Point-Slope Form

Point-Slope Form

In this chapter, you will learn about the point slope form equation, the slope of a line equation, the point-slope formula, and point-slope examples. Check-out the interactive simulations to know more about the lesson and try your hand at solving a few interesting practice questions on Point-Slope Form at the end of the page.

Now, that we're done with the formalities, here's a bit of info for you. Did you know that the equation of a line is an equation that is satisfied by each and every point on the line? This means that a linear equation in two variables represents a line.

Standard form of the equation of a line: A x plus B y is equal to C.

The equation of a line can be found through various methods depending on the available information.

Some of the methods are:

  • Point-slope form
  • Slope-intercept form
  • Intercept form
  • Two-point form

We will study the point-slope form on this page in detail.

Lesson Plan

What is Point-Slope Form?

The formula of point-slope form is used to find the equation of a line.

This formula is used only when we know the slope of the line and a point on the line.

The equation of a line whose slope is \(m\) and which passes through a point \((x_1, y_1\)) is found using the point-slope form.

The equation of the point-slope form is:

\({y-y_1=m(x-x_1)}\)

where \((x, y)\) is a random point on the line.

Point slope formula: y minus y1 is equal to m times x minus x1


How to Find Point-Slope Form?

Let us see how to find the point-slope form (i.e. the proof of the formula of point-slope form).

We will derive this formula using the equation for the slope of a line.

Let us consider a line whose slope is \(m\).

Let us assume that \((x_1, y_1\)) is a known point on the line.

Let \((x,y)\) be any other random point on the line whose coordinates are not known.

Proof of point slope formula: the difference in y coordinates is y minus y1; the difference in x coordinates is x minus x1

We know that the equation for the slope of a line is:

\[ \begin{aligned}\text{Slope }&= \dfrac{ \text{Difference in y-coordinates}}{ \text{Difference in x-coordinates}}\\[0.2cm]m&= \dfrac{y-y_1}{x-x_1} \end{aligned} \]

Multiplying both sides by \((x-x_1)\),

\[m(x-x_1) = y-y_1\]

This can be written as,

\[y-y_1=m(x-x_1)\]

Hence the point-slope formula is proved.


Examples of Point-Slope Form

Some examples of the point-slope form are shown here.

  1. The equation of a line with slope \(-1\) and a point \((1, 2\)) is found using: \[y-2=-1(x-1)\]
  2. The equation of a line with slope \( \dfrac{3}{2}\) and a point \( \left( \dfrac{-1}{2}, \dfrac{2}{3} \right)\) is found using: \[y- \dfrac{2}{3}= \dfrac{3}{2} \left(x- \left(- \dfrac{1}{2} \right) \right)\]
  3. The equation of a line with slope \(0\) and a point \((3, -2\)) is found using: \[y-(-2) =0(x-3)\]

In each of these cases, we can simplify the equation further and get it to the form \(y=mx+b\).


Point-Slope Form Calculator

Here is the point-slope form calculator.

Here, you can find the point-slope form of a line by changing the values of \(m, x_1\), and \(y_1\)

Also, you can see how the graph changes accordingly.

 
important notes to remember
Important Notes
  1. The equation of the point-slope form of a line whose slope is \(\mathbf{m}\) and that passes through a point \(\mathbf{(x_1, y_1)}\) is \( \mathbf{y-y_1=m(x-x_1)}\)
  2. The equation of a horizontal line passing through \(\mathbf{(a,b)}\) is of the form \(\mathbf{y=b}\)
  3. The equation of a vertical line passing through \(\mathbf{(a,b)}\) is of the form \(\mathbf{x=a}\)
    This is an exceptional case where the point-slope form cannot be used.

Solved Examples

We can find more examples of the point-slope form here.

Example 1

 

 

Find the equation of a line that passes through a point \((2, -3\)) and whose slope is \( \dfrac{-1}{2}\)

Solution:

The point on the given line is:

\[(x_1, y_1) = (2, -3)\]

The slope of the line is:

\[m=\dfrac{-1}{2}\]

The equation of the line is found using the point-slope form:

\[ \begin{aligned}y-y_1&=m(x-x_1)\\[0.2cm]y-(-3)&=\dfrac{-1}{2}(x-2)\\[0.2cm] y+3&= \dfrac{-1}{2}x+1\\[0.2cm] \text{Subtracting 3 } & \text{from both sides},\\[0.2cm] y&=\dfrac{-1}{2}x-2 \end{aligned}\]

Thus, the equation of the required line is

\( \therefore\) \({y=\dfrac{-1}{2}x-2}\)
Example 2

 

 

Find the equation of the following line.

Example of the point slope formula: a graph showing a vertical line

Solution:

We know that the slope of a vertical line is undefined, i.e. \(m=\infty\)

Thus, the equation of the given line can't be found using the point-slope form.

