# Distance between Two Points

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**Definition of Distance between Two Points**

The distance between any two points is the length of the line segment joining the points.

For example, if \(A\) and \(B\) are two points and if \(\overline{AB}=10\) cm, it means that the distance between \(A\) and \(B\) is \(10\) cm.

The distance between two points is the length of the line segment joining them (but this CANNOT be the length of the curve joining them).

How do we find the distance between two points if their coordinates are given?

Let's learn more about this.

**Formula for Distance Between Two Points **

The formula for the distance, \(d\), between two points whose coordinates are \((x_1, y_1)\) and \((x_2, y_2\)) is:

\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] |

This is called the** Distance Formula.**

Let's learn how to derive this formula next.

**Proof of Distance Formula **

Let us assume that:

\[A = (x_1, y_1)\\[0.2cm]B=(x_2, y_2)\]

Next, we will assume that \(\overline{AB}=d\)

Now, we will plot the given points on the coordinate plane and join them by a line.

Next, we will construct a right-angled triangle with \(\overline{AB}\) as the hypotenuse.

Applying Pythagoras theorem for the \(\triangle ABC\):

\[ \begin{aligned} AB^2 &\!=\! AC^2\!+\!BC^2\\d^2 &\!=\! (x_2\!-\!x_1)^2\!+\!(y_2\!-\!y_1)^2 \,\, \text{[Values from the figure] }\\d &\!=\! \sqrt{(x_2\!-\!x_1)^2\!+\!(y_2\!-\!y_1)^2} \, [ \text{Taking square root on both sides} ] \end{aligned} \]

Thus, the distance formula is proved.

- The distance, \(d\), between two points whose coordinates are \((x_1, y_1)\) and \((x_2, y_2\)) is:

\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] - Distance of a point \((a,b)\) from:

(i) x - axis is \(|b|\)

(ii) y - axis is \(|a|\)

Here, we have used the absolute value signs because distance can never be negative.

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**Solved Examples**

Example 1 |

Find the distance between the two points \((2, -6\)) and \((7, 3\))

**Solution:**

Let us assume the given points to be:

\[ \begin{aligned} (x_1,y_1)&=(2, -6)\\[0.2cm](x_2, y_2)&= (7,3) \end{aligned} \]

The formula to find the distance between two points is:

\[ \begin{aligned} d &=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\[0.3cm] d &= \sqrt{(7-2)^2+(3-(-6))^2}\\[0.3cm] d&= \sqrt{5^2 + 9^2}\\[0.3cm] d&= \sqrt{25+81}\\[0.3cm] d &= \sqrt{106} \end{aligned} \]

\( \therefore\) Distance \(= \sqrt{106} \) |

Example 2 |

Show that the points \((2, -1), (0, 1)\) and \((2, 3\)) are the vertices of a right-angled triangle.

**Solution:**

Let us assume the given points to be:

\[ \begin{aligned} A &=(2,-1)\\[0.2cm] B&=(0,1)\\[0.2cm] C&=(2,3)\\[0.2cm] \end{aligned} \]

We will find the distance between every two points using the **distance formula.**

\[\begin{aligned} AB&\! =\! \sqrt{(0-2)^2+(1-(-1)^2}\!=\! \sqrt{(-2)^2+(2)^2}\!=\! \sqrt{4+4}\!=\! \sqrt{8}\\ BC &\!=\! \sqrt{(2-0)^2+(3-1)^2}\!=\! \sqrt{(2)^2+(2)^2}\!=\! \sqrt{4+4}\!=\! \sqrt{8}\\ CA &\!=\! \sqrt{(2-2)^2+(3-(-1))^2}\! =\! \sqrt{0^2+4^2}\!=\! \sqrt{16}\!=\!4\end{aligned} \]

Now that we know the lengths of all three sides,

\[ \begin{aligned} AB^2+BC^2 &= CA^2 \\[0.3cm] (\sqrt{8})^2 +(\sqrt{8})^2 &= 4^2 \\[0.3cm] 8+8&=16\\[0.3cm] 16&=16 \end{aligned} \]

Thus, \(A, B\) and \(C\) satisfy the Pythagoras theorem.

So \(\triangle ABC\) is a right-angled triangle.

We can prove the same by marking all the coordinates on a graph:

Thus,

The given points form a right-angled triangle. |

Example 3 |

Find a point on the y-axis that is equidistant from the points \((-1, 2\)) and \((2, 3)\)

**Solution:**

We know that the x-coordinate of any point on the y-axis is \(0\)

Hence, we assume the point that is equidistant from the given points to be \((0, k\)). i.e.,

Distance between \((0, k)\) and \((-1, 2)\) \(=\) Distance between \((0, k)\) and \((2, 3)\)

\[ \begin{aligned} \sqrt{(-1-0)^2+(2-k)^2}&= \sqrt{(2-0)^2+(3-k)^2}\\ \end{aligned} \]

Squaring on both sides

\[ \begin{aligned}(-1-0)^2+(2-k)^2 &=(2-0)^2+(3-k)^2 \\1 + 4+k^2-4k &= 4+ 9 +k^2-6k \\2k &=8 \\ k&=4 \end{aligned} \]

Therefore, the required point is, \((0, k)=(0, 4)\)

\( \therefore\) Required Point \(= (0, 4)\) |

**Distance between Two Points Calculator**

Here is the "Distance Between Two Points Calculator".

Here, you can enter the coordinates of two points and it will show you the distance between them with the step-by-step explanation.

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**Practice Questions**

- The vertices of a rectangle are \((-4, -3), (4, -3), (4, 3),\) and \((-4, 3)\). What is its area?
- The vertices of a right-angled triangle are \((-3, 6), (1, 6)\) and \((1, -1)\). What is its area?

**Maths Olympiad Sample Papers**

IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. It encourages children to develop their math solving skills from a competition perspective.

You can download the FREE grade-wise sample papers from below:

- IMO Sample Paper Class 1
- IMO Sample Paper Class 2
- IMO Sample Paper Class 3
- IMO Sample Paper Class 4
- IMO Sample Paper Class 5
- IMO Sample Paper Class 6
- IMO Sample Paper Class 7
- IMO Sample Paper Class 8
- IMO Sample Paper Class 9
- IMO Sample Paper Class 10

To know more about the Maths Olympiad you can **click here**

**Frequently Asked Questions (FAQs)**

## 1. What is the definition of the distance between two points?

The distance between any two points is the length of the line segment joining the points.

For example, if \(A\) and \(B\) are two points and if \(\overline{AB}=10\) cm, it means that the distance between \(A\) and \(B\) is \(10\) cm.

## 2. What is the formula for the distance between two points?

The distance, \(d\), between two points whose coordinates are \((x_1, y_1)\) and \((x_2, y_2\)) is:

\[ d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \] |

This is called the** distance formula.**