# Two Point Form – Definition, Derivation, Formula, Examples

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## What Is Two Point Form in Math?

Two point form in math is one of the important methods to find the equation of a line when coordinates of any two points on the line are known. The equation of a line is used to represent each and every point on the line, we can say that it is satisfied by each point on the line.

Let $A(x_1,\; y_1)$ and $B (x_2,\; y_2)$ be any two distinct points on the line l.

The two-point form is given by

$\frac{y\;-\;y_1}{x\;-\;x_1} = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}$

### Two Point Form: Definition

If $(x_1,\; y_1)$ and $(x_2,\; y_2)$ be the coordinates of any two distinct points on the line, then the equation of line in two-point form is

$\frac{y\;-\;y1}{x\;-\;x_1} = \frac{y2\;-\;y_1}{x_2\;-\;x_1}$

OR

$(y\;-\;y_2) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_2)$

## Two Point Form Formula

The two points formula of two point form of an equation is given as-

Let $(x_1,\; y_1)$ and $(x_1,\; y_1)$ be the two points such that the equation of line passing through these two points is given by the formula mentioned below-

$(y\;-\;y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

Here, $(x,\; y)$ is just an arbitrary point on the line.

## Derivation of Two Point Form Formula

Let $M (x_1,\; y_1)$ and $N (x_2,\; y_2)$ be the two given points on the line L.

Let $P (x,\; y)$ be a random point on the line L.

From the figure, we can observe that the three points $P_1,\; P_2$ and P lie on the same line. They are collinear.

Slope of line MP $=$ Slope of line NP

$\frac{y\;-\;y1}{x\;-\;x1} = \frac{y2\;-\;y1}{x2\;-\;x1}$

Or

$(y\;-\;y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

## Equation of a Line in Two-Point Form

The equation of a line in the two-point form is given by

$(y\;-\;y1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

or

$(y\;-\;y_2) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_2)$

where

$(x,\; y)$: coordinates of any point on a line

$(x_1,\; y_1)$ and $(x_2,\; y_2)$: coordinates of two points on the line

## How To Write Equation of a Line in Two Point Form

Let’s understand the steps to find the equation of a line using two-point form.

• Name the coordinates of the given two points as $(x_1,\; y_1)$ and $(x_2,\; y_2)$.
• Substitute the values in the Two-Point formula given by

$(y \;-\; y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

• Simplify.

## Finding Equation of a Line Using Two Point Form

To write equation of a line in two-point form, simply substitute the coordinates of the given two points in the equation $(y\;-\;y_2) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_2)$.

Example: Find the equation of a line passing through the points $(1,\; 2)$ and $(3,\; 4)$.

Here, $(x_1,\; y_1) = (1,\; 2)$

$(x_2,\; y_2) = (3,\; 4)$

Substitute the values in $(y\;-\;y_2) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_2)$.

$(y\;-\;4) = \frac{4\;-\;2}{3\;-\;1} (x\;-\;3)$

$(y\;-\;4) = (x\;-\;3)$

$y = x + 1$

## Deriving Two Point Form Using Point Slope Form

Let $A (x_1,\; y_1)$ and $B (x_2,\; y_2)$ be two given distinct points on a line as given below:

The two-point slope form of the straight line passing through these points is given by “m”:

$m = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}$ __________(1)

The Point-Slope form equation of a line is given by:

$(y\;-\;y_1) = m (x\;-\;x_1)$

Substituting 1 in 2:

$(y\;-\;y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

## Facts about Two Point Form!

• The two-point form of a line can also be written as given:

$\frac{y\;-\;y_1}{x\;-\;x_1} = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}$

Or

$\frac{y\;-\;y_2}{x\;-\;x_2} = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}$

• An exceptional case where the two-point form cannot be used is the equation of a vertical line passing through point (a, b) given by $x = a$. Slope of the vertical line is not defined.

## Conclusion

In this article, we learned about two-point forms of a line. Let us now look at some examples and practice some problems to understand the concept better.

## Solved Examples On Two Point Form

1. Find the equation of the line passing through the points:

$(\;-\;2,\; 3)$ and $(3,\; 5)$.

