Slope Intercept Form Equation of a Straight Line | Proof, Solved Examples, Important Questions

Class 11 Mathematics | Written by Neeraj Anand

Published by ANAND TECHNICAL PUBLISHERS

Slope-intercept form is a special case of point-slope form where the fixed point on the straight line lies on X-axis or Y-axis.

When a straight line passes through any of the axes, it cuts the axis at a fixed point at a certain distance from the origin. This distance is called the intercept of the line.

If the line is said to cut the Y-axis on a distance of c₁ from the origin, then its cutting point is on Y-axis and its coordinate is (0, c₁), which will be the fixed point of the point-slope form.

Similarly, if the line cuts the X-axis on a distance of c₂ from the origin, the cutting point is on X-axis and its coordinate will be (c₂, 0).

Depending on from which axis the line is said to cut an intercept, the two cases are the following:

Case 1: If y-intercept is given

Let a straight line with slope m which is cut the Y-axis on a distance of c₁ from the origin. The line cuts Y-axis on a fixed cutting point (0, c₁). Putting slope = m, x₁ = 0 and y₁ = c₁ in the point-slope form we get,

(y – y₁) = m (x – x₁)

Putting the values we get,

⇒ y – c₁ = m (x – 0)

⇒ y – c₁ = mx

⇒ y = mx + c₁               …(3),

which is the equation of the slope-intercept form.

Slope Intercept Form of Line Passing Through Origin

The intercept form of the equation can also be made to pass through the origin (0, 0). If the line passes through the origin then the y-intercept made by the line is 0, i.e. c = 0 in this case. Suppose a line with slope m passes through the origin, then its equation of the line in slope-intercept form is,

y = mx

Problem 1: Find the equation of the straight line with slope -1 and y-intercept 3.

Solution: 

Here slope of the line is -1 or m = -1 and y-intercept is 3, hence the line cuts the Y-axis on the fixed point (0, 3) or (x₁, y₁) = (0, 3)

Putting the values of m, x₁ and y₁ in the point-slope form we get,

(y – y₁) = m(x – x₁)

⇒ y – 3 = (-1)(x – 0)

⇒ y – 3 = (-1)x

⇒ y – 3 = -x

⇒ y + x – 3 = 0,

which is the required equation

Problem 2: Find the equation of the straight line with slope 4 and y-intercept -5

Solution: 

Here slope of the line is 4 or m = 4 and y-intercept is -5 or c₁ = -5

Putting the values of m and c₁ in the slope-intercept form we get,

Slope-intercept form: y = mx + c₁

⇒ y = 4x + (-5)

⇒ y = 4x – 5

⇒ y – 4x + 5 = 0, which is the required equation

Case 2: If x-intercept is given

Let a straight line with slope m which is cut the X-axis on a distance of c₂ from the origin. The line cuts X-axis on a fixed cutting point which is (c₂, 0). Putting slope = m, x₁ = c₂ and y₁ = 0 in the point-slope form we get,

(y – y₁) = m (x – x₁)

Putting the values we get,

⇒ y – 0 = m (x – c₂)

⇒ y = m (x – c₂)               …(4),

which is the equation of the slope-intercept form.

Problem 1: Find the equation of the straight line with slope 2 and x-intercept 1

Solution: 

Here slope of the line is 2 or m = 2 and x-intercept is 1, hence the line cuts the X-axis on the fixed point (1, 0) or (x₁, y₁) = (1, 0)

Putting the values of m, x₁ and y₁ in the point-slope form we get,

(y – y₁) = m(x – x₁)

⇒ y – 0 = 2(x – 1)

⇒ y = 2(x – 1)

⇒ y = 2x – 2

⇒ y – 2x + 2 = 0,

which is the required equation

Problem 2: Find the equation of the straight line with slope -3 and x-intercept -7

Solution: 

Here slope of the line is -3 or m = -3 and x-intercept is -7 or c₂ = -7

Putting the values of m and c₂ in the slope-intercept form we get,

Slope-intercept form:

y = m(x – c₂)

⇒ y = (-3)(x – (-7))

⇒ y = (-3)(x + 7)

⇒ y = -3x – 21

⇒ y + 3x + 21 = 0, which is the required equation

Standard Form of Slope Intercept Form of Equation of a Straight Line

Consider a line L with slope m cuts the y-axis at a distance of c units from the origin.

Here, the distance c is called the y-intercept of the given line L.

So, the coordinate of a point where the line L meets the y-axis will be (0, c).

That means, line L passes through a fixed point (0, c) with slope m.

We know that, the equation of a line in point slope form, where (x₁, y₁) is the point and slope m is:

(y – y₁) = m(x – x₁)

Here, (x₁, y₁) = (0, c)

Substituting these values, we get;

y – c = m(x – 0)

y – c = mx

y = mx + c

Therefore, the point (x, y) on the line with slope m and y-intercept c lies on the line if and only if y = mx + c

Where, c = y-intercept of the line

Note: The value of c can be positive or negative based on the intercept is made on the positive or negative side of the y-axis.

Solved Examples

Example 1:

Find the equation of the straight line that has slope m = 3 and passes through the point (–2, –5).

Solution:

By the slope-intercept form we know;

y = mx+c

Given,

m = 3

As per the given point, we have;

y = -5 and x = -2

Hence, putting the values in the above equation, we get;

-5 = 3(-2) + c

-5 = -6+c

c = -5 + 6 = 1

Hence, the required equation will be;

y = 3x+1

Example 2:

Find the equation of the straight line that has slope m = -1 and passes through the point (2, -3).

Solution:

By the slope-intercept form we know;

y = mx+c

Given,

m = -1

As per the given point, we have;

y = -3 and x = 2

Hence, putting the values in the above equation, we get;

-3 = -1(2) + c

-3 = -2 + c

c = -3+2 = -1

Hence, the required equation will be;

y = -x-1

Example 3:

Find the equation of the lines for which tan θ = 1/2, where θ is the inclination of the line such that: 

(i) y-intercept is -5 

(ii) x-intercept is 7/3

Solution:

Given, tan θ = 1/2

So, slope = m = tan θ = 1/2

(i) y-intercept = c = -5

Equation of the line using slope intercept form is:

y = mx + c

y = (1/2)x + (-5)

Or

2y = x – 10

x – 2y – 10 = 0

(ii) x-intercept = d = 7/3

Equation of slope intercept form with x-intercept is:

y = m(x – d)

y = (1/2)[x – (7/3)]

Or

2y = (3x – 7)/3

6y = 3x – 7

3x – 6y – 7 = 0

Practice Problems

1. Find the slope of the line y = 5x + 2.
2. Find the slope of the line which crosses the line at point (-2,6) and have an intercept of 3.
3. What is the equation of the line whose angle of inclination is 45 degrees and x-intercept is -⅗?
4. Write the equation of the line passing through the point (0, 0) with slope -4.

Applications of Slope-Intercept Form

• Used in physics to model motion and velocity.
• Essential in economics for demand and supply analysis.
• Helps in predicting trends in data science and statistics.

📥 Download PDF

Students preparing for CBSE Board & JEE Mains/Advanced can download the full PDF from ANAND CLASSES for detailed explanations and additional solved examples.