Equation of a straight line in terms of
its slope (m) and a point (x_{1}, y_{1}) through which it
passes is called the **point slope form**
equation. It is written as

**y – y _{1 }= m(x – x_{1})**

To derive this equation, let us suppose a
straight line AB whose slope is m and passes through the point C (x_{1},
y_{1}). If P (x, y) be any point on the line AB, then

But slope of AB = m

Since AB
and CP represents the same line,

Slope of line AB = Slope of CP

or, y –
y_{1} = m(x – x_{1})

which is the required equation in **point slope form**.

** Corollary:** If the straight line
passes through the two given points (x

_{1}, y

_{1}) and (x

_{2}, y

_{2}) then its slope can be obtained by

And hence the equation of the
line can be obtained by the formula,

y – y_{1} = m (x – x_{1})

This is called **two point form** equation.

*Workout
Examples*

*Workout Examples*

*Example 1: **Find the equation of a straight line passing through the point (3,
2) and having the slope equal to 3/5.*

*Solution:** Here,*

* Slope (m) = 3/5*

* Point = (3, 2)*

* **∴** x _{1} = 3 and y_{1} = 2*

* The equation
of the line is,*

* y – y _{1}
= m(x – x_{1})*

* i.e. y – 2 = 3/5 (x – 3)*

* or, 3(x – 3) = 5(y – 2)*

* or, 3x – 9 = 5y – 10 *

* or, 3x – 5y – 9 + 10 = 0*

* or, 3x – 5y + 1 = 0*

* Which is the
required equation of the line.*

*Example 2: **Find the equation of a straight line passing through the point (-2,
3) and having an angle of inclination 60° with x-axis.*

*Solution:** Here,*

* Angle of inclination (θ) = 60°*

* **∴** Slope (m) = tanθ = tan60° = √3*

* Point
= (–2, 3)*

* **∴** x _{1} = –2 and y_{1} = 3*

* The equation
of the line is,*

* y – y _{1}
= m(x – x_{1})*

* i.e. y – 3 = √3(x + 2)*

* or, √3(x + 2) = y – 3*

* or, √3x + 2√3 – y + 3 = 0 *

* or, √3x – y + 3 + 2√3 = 0*

* or, 3x – 5y + 1 = 0*

* Which is the
required equation of the line.*

*Example 3: **Find the equation of a straight line passing through the points (-2,
3) and (2, 4).*

*Solution:** Here,*

* The given points are (-2, 3)
and (2, 4)*

* **∴** x _{1} = –2 and y_{1} = 3*

* x _{2} = 2 and y_{2}
= 4*

* The equation
of the line is,*

* or, x + 2 = 4(y – 3)*

* or, x + 2 = 4y – 12 *

* or, x – 4y + 2 + 12 = 0*

* or, x – 4y + 14 = 0*

* Which is the
required equation of the line.*

*You can comment your
questions or problems regarding the point
slope form equation here.*

## No comments: