Straight line in Two-point Form

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

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