# Equation of a line given two points

This online calculator finds the equation of a line given two points on that line, in slope-intercept and parametric forms

You can find an equation of a straight line given two points laying on that line. However, there exist different forms for a line equation. Here you can find two calculators for an equation of a line:

• first calculator finds the line equation in slope-intercept form, that is,
$y=ax+b$
It also outputs slope and intercept parameters and displays the line on a graph.

• second calculator finds the line equation in parametric form, that is,
$x=at+x_0\\y=bt+y_0$
It also outputs a direction vector and displays line and direction vector on a graph.

Also, the text and formulas below the calculators describe how to find the equation of a line from two points manually.

#### Second point

Line equation

Slope

Intercept

Digits after the decimal point: 2

#### Second point

Equation for x

Equation for y

Direction vector

Digits after the decimal point: 2

## How to find the equation of a line in slope-intercept form

Let's find slope-intercept form of a line equation from the two known points $(x_0, y_0)$ and $(x_1, y_1)$.
We need to find slope a and intercept b.
For two known points we have two equations in respect to a and b
$y_0=ax_0+b\\y_1=ax_1+b$

Let's subtract the first from the second
$y_1 - y_0=ax_1 - ax_0+b - b\\y_1 - y_0=ax_1 - ax_0\\y_1 - y_0=a(x_1 -x_0)$
And from there
$a=\frac{y_1 - y_0}{x_1 -x_0}$

Note that b can be expressed like this
$b=y-ax$
So, once we have a, it is easy to calculate b simply by plugging $x_0, y_0, a$ or $x_1, y_1, a$ to the expression above.

Finally, we use the calculated a and b to write the result as
$y=ax+b$

### Equation of a vertical line

Note that in the case of a vertical line, the slope and the intercept are undefined because the line runs parallel to the y-axis. The line equation, in this case, becomes $x=x_1$

### Equation of a horizontal line

Note that in the case of a horizontal line, the slope is zero and the intercept is equal to the y-coordinate of points because the line runs parallel to the x-axis. The line equation, in this case, becomes $y=y_1$

### How to find the slope-intercept equation of a line example

Problem: Find the equation of a line in the slope-intercept form given points (-1, 1) and (2, 4)
Solution:

1. Calculate the slope a:
$a=\frac{y_1 - y_0}{x_1 -x_0} = \frac{4 - 1}{2 - (-1)} = \frac{3}{3} = 1$
2. Calculate the intercept b using coordinates of either point. Here we use the coordinates (-1, 1):
$b=y_0 - a x_0 = 1 - 1\cdot(-1)=2$
3. Write the final line equation (we omit the slope, because it equals one):
$y=x+2$

And here is how you should enter this problem into the calculator above: slope-intercept line equation example

## Parametric line equations

Let's find out parametric form of a line equation from the two known points $(x_0, y_0)$ and $(x_1, y_1)$.
We need to find components of the direction vector also known as displacement vector.
$D=\begin{vmatrix}d_1\\d_2\end{vmatrix}=\begin{vmatrix}x_1-x_0\\y_1-y_0\end{vmatrix}$
This vector quantifies the distance and direction of an imaginary motion along a straight line from the first point to the second point.

Once we have direction vector from $x_0, y_0$ to $x_1, y_1$, our parametric equations will be
$x=d_1t+x_0\\y=d_2t+y_0$
Note that if $t = 0$, then $x = x_0, y = y_0$ and if $t = 1$, then $x = x_1, y = y_1$

### Equation of a vertical line

Note that in the case of a vertical line, the horizontal displacement is zero because the line runs parallel to the y-axis. The line equations, in this case, become
$x=x_0\\y=d_2t+y_0$

### Equation of a horizontal line

Note that in the case of a horizontal line, the vertical displacement is zero because the line runs parallel to the x-axis. The line equations, in this case, become
$x=d_1t+x_0\\y=y_0$

### How to find the parametric equation of a line example

Problem: Find the equation of a line in the parametric form given points (-1, 1) and (2, 4)
Solution:

1. Calculate the displacement vector:
$D=\begin{vmatrix}d_1\\d_2\end{vmatrix}=\begin{vmatrix}x_1-x_0\\y_1-y_0\end{vmatrix}=\begin{vmatrix}2-(-1)\\4-1\end{vmatrix}=\begin{vmatrix}3\\3\end{vmatrix}$
2. Write the final line equations:
$x=3t-1\\y=3t+1$
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