In this mini-lesson, you will learn about the two-point form of a line.

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

The two-point form is one of the methods.

This is used to find the equation of a line when two points are given.

You can explore the two-point form using the following two-point form calculator.

Enter any two points on the line to get its equation using the two-point form.

**Lesson Plan**

**What Is the Equation of a Line in Two-Point Form? **

The equation of a line represents each and every point on the line, i.e., it is satisfied by each point on the line.

The two-point form of a line is used for finding the equation of a line given two points \((x_1,y_1)\) and \((x_2,y_2)\) on it.

The two-point form of a line passing through these two points is:

\[y-y_1= \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\](or)\[y-y_2= \dfrac{y_2-y_1}{x_2-x_1}(x-x_2)\] |

Here, \((x, y)\) represents any random point on the line and we keep \(x\) and \(y\) as variables.

**How Do You Find the Equation of a Line That Passes Through Two Points?**

How do we derive the two-point form of a line? Let's see.

Let us consider two fixed points \(A\,(x_1,y_1)\) and \(B\,(x_2,y_2)\) on the line.

Let us assume that \(C\,(x, y)\) is any random point on the line.

Since \(A, B\), and \(C\) lie on the same line:

Slope of \(\overleftrightarrow{AC}\) = Slope of \(\overleftrightarrow{AB}\)

Using the two point slope formula,

\[ \begin{align} \dfrac{y-y_1}{x-x_1}&= \dfrac{y_2-y_1}{x_2-x_1}\\[0.2cm] \text{Multiplying both }& \text{ sides by }(x-x_1),\\[0.2cm] y-y_1&= \dfrac{y_2-y_1}{x_2-x_1}(x-x_1) \end{align}\]

Hurray! We derived the two-point form. It is used to find the equation of a line that passes through two points.

Can you try deriving another formula of two-point form, which is, \(y-y_2= \dfrac{y_2-y_1}{x_2-x_1}(x-x_2)\) now?

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

\[\frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\] \[\text{(or)} \]\[\frac{y-y_{2}}{x-x_{2}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\] - T
**he equation of a vertical line passing through \(\mathbf{(a,b)}\) is of the form \(\mathbf{x=a}\).**

This is an exceptional case where the two-point form cannot be used.

**Solved Examples**

Example 1 |

Can we help Mary find the equation of a straight line passing through the points A (1, 2) and B (-1, 3)?

**Solution**

The following figure shows the line passing through the given points:

The given two points are: \[ A(1,2)=(x_1,y_1)\\[0.2cm] B(-1,3)= (x_2, y_2)\]

Since we know two points on the line, we use the two-point form to find its equation.

\[ \begin{align}y-y_1&= \dfrac{y_2-y_1}{x_2-x_1}(x-x_1) \\[0.2cm] y-2 &= \dfrac{3-2}{-1-1}(x-1) \\[0.2cm] y-2 &= \dfrac{1}{-2}(x-1)\\[0.2cm] \text{Multiplying } &\text{both sides by -2},\\[0.2cm] -2(y-2) &= x-1\\[0.2cm] -2y+4 &= x-1\\[0.2cm] x+2y&-5=0 \end{align}\]

Therefore, the equation of the line is,

\(\therefore\) \(x+2 y-5=0\) |

Example 2 |

Can we help Margarette find the *y*-intercept of the line passing through the points A (3, -2) and B (-1, 3)?

**Solution**

The following figure shows the line passing through the given points:

The given two points are: \[ A(3,-2)=(x_1,y_1)\\[0.2cm] B(-1,3)= (x_2, y_2)\]

Since we know two points on the line, we use the two-point form to find its equation.

\[ \begin{align}y-y_1&= \dfrac{y_2-y_1}{x_2-x_1}(x-x_1) \\[0.2cm] y+2 &= \dfrac{3+2}{-1-3}(x-3) \\[0.2cm] y+2 &= \dfrac{5}{-4}(x-3)\\[0.2cm] \text{Multiplying } &\text{both sides by -4},\\[0.2cm] -4(y+2) &= 5(x-3)\\[0.2cm] -4y-8 &= 5x-15\\[0.2cm] -4y&= 5x-7\\[0.2cm] y&=\left(-\frac{5}{4}\right) x+\frac{7}{4}\end{align}\]

The final equation is in the slope-intercept form, \(y=mx+b\).

