We will learn how to find the equation of a straight line in two-point form or the equation of the straight line through two given points.

The equation of a line passing through two points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) is y - y\(_{1}\) = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)(x - x1)

Let the two given points be (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)).

We have to find the equation of the straight line joining the above two points.

Let the given points be A (x\(_{1}\), y\(_{1}\)), B (x\(_{2}\), y\(_{2}\)) and P (x, y) be any point on the straight line joining the points A and B.

Now, the slope of the line AB is \(\frac{y_{1}
- y_{2}}{x_{1} - x_{2}}\)

And the slope of the line AP is \(\frac{y - y_{1}}{x - x_{1}}\)

But the three points A, B and P are collinear.

Therefore, slope of the line AP = slope of the line AB

⇒ \(\frac{y - y_{1}}{x - x_{1}}\) = \(\frac{y_{1} - y_{2}}{x_{1} - x_{2}}\)

⇒ y - y\(_{1}\) = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) (x - x\(_{1}\))

The above equation is satisfied by the co-ordinates of any point P lying on the line AB and hence, represents the equation of the straight line AB.

Solved examples to find the equation of a straight line in two-point form:

**1.** Find the equation of the straight line
passing through the points (2, 3) and (6, - 5).

**Solution:**

The equation of the straight line passing through the points (2, 3) and (6, - 5) is

\(\frac{ y - 3}{ x + 2}\) = \(\frac{3 + 5}{2 - 6}\),[Using the form, \(\frac{y - y_{1}}{x - x_{1}}\) = \(\frac{y_{1} - y_{2}}{x_{1} - x_{2}}\)]

⇒ \(\frac{ y - 3}{ x + 2}\) = \(\frac{ 8}{ -4}\)

⇒ \(\frac{ y - 3}{ x + 2}\) = -2

⇒ y - 3 = -2x - 4

⇒ 2x + y + 1 = 0, which is the required equation

**2.** Find the equation of the straight line
joining the points (- 3, 4) and (5, - 2).

**Solution:**

Here the given two points are (x\(_{1}\), y\(_{1}\)) = (- 3, 4) and (x\(_{2}\), y\(_{2}\)) = (5, - 2).

The equation of a line passing through two points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) is y - y\(_{1}\) = [\(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)](x - x\(_{1}\)).

So the equation of the straight line in two point form is

y - y\(_{1}\) = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) (x - x\(_{1}\))

⇒ y - 4 = \(\frac{-2 - 4}{5 - (-3)}\)[x - (-3)]

⇒ y - 4 = \(\frac{ -6}{ 8}\)(x + 3)

⇒ y - 4 = \(\frac{ -3}{ 4}\)(x + 3)

⇒ 4(y - 4) = -3(x + 3)

⇒ 4y - 16 = -3x - 9

⇒ 3x + 4y - 7 = 0, which is the required equation.

**●**** The Straight Line**

**Straight Line****Slope of a Straight Line****Slope of a Line through Two Given Points****Collinearity of Three Points****Equation of a Line Parallel to x-axis****Equation of a Line Parallel to y-axis****Slope-intercept Form****Point-slope Form****Straight line in Two-point Form****Straight Line in Intercept Form****Straight Line in Normal Form****General Form into Slope-intercept Form****General Form into Intercept Form****General Form into Normal Form****Point of Intersection of Two Lines****Concurrency of Three Lines****Angle between Two Straight Lines****Condition of Parallelism of Lines****Equation of a Line Parallel to a Line****Condition of Perpendicularity of Two Lines****Equation of a Line Perpendicular to a Line****Identical Straight Lines****Position of a Point Relative to a Line****Distance of a Point from a Straight Line****Equations of the Bisectors of the Angles between Two Straight Lines****Bisector of the Angle which Contains the Origin****Straight Line Formulae****Problems on Straight Lines****Word Problems on Straight Lines****Problems on Slope and Intercept**

**11 and 12 Grade Math**

**From Straight line in Two-point Form to HOME PAGE**

**Didn't find what you were looking for? Or want to know more information
about Math Only Math.
Use this Google Search to find what you need.**

## New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.