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.

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*

*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) = (y _{2}
– y_{1})/(x_{2} – x_{1})*

* = (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.

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