Equation of Line: Two Point Form

Science > Mathematics > Coordinate Geometry > Straight Lines > Equation of Line: Two Point Form

In this article, we shall study to find the equation of a line using a two-point form.

Example 01:

Find the equation of a line passing through the points

• (5, 4) and (3, -2)

Solution:

Let the points be A(5, 4) ≡ (x₁, y₁) and (3, -2) ≡ (x₂, y₂)

By two-point form

∴ (y – 4)/(4 + 2) = (x – 5)/(5 – 3)

∴ (y – 4)/(6) = (x – 5)/(2)

∴ 2(y – 4) = 6 (x – 5)

∴ 2y – 8 = 6 x – 30

 ∴ 6x – 2y – 30 + 8 =0

∴ 6x – 2y – 22 =0

Dividing throughout by 2

∴ 3x –y – 11 =0

Ans: The equation of the line is 3x –y – 11 =0

• (2, -3) and (-4, 6)

Solution:

Let the points be A(2, -3) ≡ (x₁, y₁) and (-4, 6) ≡ (x₂, y₂)

By two-point form

∴ (y + 3)/(-3 – 6) = (x – 2)/(2 + 4)

∴ (y + 3)/(-9) = (x – 2)/(6)

∴ 6(y + 3) = -9(x – 2)

∴ 6y + 18 =  – 9x + 18

 ∴ 9x + 6y =0

Dividing equation throughout by 3

∴ 3x + 2y =0

Ans: The equation of the line is 3x + 2y =0

• (0, 0) and (6, 4)

Solution:

Let the points be A(0, 0) ≡ (x₁, y₁) and (6, 4) ≡ (x₂, y₂)

By two-point form

∴ (y – 0)/(0 – 4) = (x – 0)/(0 – 6)

∴ (y)/(-4) = (x)/(-6)

∴ – 6y = -4x

∴ 4x – 6y = 0

Dividing equation throughout by 2

∴ 2x – 3y =0

Ans: The equation of the line is 2x – 3y =0

• (3, 4) and (4, 3)

Solution:

Let the points be A(3, 4) ≡ (x₁, y₁) and (4, 3) ≡ (x₂, y₂)

By two-point form

∴ (y – 4)/(4 – 3) = (x – 3)/(3 – 4)

∴ (y – 4)/(1) = (x – 3)/(-1)

∴ y – 4 =-x + 3

∴ x + y -4 -3 = 0

∴ x + y – 7 = 0

Ans: The equation of line is x + y – 7 = 0

• (1, -2) and (2, 3)

Solution:

Let the points be A(1, -2) ≡ (x₁, y₁) and (2, 3) ≡ (x₂, y₂)

By two-point form

∴ (y + 2)/(-2 – 3) = (x – 1)/(1 – 2)

∴ (y + 2)/(-5) = (x – 1)/(-1)

∴ -1(y + 2) = -5(x – 1)

∴ – y – 2 = -5x + 5

∴ 5x – y – 2 – 5 = 0

∴ 5x – y – 7 = 0

Ans: The equation of the line is 5x – y – 7 = 0

Example 02:

Find the equation of a line passing through the points (-4, 6) and (8, -3) and also find the slope and its intercepts on coordinate axes.

Solution:

Let the points be A(-4, 6) ≡ (x₁, y₁) and (8, -3) ≡ (x₂, y₂)

By two-point form

∴ (y – 6)/(6 + 3) = (x + 4)/(-4 – 8)

∴ (y – 6)/(9) = (x + 4)/(-12)

∴ – 12(y – 6) = 9(x + 4)

∴ – 12y + 72 = 9x + 36

∴ 9x + 12y + 36 – 72 = 0

∴ 9x + 12y – 36 = 0

Dividing equation throughout by 3

∴ 3x + 4y – 12 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (3/4) = – 3/4

To find x-intercept of line substitute y = 0 in the equation of line

∴ 3x + 4(0) – 12 = 0

∴ 3x = 12

∴ x = 4

To find y-intercept of line substitute x = 0 in the equation of line

∴ 3(0) + 4 y -12 = 0

∴  4y = 12

∴  y = 3

Ans: The equation of line is 3x + 4y – 12 = 0, Its lope is – 3/4, x-intercept = 4, and y-intercept = 3

Example 03:

Find equation of a line passing through the points (2, 6) and (5, 1) and also find the slope and its intercepts on coordinate axes.

Solution:

Let the points be A(2, 6) ≡ (x₁, y₁) and (5, 1) ≡ (x₂, y₂)

By two-point form

∴ (y – 6)/(6 – 1) = (x – 2)/(2 – 5)

∴ (y – 6)/(5) = (x – 2)/( – 3)

∴ – 3(y – 6) = 5(x – 2)

∴ – 3y + 18 = 5x – 10

∴ 5x + 3y – 10 – 18 = 0

∴ 5x + 3y – 28 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (5/3) = – 5/3

To find x-intercept of line substitute y = 0 in the equation of line

∴ 5x + 3(0) – 28 = 0

∴ 5x = 28

∴ x = 28/5

To find y-intercept of line substitute x = 0 in the equation of line

∴ 5(0) + 3y – 28 = 0

∴  3y = 28

∴  y = 28/3

Ans: The equation of line is 5x + 3y – 28 = 0, Its lope is – 5/3, x-intercept = 28/5, and y-intercept = 28/3.

Example 04:

Find ‘k’ if the point (7, k) lies on a line passing through the points (3, 6) and (-5, -2).

Solution:

Let the points be A(3, 6) ≡ (x₁, y₁) and (-5, -2) ≡ (x₂, y₂)

By two-point form

∴ (y – 6)/(6 + 2) = (x – 3)/(3 + 5)

∴ (y – 6)/(8) = (x – 3)/(8)

∴ y – 6 = x  – 3

∴ x – y – 3 + 6 = 0

∴ x – y + 3 = 0  ………. (1)

Thus the equation of the line is x – y + 3 = 0

Now point (7, k) lies on this line

Substituting x = 7 and y = k in equation (1)

7 – k + 3 = 0

∴ k = 10

Example 05:

Find ‘k’ if the point (2, b) lies on a line passing through the points (5, 4) and (3, -2).

Solution:

Let the points be A(5, 4) ≡ (x₁, y₁) and (3, -2) ≡ (x₂, y₂)

By two-point form

∴ (y – 4)/(4 + 2) = (x – 5)/(5 – 3)

∴ (y – 4)/(6) = (x – 5)/(2)

∴ 2(y – 4) = 6 (x – 5)

∴ 2y – 8 = 6 x – 30

 ∴ 6x – 2y – 30 + 8 =0

∴ 6x – 2y – 22 =0

Dividing throughout by 2

∴ 3x –y – 11 =0 ………. (1)

Thus the equation of the line is 3x –y – 11 = 0

Now point (2, b) lies on this line

Substituting x = 2 and y = b in equation (1)

3(2) – b – 11 =0

6 – b – 11 =0

∴ b = – 5

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