Distance Formula: Derivation, Examples and Applications

Distance Formula: The shortest path between two points or places is the distance between those places or points. You can either use a scale or measure tape to measure this distance, however, this won’t always be the best method of measuring the distance. Therefore, we can use the Distance Formula when two points’ coordinates are given. In this article, let’s discuss the Distance Formula in detail.

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What is Distance Formula?

The Distance formula is a formula used in finding the distance between two points, which can be represented as \(\left({{x_1},\, {y_1}} \right)\) and \(\left({{x_2},\, {y_2}} \right)\).
The distance formula is derived with the help of Pythagoras theorem, that is \(c = \sqrt { {a^2} + {b^2}} \) where \(c\) is the longest side (the hypotenuse) of a right triangle and \(a\) and \(b\) are the other shorter sides (base and perpendicular).

Distance of a Point from the Co-ordinate Axes

To locate the position of a point on a plane, we require a pair of coordinate axes. The distance of a point from the \(Y\)-axis is called its \(x\)-coordinate or abscissa. The distance of a point from the \(X\)-axis is called its \(y\)-coordinate or ordinate.

The coordinates of a point on the \(X\)-axis are of the form \((x, 0)\), and those of a point on the \(Y\)-axis are of the form \((0, y)\). Hence, the distance of a point \((a, 0)\) from the \(Y\)-axis is \(a\) units, and the distance of a point \((0,b)\) from the \(X\)-axis is \(b\) units. And, the distance of a point \((a,b)\) from the \(X\)-axis is \(b\) units, and from the \(Y\)-axis is \(a\) units.

Distance Formula: Coordinate Geometry

Let us consider the following example:
A town \(Q\) is located \(4\, {\text{km}}\) east and \(3\, {\text{km}}\) north of town \(P\). How would you find the distance from town \(P\) to town \(Q\) without actually measuring it?
This situation can be represented graphically as shown in the figure.

You may use the Pythagoras theorem to calculate this distance. That is, \(PQ = \sqrt {{4^2} + {3^2}}  = \sqrt {16 + 9}  = \sqrt {25}  = 5\,{\text{units}}\)

Now, suppose two points lie on the \(X\)-axis. Can we find the distance between them? For instance, consider two points \(P(2,0)\) and \(Q\left({4,\,0} \right)\) in the figure. Both the points \(P\) and \(Q\) lie on the \(X\)-axis.

From the figure, you can see that \({{OP=2}}\,{\text{units}}\) and \({{OQ=4}}\,{\text{units}}\). Therefore, the distance of \(Q\) from \(P\), i.e., \(PQ = OQ – OP = 4 – 2 = 2\,{\rm{units}}.\)

So, if two points lie on the \(x\)-axis, we can easily find the distance between them. Now, let us take two points lying on the \(y\)-axis. Can you find the distance between them? If the points \(R\left({0,\,3} \right)\) and \(S\left({0,\,5} \right)\) lie on the \(y\)-axis, similarly, we find that \(RS = 5 – 3 = 2\,{\text{units}}\).

Next, can you find the distance of \(P\) from \(R\)? Since \(OP = 2\,{\text{units}}\) and \(OR = 3\), the distance of \(P\) and \(R\), i.e., \(P {R^2} = {2^2} + {3^2} = \sqrt {13} \, {\text{units}}\).
Similarly, you can find the distance of \(Q\) from \(S = \) \(QS = \sqrt { {4^2} + {5^2}} = \sqrt {41} \, {\text{units}}\).
If we consider two points not lying on the coordinate axis, then we can find the distance between them.

Distance Between Any Two Points on the Coordinate Plane

Let us now find the distance between any two points \(A\left({{x_1},\,{y_1}} \right)\) and \(B\left({{x_2},\,{y_2}}\right)\). Draw \(AC\) and \(BD\) perpendicular to the \(X\)-axis. A perpendicular from the point \(A\) on \(BD\) is drawn to meet it at the point \(E\).

Then, \(OC = {x_1}\), \(OD = {x_2}\). So, \(CD = OD – OC = {x_2} – {x_1} = AE\).
Also, \(BD = {y_2}\), \(AC = ED = {y_1}\). So \(BE = {y_2}-{y_1}\).

Now, applying the Pythagoras theorem in \(\Delta ABE\), we get
\(A {B^2} = A {E^2} + B {E^2} = {\left({ {x_2} – {x_1}} \right)^2} + {\left({{y_2} – {y_1}} \right)^2}\)
Therefore, \(AB = \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \).

Note that, since the distance is always non-negative, we take only the positive square root. So, the distance between the points \( A\left({x_1,\,y_1} \right)\) and \( B\left({x_2,\,y_2} \right)\) is
\(AB = \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \).
This is called the distance formula.

What is Distance Formula Between Two Points?

The distance between two points in the plane is the length of the line segment joining them.
The distance between two points \( A\left({x_1,\,y_1} \right)\) and \( B\left({x_2,\,y_2} \right)\) is given by \(AB = \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \).
Or, \(AB = \sqrt { { {\left({{\text{Difference between the abscissae}}} \right)}^2} + {{\left({ {\text{Difference between ordinates}}} \right)}^2}} \).

