How do you find the equation of a line that passes through two points?

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

11 and 12 Grade Math

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In the last lesson, I showed you how to get the equation of a line given a point and a slope using the formula

Anytime we need to get the equation of a line, we need two things

a slope

ALWAYS!

So, what do we do if we are just given two points and no slope?

No problem -- we'll just use the two points to pop the slope using this guy:

Check it out:

Let's find the equation of the line that passes through the points

This one's a two-stepper...

STEP 1:  Find the slope

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If you know two points that a line passes through, this page will show you how to find the equation of the line.

( , )
Example: (3,2)

and the other point...

( , )
Example: (7,-4)

First, let's see it in action. Here are two points (you can drag them) and the equation of the line through them. Explanations follow.

../geometry/images/geom-line-equn.js

The Points

We use Cartesian Coordinates to mark a point on a graph by how far along and how far up it is:

Steps

There are 3 steps to find the Equation of the Straight Line :

• 1. Find the slope of the line
• 2. Put the slope and one point into the "Point-Slope Formula"
• 3. Simplify

Step 1: Find the Slope (or Gradient) from 2 Points

What is the slope (or gradient) of this line?

We know two points:

• point "A" is (6,4) (at x is 6, y is 4)
• point "B" is (2,3) (at x is 2, y is 3)

The slope is the change in height divided by the change in horizontal distance.

Looking at this diagram ...

Slope m  =  change in ychange in x  =  yA − yBxA − xB

In other words, we:

• subtract the Y values,
• subtract the X values
• then divide

Like this:

m  =   change in y change in x   =   4−3 6−2   =   1 4 = 0.25

It doesn't matter which point comes first, it still works out the same. Try swapping the points:

m  =   change in y change in x   =   3−4 2−6   =   −1 −4 = 0.25

Same answer.

Step 2: The "Point-Slope Formula"

Now put that slope and one point into the "Point-Slope Formula"

Start with the "point-slope" formula (x1 and y1 are the coordinates of a point on the line):

y − y1 = m(x − x1)

We can choose any point on the line for x1 and y1, so let's just use point (2,3):

y − 3 = m(x − 2)

We already calculated the slope "m":

m = change in ychange in x = 4−36−2 = 14

And we have:

y − 3 = 14(x − 2)

That is an answer, but we can simplify it further.

Step 3: Simplify

Start with:y − 3 = 14(x − 2)

Multiply 14 by (x−2):y − 3 = x4 − 24

Add 3 to both sides:y = x4 − 24 + 3

Simplify:y = x4 + 52

And we get:

y = x4 + 52

Which is now in the Slope-Intercept (y = mx + b) form.

Check It!

Let us confirm by testing with the second point (6,4):

y = x/4 + 5/2 = 6/4 + 2.5 = 1.5 + 2.5 = 4

Yes, when x=6 then y=4, so it works!

Another Example

Example: What is the equation of this line?

Start with the "point-slope" formula:

y − y1 = m(x − x1)

Put in these values:

• x1 = 1
• y1 = 6
• m = (2−6)/(3−1) = −4/2 = −2

And we get:

y − 6 = −2(x − 1)

Simplify to Slope-Intercept (y = mx + b) form:

y − 6 = −2x + 2

y = −2x + 8

DONE!

The Big Exception

The previous method works nicely except for one particular case: a vertical line:

A vertical line's gradient is undefined (because we cannot divide by 0):

m = yA − yBxA − xB = 4 − 12 − 2 = 30 = undefined

But there is still a way of writing the equation: use x= instead of y=, like this:

x = 2

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Equation of a Straight Line Straight Line Graph Calculator Algebra Index

What is the formula for equation of a line passing through a point?