DIFFERENT FORMS OF EQUATION OF STRAIGHT LINE

Suppose that we have the graph of a straight line and we want to find its equation.

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

(i)  Slope and y-intercept

(ii)  One point and slope

(iii) Two points

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

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

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

Different Forms of Equation of a Straight 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 , y₁)

3. Two-Points Form Equation of a Line :

Here, the two points are (x₁ , y₁) and (x₂ , y₂)

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

Practice Problems

Problem 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

So, the equation of the line L' is

x  =  3

Problem 2 :

Find the general equation of the straight line whose angle of inclination is 45° and y-intercept is 2/5.

Solution :

From the angle of inclination 45°, we can get the slope.

Slope of the line m  =  tan45°  =  1

Given : y-intercept b  =  2/5 

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

y  =  mx + b 

Substitute m  =  1 and b  =  2/5.

y  =  1x + 2/5

Multiply each side by 5. 

5y  =  5x + 2

Subtract 5y from each side.

0  =  5x - 5y + 2

or

5x - 5y + 2  =  0

So, the general equation of straight line is

5x - 5y + 2  =  0

Problem 3 :

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

Add 9 to each side.

0  =  x - 3y + 11

So, the general equation of straight line is

x - 3y + 11 = 0

Problem 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  

So, the general equation of straight line is

5x + 3y + 2  =  0

Problem 5 :

The vertices of a triangle ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.

Solution :

Median is a straight line joining a vertex and the midpoint of the opposite side.

Let D be the midpoint of BC.

The median through A is nothing but the line joining two points A (2,1) and D(1, 4). 

So, the equation of the median through A is 

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

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

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

Simplify. 

(y - 1) / 3  =  (x - 2) / (-1)

Cross multiply.

-1(y - 1)  =  3(x - 1)

Simplify.

-y + 1 = 3x - 3

3x + y - 4 = 0     

So, the equation of the median through A is 

3x + y - 4  =  0

Problem 6 :

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  

So, the general equation of straight line is

9x + 8y - 6  =  0

Problem 7 :

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

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

Problem 8 :

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. 

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

Problem 9 :

Find the slope and y-intercept of the straight line whose equation is 4x - 2y + 1  =  0. 

Solution :

Since we want to find the slope and y-intercept, let us write the given equation 4x - 2y + 1  = 0 in slope-intercept form. 

4x - 2y + 1  = 0

4x + 1  =  2y

Divide each side by 2. 

(4x + 1)/2  =  y

2x + 1/2  =  y 

or

y  =  2x + 1/2

The above form is slope intercept form. 

If we compare y  =  2x + 1/2  and  y  =  mx + b, we get  

m  =  2     and     b  =  1/2

So, the slope is 2 and y-intercept is 1/2.

Problem 10 :

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

So, the general equation of the required line is

5x - y - 2  =  0

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