Distance of a Point in Two Dimensions from the Origin

In particular, the distance of a point \(P\left({x,\,y} \right)\) from the origin \(O\left({0,\,0} \right)\) is given by \(OP = \sqrt { { {\left({0 – x} \right)}^2} + { {\left({0 – y} \right)}^2}} = \sqrt { {x^2} + {y^2}} \).

Three-dimensional Space

Unlike two-dimensional space, which has a single plane, the \(XY\)-plane, three-dimensional space has an infinite number of planes, just as two-dimensional space has an infinite number of lines. Three planes are of particular importance: the \(XY\)- plane, which contains the \(X\)-axes and \(Y\)-axes; the \(YZ\)-plane, which contains the \(Y\)-axes and \(Z\)-axes; and the \(XZ\)-plane, which contains the \(X\)-axes and \(Z\)-axes.

Distance Formula for 3 Dimensions

The distance formula states that the distance between two points in \(XYZ\) space is the square root of the sum of the squares of the differences between corresponding coordinates. That is, given \(A\left({{x_1},\, {y_1},\, {z_1}} \right)\) and \(B\left({{x_2},\, {y_2},\, {z_2}} \right)\) the distance between \(A\) and \(B\) is given by \(AB = \sqrt { { {\left({{x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2} + { {\left( { {z_2} – {z_1}} \right)}^2}} \).
Similarly, the distance of a point \(P\left({x,\,y,\,z} \right)\), in three dimensions from the origin \(O\left({0,\,0,\,0}\right)\) can be given by \(OP = \sqrt { { {\left({0 – x} \right)}^2} + { {\left({0 – y} \right)}^2} + { {\left({0 – z} \right)}^2}} = \sqrt { {x^2} + {y^2} + {z^2}} \).

Applications of Distance Formula

Distance formulas have numerous applications in everyday life. For example, to determine the distance between two points on a map, simply get the coordinates of the two points and apply the formula.

Some Useful Points on Distance Formula

The distance formula is generally used in the following cases in geometry:

1. To prove that a given figure is a square, prove that the four sides are equal, and the diagonals are also equal.
2. To prove that a given figure is a rhombus, prove that the four sides are equal.
3. To prove that a given figure is a rectangle, prove that opposite sides are equal, and the diagonals are also equal.
4. To prove that a given figure is a parallelogram, prove that the opposite sides are equal.
5. To prove that a given figure is a parallelogram but not a rectangle, prove that its opposite sides are equal, but the diagonals are not equal.
6. To prove that a given figure is a rhombus but not a square, prove that it’s all sides are equal, but the diagonals are not equal.
7. To prove that for three points to be collinear, prove that the sum of the distances between two pairs of points is equal to the third pair of points.

Solved Examples – Distance Formula

Question-1: Find the distance between the coordinates \(A\left({2,\,3} \right)\) and \(B\left({-4,\,5} \right)\)

Solution: Given
Coordinates \(A\left({2,\,3} \right)\) and \(B\left({-4,\,5} \right)\)
Here, \( {x_1} = 2\), \( {y_1} = 3\), \( {x_2} = -4\) and \( {y_2} = 5\).
We know that Distance \(= \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \)
Now, \(AB = \sqrt { { {\left({ – 4 – 2} \right)}^2} + { {\left({5 – 3} \right)}^2}} = \sqrt {36 + 4} = \sqrt {40} \, {\text{units}}\)
Hence, the distance between two given coordinates is \(\sqrt {40} \, {\text{units}}\).

Question-2: Find the distance between the coordinates \(P\left({-2,\,-3} \right)\) and the origin.

Solution: Given: The coordinate \(P\left({ – 2,\, – 3} \right)\) and the origin \(O\left({0,\,0} \right)\).
Here, \( {x_1} = – 2\), \( {y_1} = – 3\), \( {x_2} = 0\) and \( {y_2} = 0\).
We know that Distance \(= \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \)
Now, \(AB = \sqrt { { {\left({0 – \left({ – 2} \right)} \right)}^2} + { {\left({0 – \left({ – 3} \right)} \right)}^2}} = \sqrt { {2^2} + {3^2}} = \sqrt {4 + 9} = \sqrt {13} \, {\text{units}}\)
Hence, the distance between the given coordinate and origin is \(\sqrt {13} \, {\text{units}}\)

Question-3: The distance between two coordinates is \(\sqrt {40} \, {\text{units}}\) and \(R\left({6,\, – 2} \right)\). If the point \(S\) lies on \(x\)-axis, then find the coordinates of the point \(S\).