We will use the form listed under the "important notes" section of this page which is:

The equation of vertical line passing through \((a,b)\) is of the form \(x=a\)

Let us take any point on the line:

\[(a,b)=(4,1)\]

The equation of the line using the above form is,

\( \therefore\) \({x=4}\)
Example 3

 

 

Find the equation of a horizontal line that passes through a point \((3, 2)\)

Solution:

Method 1:

We know that the slope of a horizontal line is \(m=0\)

The line passes through the point \((x_1,y_1)=(3,2)\)

The equation of the line using the point-slope form is:

\[ \begin{aligned}y-y_1&=m(x-x_1)\\[0.2cm] y-2&=0(x-3)\\[0.2cm] y-2&=0\\[0.2cm] y&=2 \end{aligned}\]

Method 2:

From the formulas listed under the "important notes" section of this page, we know that:

The equation of a horizontal line passing through \((a,b)\) is of the form \(y=b\)

A point on the given line is,

\[(a,b)=(3, 2)\]

Thus, the equation of the required horizontal line is:

\[y=2\]

From both the methods, the equation of the given line is:

\( \therefore\) \({y=2}\)
Example 4

 

 

Find the equation of a line that passes through two points \((1, 3)\) and \((-2, 4)\) using the point-slope form and express the answer in the standard form.

Solution:

The given two points are:

\[ \begin{aligned} (x_1, y_1)&=(1,3)\\[0.2cm] (x_2,y_2)&=(-2, 4) \end{aligned}\]

To use the point-slope form, we first need to find the slope of the line.

The slope of the given line is found using:

\[ \begin{aligned} m &= \dfrac{y_2-y_1}{x_2-x_1}\\[0.2cm]&= \dfrac{4-3}{-2-1}\\[0.2cm]&= \dfrac{1}{-3}\\[0.2cm]&=- \dfrac{1}{3} \end{aligned}\] The equation of the line is found using the point-slope form. \[ \begin{aligned} y-y_1&=m(x-x_1)\\[0.2cm]y-3&=- \dfrac{1}{3}(x-1)\\[0.2cm] y-3&=- \dfrac{1}{3} x+ \dfrac{1}{3}\\[0.2cm] \text{Adding 3 }& \text{ on both sides},\\[0.2cm] y&= - \dfrac{1}{3}x+ \dfrac{10}{3} \end{aligned}\]

To express this in the standard form, first we multiply both sides by \(3\)

\[3y=-x+10\]

Adding \(x\) on both sides,

\[x+3y=10\]

Thus, the equation of the given line in the standard form is:

\( \therefore\) \({x+3y=10}\)
 
Thinking out of the box
Think Tank
  1. Which of the following graphs may represent the equation \(\mathbf{y+2= -2(x-1)}\)?

Question on the point-slope formula. There is a point (1,-2) on the line.Question on the point-slope formula. There is a point (1,-2) on the line.

Interactive Questions

Here are a few activities for you to practice. Select/type your answer and click the "Check Answer" button to see the result.

 
 
 
 
 
 

Let's Summarize

We hope you enjoyed learning about Point-Slope Form with the simulations and practice questions. Now you will be able to easily solve problems on Point Slope form equation, slope of a line equation and also how to find point-slope form.

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Frequently Asked Questions (FAQs)

1. How do you find the point-slope form with a point-slope?

The point-slope formula is a formula that is used to find the equation of a line.

This formula is used only when we know the slope of a line and a point on the line.

The equation of a line whose slope is m and which passes through a point \((x_1, y_1\)) is found using the point-slope formula.

The equation of the point-slope form is:

\( \mathbf{y-y_1=m(x-x_1)}\)

Here \((x, y)\) is a random point on the line.

2. How do you change point-slope form into slope-intercept form?

The point-slope form is of the form:

\[y-y_1=m(x-x_1)\]

We will solve this equation for \(y\) which gives an equation of the form \[y=mx+b\]

This is called the slope-intercept form.

Here is an example.

Example:

The equation of a line with slope -1 and a point \((1, 2\)) is found using: \[y-2=-1(x-1)\]

For changing this into the slope-intercept form, we will solve it for \(y\).

\[ \begin{align} y-2&=-x+1\\[0.2cm] y&= -x+1+2\\[0.2cm] y&=-x+3 \end{align}\]

3. How do you find the point-slope form of a graph?

To find the point-slope form of a line, we just find the slope and a point on the line.

A point on the line can be easily found by looking at its graph.

The slope of a line is found by first finding any two points on the line from its graph and then applying the formula:

\[\text{Slope }= \dfrac{ \text{Difference in y-coordinates}}{ \text{Difference in x-coordinates}}\]

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