Point Slope Form

Equation of a straight line in terms of its slope (m) and a point (x₁, y₁) through which it passes is called the point slope form equation. It is written as

                    y – y₁ = m(x – x₁)

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To derive this equation, let us suppose a straight line AB whose slope is m and passes through the point C (x₁, y₁). 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 – y₁ = m(x – x₁)

which is the required equation in point slope form.

Corollary: If the straight line passes through the two given points (x₁, y₁) and (x₂, y₂) then its slope can be obtained by                 

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

                    y – y₁ = m (x – x₁)

This is called two point form equation.

 

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)

                   ∴ x₁ = 3    and    y₁ = 2

          The equation of the line is,

                   y – y₁ = m(x – x₁)

          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)

                   ∴ x₁ = –2     and     y₁ = 3

          The equation of the line is,

                   y – y₁ = m(x – x₁)

          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)

                   ∴ x₁ = –2     and      y₁ = 3

                     x₂ = 2       and      y₂ = 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.

 

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