Slope Intercept Form

Equation of a straight line in terms of slope (m) and y-intercept (c) of the line is called the equation in slope intercept form. It is written as y = mx+c.

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To find the equation of a straight line which makes an angle θ with the x-axis in positive direction and cuts the y-axis at a distance of c from the origin, proceed to the following:

Let line AB intersect x-axis at A and y-axis at C such that ∠BAX = θ and y-intercept ( OC) = c.

            ∴ slope of line AB (m) = tanθ

Let P (x, y) be any point on the line AB, draw PM⊥OX and CN⊥PM, then

            CN = OM = x

            PN = PM – MN = PM – OC = y – c

From right angled triangle PNC,

                        tanθ = PN/CN

            or,        m = (y – c)/x

     or,        y – c = mx

     or,        y = mx + c

This relation is true for every point on the line AB. So y = mx + c represents the equation of the straight line AB in slope intercept form.

Note:

1. If the line passes through the origin, c = 0, so the equation of the line passing through the origin is y = mx.

2. If the line be parallel to x-axis, i.e. its slope equals zero (m = 0), the equation becomes y = c.

3. If the line be parallel to y-axis, the equation cannot be written in this form.

 

Workout Examples

Example 1: Find the equation of a straight line making an angle of 60° with the x-axis and cutting an intercept 3 from the y-axis.

Solution: Here,

                        Θ = 60°

                        y-intercept (c) = 3

                        slope (m) = tan60° = √3

                        ∴ The equation of the line is,

                        y = mx + c

            i.e.       y = √3x + 3

            ∴ The required equation of the line is y = √3x + 3.

 

Example 2: Find the equation of a straight line which is inclined to the x-axis at an angle 30° and cutting an intercept 2 from y-axis.

Solution: Here,

                        Θ = 30°

                        y-intercept (c) = 2

                        slope (m) = tan30° = 1/√3

                        ∴ The equation of the line is,

                        y = mx + c

            i.e.       y = (1/√3)x + 2

            or,        y = (x + 2√3)/√3

            or,        x + 2√3 = √3y

            or,        x  - √3y + 2√3 = 0

            ∴ The required equation of the line is x - √3y + 2√3 = 0.

 

Example 3: Find the equation of straight line passing through the points (0, -2) and (2, 1).

Solution: Here,

                        The straight line passes through the points (0, -2) and (2, 1).

                        ∴ slope (m) = (y₂ – y₁)/(x₂ – x₁)

                                           = (1 + 2)/(2 – 0)

                                           = 3/2

                         The point (0, -2) is on y-axis.

                        ∴ y-intercept (c) = –2

                        The equation of the straight line is,

                        y = mx + c

            i.e.       y = (3/2)x – 2

            or,        y = (3x – 4)/2

            or,        3x – 4 = 2y

            or,        3x – 2y – 4 = 0

            ∴ The required equation of the line is 3x – 2y – 4 = 0.

You can comment your questions or problems regarding slope intercept form of the equation here.