Comparing the last two equations, we get the y-intercept to be

\(\therefore\) \(b=\dfrac{7}{4}\) |

Example 3 |

Find the equation of a straight line whose *x*-intercept is \(a\) and *y*-intercept is \(b\), as shown in the following figure.

**Solution**

The given line passes through the points: \[ A(a, 0)=(x_1,y_1)\\[0.2cm] B(0, b)= (x_2, y_2)\]

Since we know two points on the line, we use the two-point form to find its equation.

\[ \begin{align}y-y_1&= \dfrac{y_2-y_1}{x_2-x_1}(x-x_1) \\[0.2cm] y-0&= \dfrac{b-0}{0-a}(x-a) \\[0.2cm] y&= \dfrac{b}{-a}(x-a)\\[0.2cm] \text{Multiplying } &\text{both sides by }-a,\\[0.2cm] -ay &= b(x-a)\\[0.2cm] -ay&=bx-ab\\[0.2cm] bx+ay&=ab \\[0.2cm] \text{Dividing } &\text{both sides by }ab,\\[0.2cm] \frac{x}{a}+\frac{y}{b}&=1\end{align}\]

Thus, the equation of the given line is:

\(\therefore\) \(\frac{x}{a}+\frac{y}{b}=1\) |

Note: This is also called the intercept-form of a line.

- Which of the following graphs represents the equation \(y-2=\dfrac{1-2}{-2-1}(x-1)\)?

**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**

The mini-lesson targeted the fascinating concept of Two-Point Form. The math journey around Two-Point Form starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.

**About Cuemath**

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

**FAQs on Two-Point Form**

### 1. What is the two-point form?

The two-point form of a line is used for finding the equation of a line given two points \((x_1,y_1)\) and \((x_2,y_2)\) on it.

The two point-form of a line is:\[y-y_1= \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)\] \[\text{(or)} \]\[y-y_2= \dfrac{y_2-y_1}{x_2-x_1}(x-x_2)\]

### 2. How do you write an equation with two given points?

We substitute the points in the two-point form to find the equation.

For more information, refer to the "Solved Examples" section of this page.

### 3. How do you find the intercept form?

The intercept form of a line whose *x *and *y *intercepts are \(a\) and \(b\) respectively is \(\frac{x}{a}+\frac{y}{b}=1\).

To know how to find this, refer to Example 3 under the "Solved Examples" section of this page.

### 4. What is the example of a point slope form?

The equation of a line with slope, \(m= 1\) that passes through a point \((x_1,y_1)= (-2, 3)\) using the point-slope form is:

\[\begin{align} y-y_1&=m(x-x_1)\\[0.2cm]

y-3&=1(x+2) \\[0.2cm]

y&=x+5 \end{align}\]

### 5. What is the Y intercept in an equation?

To find the y-intercept in an equation, substitute x = 0 and solve for y.

### 6. How do you find the Y intercept with two points and slope?

- First, find the equation of the line using the two-point form and solve it for \(y\).
- Compare it with \(y=mx+b\)
- Here, \(b\) is the y-intercept.

To know more about this, refer to Example 2 under the "Solved Examples" section of this page.

### 7. What is the normal form of a line?

The normal form of a line is \(x \cos \alpha+y \sin \alpha=p\).

Here, \( \alpha\) is the angle made by the line with the positive direction of the x-axis and

\(p\) is the perpendicular distance of the line from the origin.

### 8. How do you find the equation of a line with only one point?

We can't find the equation of a line just with one point.

To find the equation of a line one of the following information should be available.

- Two points on a line.
- One point and the slope of a line.
- One point and one intercept of a line.
- Two intercepts of a line.

### 9. How do you determine if a point lies on a line?

Every point on a line satisfies its line equation.

For example, to see whether (1, 2) lies on a line \(y= 2x\), we substitute \(x=1\) and \(y=2\) in the given equation. Then we get:

\[ 2= 2(1)\\[0.1cm] 2=2\]

The equation is satisfied and hence the point (1, 2) lies on the line \(y=2x\).

### 10. How do you find the slope of a line with two given points?

To find the slope of a line with two given points:

- First find the equation of line using the two-point form and solve it for \(y\).
- Compare it with \(y=mx+b\)
- Here, \(m\) is the slope of the line.