Solution: Given
The distance between two coordinates \(=\sqrt {40} \, {\text{units}}\)
The coordinates are \(R\left({6,\, – 2} \right)\) and \(S\left({x,\, 0} \right)\).
Here, \( {x_1} = 6\), \( {y_1} = -2\), \( {x_2} = x\) and \( {y_2} = 0\).
We know that Distance \(= \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \)
Now, \(RS = \sqrt { { {\left({x – 6} \right)}^2} + { {\left({0 – \left({ – 2} \right)} \right)}^2}} \)
\(\Rightarrow \sqrt {40} = \sqrt { {x^2} – 12x + 36 + 4} \)
By squaring on both side, we get
\( \Rightarrow 40 = {x^2} – 12x + 40\)
\(\Rightarrow {x^2} – 12x = 0\)
\( \Rightarrow {x^2} = 12x \Rightarrow x\left({x – 12} \right) = 0 \Rightarrow x = 0,\,12\)
Hence, the coordinates of the point \(S\) is \((0,\,0)\) or \((12,\,0)\).

Question-4: Find the distance between the coordinates \(A( – 4,\,0)\) and \(B(0,\, – 4)\).

Solution: Given
The coordinates \(A( – 4,\,0)\) and \(B(0,\, – 4)\)
Here, \( {x_1} = 4\), \( {y_1} = 0\), \( {x_2} = 0\) and \( {y_2} = -4\).
We know that Distance \(= \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \)
Now, \(AB = \sqrt { { {\left({0 – \left({ – 4} \right)} \right)}^2} + { {\left({ – 4 – 0} \right)}^2}} \)
\( = \sqrt {16 + 16} = 4\sqrt 2 \, {\text{units}}\)
Hence, the distance between the given coordinates is \(4\sqrt 2 \, {\text{units}}\)

Question-5: Find a relation between \(x\) and \(y\) such that the point \(P(x,\,y)\) is equidistant from the points \(Q(2,\,1)\) and \(R(1,\,3)\).

Solution: Given
The coordinates \(Q(2,\,1)\) and \(R(1,\,3)\) are equidistant from the point \(P(x,\,y)\). That is, \(PQ = PR\)
We know that Distance \(= \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \)
\( PQ = \sqrt { { {(2 – x)}^2} + { {(1 – y)}^2}} \)
Now, \(\sqrt { { {(2 – x)}^2} + { {(1 – y)}^2}} = \sqrt { { {(1 – x)}^2} + { {(3 – y)}^2}} \)
By squaring on both the sides, we get \(\Rightarrow 4 + {x^2} – 4x + 1 + {y^2} – 2y = 1 + {x^2} – 2x + 9 + {y^2} – 6y\)
\(\Rightarrow 5 – 4x – 2y = 10 – 2x – 6y\)
\( \Rightarrow 2x – 4y – 5 = 0\)
Hence, the relation between \(x\) and \(y\) is \( 2x – 4y – 5 = 0\).

Summary

In this article, we learnt about, the distance of a point from the coordinate axes (\(X\) and \(Y\) axis), the definition of the distance formula between two points, distance of a point in two dimensions from the origin, the distance formula for \(3\) (three) dimensions, applications of and solved examples on distance formula.

The learning outcome of this article is that we can find the distance between two coordinate points using the distance formula. In real life, the distance formula can be used as a strategy for easy navigation and distance estimation.

Frequently Asked Questions About Distance Formula

Let’s look at some of the commonly asked questions about Distance Formula:

Q.1. What is the distance between \((0,\,0)\) and \((x,\,y)\)?
Ans: The distance of a point \((x,\,y)\) from the origin \((0,\,0)\) is \(\sqrt { {x^2} + {y^2}} \)

Q.2. Can we consider the negative values in the distance formula in geometry?
Ans: The distance is always non-negative, we take only the positive square root. So, the distance between the points \(A(x_1,\,y_1)\) and \(B(x_2,\,y_2)\) is given by
\(AB = \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \)
\( {\text{AB}} = \sqrt { { {( {\text{Difference of abscissae}})}^2} + { {( {\text{Difference of ordinates}})}^2}} \)

Q.3. What is distance formula and section formula?
Ans: Distance formula is used to find distance between two coordinates, we have
Distance \(= \sqrt { { {\left({ {x_2} – {x_1}} \right)}^2} + { {\left({{y_2} – {y_1}} \right)}^2}} \)
Section formula is used to find the coordinates of any point that divides the line segment internally in the ratio \(m:n\) are \(\left({\frac{ {m {x_2} + n {x_1}}}{ {m + n}},\,\frac{ {m {y_2} – n {y_1}}}{ {m + n}}} \right)\)
The coordinates of any point, divides the line segment externally in the ratio \(m:n\) are
\(\left({\frac{ {m {x_2} – n {x_1}}}{ {m + n}},\,\frac{ {m {y_2} – n {y_1}}}{ {m + n}}} \right)\)

Q.4. How do you find the distance formula?
Ans: The Distance formula is derived from the Pythagorean theorem, that is \(c = \sqrt { {a^2} + {b^2}} \) where \(c\) is the longest side (the hypotenuse) of a right triangle and \(a\) and \(b\) are the other shorter sides (base and perpendicular).

Use Pythagoras Theorem to Prove Distance Formula 3D

Now that you are provided with all the necessary information on the distance formula and we hope this detailed article is helpful to you. If you have any queries on the distance formula or in general about this article, reach us through the comment box below, and we will get back to you as soon as possible.