Solution:

The given points are:

$(x_1,\; y_1) = (2,\; 3)$ and $(x_2,\; y_2) = (3,\; 5)$

The equation of a straight line passing through the points $(x_1,\; y_1)$ and $(x_2,\; y_2)$ can be written using two-point form as:

$(y\;-\;y_2)= \frac{y_2\;-\;y_1}{x_2\;-\;x_1}(x\;-\;x_2)$

Substituting the values, we get:

$(y\;-\;3) = \frac{5\;-\;3}{3\;-\;(\;-\;2)}(x\;-\;2)$

$(y\;-\;3)= \frac{2}{5} (x\;-\;2)$

$5 (y\;-\;3) = 2 (x\;-\;2)$

$5y \;–\; 15 = 2x + 4$

$2x + 4 \;–\; 5y + 15 = 0$

$2x \;–\; 5y + 19 = 0$

Therefore, this is the required equation of a line passing through the given points.

2. What is the equation of a straight line passing through the points (3, 0) and (0, 3)?

Solution:

Let’s write the coordinates as

$(x_1,\; y_1) = (3,\; 0)$ and $(x_2,\; y_2) = (0, 3)$

The equation of a straight line passing through the points $(x_1,\; y_1)$ and $(x_2,\; y_2)$ can be written as:

$(y\;-\;y2) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_2)$

Substituting the values, we get

$(y\;-\;0) = \frac{3\;-\;0}{0\;-\;3} (x\;-\;3)$

$y = \frac{3\;-\;}{3} (x\;-\;3)$

$y =\;-\;1(x \;–\; 3)$

$y =\;-\;x + 3$

$x + y \;–\; 3 = 0$

Therefore, the equation of a line passing through the given points $(3,\; 0)$ and $(0,\; 3)$ is $x + y \;–\; 3 = 0$

3. Find the equation of a line with two points $(3,\; 5)$ and $(2,\; 3)$.

Solution:

Given : Two points on the straight line are $(3,\; 5)$ and $(2,\; 3)$.

The equation of a straight line passing through the points $(x_1,\; y_1)$ and $(x_2\;, y_2)$ can be written as:

$(y \;-\; y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

Substitute $(x_1,\; y_1) = (3,\; 5)$ and $(x_2,\; y_2) = (2,\; 3)$

$(y\;-\;5) = \frac{3\;-\;5}{2\;-\;3} (x\;-\;3)$

$(y\;-\;5) =\frac{\;-2}{\;-1} (x\;-\;3)$

$y\;-\;5 = 2(x\;-\;3)$

$y\;-\;5 = 2x\;-\;6$

$2x\;-\;y = 6\;-\;5$

$2x\;-\;y = 1$

$2x \;-\; y \;-\;1 = 0$

4. Find the equation of a straight line whose x-intercept is “a” and y-intercept is “b.”

Solution: The line passes through the points $(a,\; 0)$ and $(0,\; b)$.

Substituting the values $(x_1,\; y_1) = (a,\; 0)$ and $(x_2,\; y_2) = (0,\; b)$ in the two point form we get

$(y\;-\;y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

$(y\;-\;0) = \frac{b\;-\;0}{0\;-\;a} (x\;-\;a)$

$y = \frac{b}{-a} (x\;-\;a)$

$-ay = b(x \;-\; a)$

$-ay = bx \;-\; ba$

$bx + ay = ab$

Dividing both sides by ab, we get:

$\frac{bx + ay}{ab} = \frac{ab}{ab}$

$\frac{x}{a} + \frac{y}{a} = 1$

Thus, the equation of the given line is given as:

$\therefore \frac{x}{a} + \frac{y}{a} = 1$

This is also known as the intercept-form of a line.

5. Derive the y-intercept of the line with the coordinates given by: $A\; (3, \;-\;2)$ and B $(\;-\;1,\; 3)$ passing through it and also find the slope m of the line.

