Suppose that we are given the slope m of a line, and one point \(A\left( {{x_0},{y_0}} \right)\) lying on it:

Slope of a line

Clearly, the line is uniquely determined. How can we find the equation of the line in such a scenario?

Let \(P\left( {x,y} \right)\) be any point lying on this line. Observe the following figure carefully:

Point-slope form of a line’s equation

Clearly, \(\angle PAB = \theta \) , and so

\[\begin{align}&\tan \theta  = m = \frac{{PB}}{{AB}}\\&\Rightarrow \,\,\,m = \frac{{y - {y_0}}}{{x - {x_0}}}\\&\Rightarrow \,\,\,y - {y_0} = m\left( {x - {x_0}} \right)\end{align}\]

Thus, the equation of a line with slope m and passing through the point \(A\left( {{x_0},{y_0}} \right)\) is

\[y - {y_0} = m\left( {x - {x_0}} \right)\]

This is known as the point-slope form of a line’s equation, as we have used the slope of the line and a point lying on the line to specify it.

Example 1: A line is inclined at an angle of 300 to the horizontal, and passes through the point \(\left( {1, - 1} \right)\) . Determine its equation.

Solution: The following figure depicts this line:

Determining line's equation

We have:

\[m = \tan {30^0} = \frac{1}{{\sqrt 3 }}\]

Using the point-slope form, the equation of the line is

\[\begin{align}&y - {y_0} = m\left( {x - {x_0}} \right)\\&\Rightarrow \,\,\,y - \left( { - 1} \right) = \frac{1}{{\sqrt 3 }}\left( {x - 1} \right)\\&\Rightarrow \,\,\,y + 1 = \frac{1}{{\sqrt 3 }}\left( {x - 1} \right)\\&\Rightarrow \,\,\,\sqrt 3 y + \sqrt 3  = x - 1\\&\Rightarrow \,\,\,x - \sqrt 3 y - 1 - \sqrt 3  = 0\end{align}\]

Example 2: What is the y-intercept of the line which is inclined at an angle of 1500 with the horizontal and passes through the point \(\left( {2,0} \right)\)?

Solution: The following figure depicts this line:

Line inclined at 150 degrees

We have:

\[m = \tan {150^0} =  - \tan {30^0} =  - \frac{1}{{\sqrt 3 }}\]

The equation of the line using the point-slope form is

\[\begin{align}&y - {y_0} = m\left( {x - {x_0}} \right)\\&\Rightarrow \,\,\,y - 0 =  - \frac{1}{{\sqrt 3 }}\left( {x - 2} \right)\\&\Rightarrow \,\,\,y = \left( { - \frac{1}{{\sqrt 3 }}} \right)x + \left( {\frac{2}{{\sqrt 3 }}} \right)\end{align}\]

Clearly, the y-intercept is

\[c = \frac{2}{{\sqrt 3 }}\]

Download practice questions along with solutions for FREE:
Coordinate Geometry
Grade 9 | Questions Set 1
Coordinate Geometry
Grade 9 | Answers Set 1
Coordinate Geometry
Grade 10 | Questions Set 1
Coordinate Geometry
Grade 10 | Answers Set 1
Download practice questions along with solutions for FREE:
Coordinate Geometry
Grade 9 | Questions Set 1
Coordinate Geometry
Grade 9 | Answers Set 1
Coordinate Geometry
Grade 10 | Questions Set 1
Coordinate Geometry
Grade 10 | Answers Set 1
Learn math from the experts and clarify doubts instantly

  • Instant doubt clearing (live one on one)
  • Learn from India’s best math teachers
  • Completely personalized curriculum