15  =  b

Substitute 15 in (1).

(1)---->  y  =  -5x + 15

5x + y  =  15

Subtract 15 from each side.

5x + y - 15  =  0

So, the general equation of straight line is 5x + y - 15 = 0.

Example 4 :

Find the general equation of the straight line passing through the points (-1, 1) and (2, -4).

Solution :

Given : Two points on the straight line : (-1, 1) and  (2, -4).

So, the equation of the straight line in two-points form is

(y - y₁) / (y₂ - y₁)  =  (x - x₁) / (x₂ - x₁)

Substitute (x1 , y1)  =  (-1, 1) and (x2, y2)  =  (2, -4).

(y - 1) / (-4 - 1)  =  (x + 1) / (2 + 1)

Simplify.

(y - 1) / (-5)  =  (x + 1) / 3

Cross multiply.

3(y - 1)  =  -5(x + 1)

3y - 3  =  -5x - 5

5x + 3y + 2  =  0

Hence the general equation of straight line is

5x + 3y + 2  =  0

Example 5 :

If the x-intercept and y-intercept of a straight line are 2/3 and 3/4 respectively, find the general equation of the straight line.

Solution :

Given :

x- intercept  "a"  =  2/3

y-intercept  "b"  =  3/4

So, the equation of the straight line in intercept form is

x/a  +  y/b  =  1

Substitute a  =  2/3  and  b  =  3/4.

x / (2/3)  +  y / (3/4)  =  1

Simplify.

3x / 2  +  4y / 3  =  1

(9x + 8y)  / 6  =  1

Multiply each side by 6.

9x + 8y  =  6

Subtract 6 from each side from 6.

9x + 8y - 6  =  0

Hence the general equation of straight line is

9x + 8y - 6  =  0

Example 6 :

Find the equations of the straight lines each passing through the point (6, -2) and whose sum of the intercepts is 5.

Solution :

Let "a" and "b" be the x-intercept and y-intercept of the required straight line respectively.

Given : Sum of the intercepts  =  5

So, we have

a + b  =  5

Subtract a from each side.

b  =  5 - a

Now, equation of the straight line in intercept form is

x/a  +  y/b  =  1

Plugging  b  =  5 - a, we get

x / a  +  y / (5-a)  =  1

Simplify.

[(5-a)x + ay ] / a(5-a)  =  1

(5-a)x + ay  =  a(5-a) -----(1)

The straight line is passing through (6, -2).

So, substitute (x, y)  =  (6, -2).

(5 - a)6 - 2a  =  a(5 - a)

30 - 6a - 2a  =  5a - a²

a² - 13a + 30  =  0

a² - 13a + 30  =  0

(a - 10)(a - 3) = 0

a  =  10 and a  =  3

When a = 10,

(1)-----> (5 - 10)x + 10y  =  10(5 - 10)

- 5x + 10y  =  - 50

5x - 10y  - 50  =  0

x - 2y - 10  =  0

When a = 3,

(1)----->(5 - 3)x + 3y  =  3(5 - 3)

2x + 3y  =  6

Hence, x - 2y - 10 = 0 and 2x + 3y - 6 = 0 are the general equations of the required straight lines.

Example 7 :

Find the general equations of the straight lines parallel to x- axis which are at a distance of 5 units from the x-axis.

Solution :

From the given information, we can sketch the two lines as given below.

One line is above the x-axis at a distance of 5 units. And another line is below the x-axis at a distance of 5 units.

Hence, y = 5 and y = -5 are the required straight lines.

Example 8 :

A straight line has the slope 5. If the line cuts y-axis at -2, find the general equation of the straight line.

Solution :

Since the line cuts y-axis at "-2", clearly y-intercept is "-2"

So, the slope m  =  5 and y-intercept b  =  -2.

Equation of a straight line in slope-intercept form :

y  =  mx + b

Substitute m  =  5  and  b  =  -2

y  =  5x - 2

Subtract y from each side.

5x - y - 2  =  0

Hence, the general equation of the required line is

5x - y - 2  =  0

After having gone through the stuff given above, we hope that the students would have understood, "Finding the Equation of a Line".

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