Solution:

Let the given points be:

$(x_1,\; y_1) = ( 3,\;-\;2)$

$(x_2,\; y_2) = (\;-\;1, 3)$

The equation of a straight line passing through the points $(x_1,\; y_1)$ and $(x_2,\; y_2)$ can be written as:

$(y \;-\; y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$

$(y\;-\;(\;-\;2)) = \frac{3\;-\;(\;-\;2)}{\;-\;1\;-\;3} (x\;-\;(\;-\;1))$

$(y + 2) = \frac{5}{-\;4} (x + 1)$

Now, multiplying both sides by $-4$ gives us,

$\;-4 (y + 2) = 5 (x + 1)$

$\;-4y \;–\; 8 = 5x + 5$

$\;-4y = 5x +13$

$5x +4y +13 =0$

$y = \frac{-5}{4}x + \frac{13}{4}$

The line equation from two points is given above.

The final equation of the slope-intercept form is written as-

$y = mx + b$

Comparing our equation to the standard form, we will get y-intercept, b as $\frac{7}{4}$.

Further, the slope of the line (m) can be given as $\frac{-5}{4}$.

## Practice Problems On Two Point Form

1

### What is the equation of the line in the slope point form?

$ax + by + c = 0$
$y = mx + c$
$(y\;-\;y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$
$(y\;-\; y_1) = (x\;-\;x_1)$
CorrectIncorrect
Correct answer is: $(y\;-\;y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_1)$
The equation of a straight line passing through the points $(x_1,\; y_1)$ and $(x_2,\; y_2)$ can be written using two-point form as $(y\;-\;y_1) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1}(x\;-\;x_1)$
2

### What is the equation of a straight line passing through the points (5, 0) and (0, 5)?

$x\;-\;y\;–\;5 = 0$
$x + y\;–\;5 = 0$
$x + y + 5 = 0$
$\;-\;x + y – 5 = 0$
CorrectIncorrect
Correct answer is: $x + y\;–\;5 = 0$
Substitute $(x_1,\; y_1) = (5,\; 0)$ and $(x_2,\; y_2) = (0,\; 5)$ in the two point form.
$(y\;-\;y_2) = \frac{y_2\;-\;y_1}{x_2\;-\;x_1} (x\;-\;x_2)$
Substituting the values, we get;
$(y\;-\;0) = \frac{5\;-\;0}{0\;-\;5} (x\;-\;5)$
$y = \frac{5}{-5} (x\;-\;5)$
$y = (\;-\;1)(x \;–\; 5)$
$y = \;-x + 5$
$x + y \;– 5 = 0$
Therefore, the equation of a line passing through the given points $(5,\; 0)$ and $(0,\; 5)$ is $x + y \;–\; 5 = 0$
3

### What is the slope of a line passing through the points $(\;-5,\; 4)$ and $(3,\; -2)$.

$\frac{-3}{4}$
$\frac{3}{4}$
$\frac{-4}{3}$
$\frac{4}{3}$
CorrectIncorrect
Correct answer is: $\frac{-3}{4}$
The slope of a straight line passing through the points $(x_1,\; y_1)$ and $(x_2,\; y_2)$ can be written as:
Slope $= \frac{y_2\;-\;y_1}{x_2\;-\;x_1}$
Substitute $(x_1,\; y_1) = (\;-\;5,\; 4)$ and $(x_2,\; y_2) = (3,\; \;-\;2)$
Slope $= \frac{-2\;-\;4}{3 + 5} = \frac{-3}{4}$
4

### What is the equation of a vertical line passing through the point $A (\;-3,\; \;-5)$?

$y = \;-5$
$x = 3$
$x = \;-3$
$\;-3x\;-\;5y = 0$
CorrectIncorrect
Correct answer is: $x = \;-3$
The equation of a vertical line passing through the point $(a,\; b)$ is $x = a$.
The equation of a vertical line passing through the point $A\; (\;-3,\; -5)$ is $x = \;-3$.

## Frequently Asked Questions On Two Point Form

Every point on a line satisfies its line equation. For example, to see whether (3, 6) lies on a line $y = 2x$, we substitute $x = 3$ and $y = 6$ in the given equation. Then we get: $6 = 2(3)$ or, $6 = 6$. The equation is satisfied and hence the point (3, 6) lies on the line $y = 2x$.

The equation of line in point slope form is given by $y \;-\; y_1 = m (x \;-\; x_1)$.

The equation of x-axis is $y = 0$.

The equation of y-axis is x = 0.