Distance Between Two Points Revision Notes for Leaving Cert Mathematics

Distance Between Two Points

What is the distance between two points?

When we have two points on a coordinate plane, we often need to find the straight-line distance between them. This is like measuring the length of a piece of string stretched directly from one point to the other, rather than following the grid lines.

The distance between any two points can be calculated using a special formula that comes from the Pythagorean theorem. This method works for any two points, regardless of which quadrant they're in or whether they have positive or negative coordinates.

The distance formula

To find the distance between two points A $(x_1, y_1)$ and B $(x_2, y_2)$ , we use the distance formula:

$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

How the formula works

The distance formula comes directly from the Pythagorean theorem. When we connect two points on a coordinate plane, we create a right-angled triangle:

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Understanding the Right Triangle Connection

When plotting two points on a coordinate plane and connecting them:

The horizontal leg has length $|x_2 - x_1|$
The vertical leg has length $|y_2 - y_1|$
The hypotenuse is the distance we want to find

Using Pythagoras' theorem: $\text{distance}^2 = (\text{horizontal leg})^2 + (\text{vertical leg})^2$

Therefore: $\text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Step-by-step method

To calculate the distance between two points, follow this systematic approach:

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The Five-Step Process

Identify the coordinates: Label your points as $(x_1, y_1)$ and $(x_2, y_2)$
Find the differences: Calculate $(x_2 - x_1)$ and $(y_2 - y_1)$
Square the differences: Calculate $(x_2 - x_1)^2$ and $(y_2 - y_1)^2$
Add the squares: Add the results from step 3
Take the square root: Find the square root of your answer from step 4

Worked example 1: Basic distance calculation

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Worked Example: Finding Distance Between Two Points

Find the distance between points A(3, 5) and B(8, 2).

Solution:

Identify coordinates: A(3, 5) means $x_1 = 3, y_1 = 5$ and B(8, 2) means $x_2 = 8, y_2 = 2$
Find differences: $(x_2 - x_1) = 8 - 3 = 5$ and $(y_2 - y_1) = 2 - 5 = -3$
Square the differences: $(8 - 3)^2 = 5^2 = 25$ and $(2 - 5)^2 = (-3)^2 = 9$
Add the squares: $25 + 9 = 34$
Take square root: $\sqrt{34} \approx 5.83$

Therefore, the distance between A and B is √34 units (approximately 5.83 units).

Worked example 2: Proving equidistant points

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Worked Example: Proving Equidistant Points

Show that point D(2, 4) is equidistant from points E(-5, 1) and F(5, -3).

Solution: Equidistant means the same distance from both points, so we need to show $|DE| = |DF|$ .

Finding $|DE|$ where D(2, 4) and E(-5, 1):

Distance $= \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
$|DE| = \sqrt{(-5 - 2)^2 + (1 - 4)^2}$
$|DE| = \sqrt{(-7)^2 + (-3)^2}$
$|DE| = \sqrt{49 + 9} = \sqrt{58}$

Finding $|DF|$ where D(2, 4) and F(5, -3):

$|DF| = \sqrt{(5 - 2)^2 + (-3 - 4)^2}$
$|DF| = \sqrt{(3)^2 + (-7)^2}$
$|DF| = \sqrt{9 + 49} = \sqrt{58}$

Since |DE| = |DF| = √58, point D is equidistant from points E and F.

Worked example 3: Finding unknown coordinates

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Worked Example: Finding Unknown Coordinates

If the distance between points (2, 3) and (5, k) is $\sqrt{10}$ , find the possible values of k.

Solution: Using the distance formula:

$\sqrt{10} = \sqrt{(5 - 2)^2 + (k - 3)^2}$
$\sqrt{10} = \sqrt{3^2 + (k - 3)^2}$
$\sqrt{10} = \sqrt{9 + (k - 3)^2}$

Squaring both sides:

$10 = 9 + (k - 3)^2$
$1 = (k - 3)^2$
$\pm 1 = k - 3$

Therefore: $k = 3 + 1 = 4$ or $k = 3 - 1 = 2$

The two possible values are k = 2 and k = 4.

Common exam tips

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Critical Points to Avoid Common Mistakes

Always square the differences: Remember that $(-3)^2 = 9$ , not $-9$
Don't forget the square root: The distance formula ends with taking the square root
Check your arithmetic: Double-check your calculations, especially when squaring negative numbers
Leave answers in surd form: Unless asked to approximate, leave answers like $\sqrt{58}$ rather than converting to decimals
Set up equations carefully: When finding unknown coordinates, set up the distance formula as an equation first

Applications in geometry

The distance formula is essential for solving many geometric problems. It allows us to work with shapes and relationships in the coordinate plane with precision and accuracy.

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Key Applications in Geometry

The distance formula is essential for:

Finding side lengths of triangles and quadrilaterals
Calculating the radius of circles when given centre and a point on the circle
Proving that triangles are isosceles (two equal sides)
Showing that points are collinear or form specific shapes

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Key Points to Remember:

The distance formula is $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
The formula comes from Pythagoras' theorem applied to coordinate geometry
Squaring eliminates the need to worry about positive or negative differences
Equidistant points are the same distance from a reference point
Always take the square root at the end - distance is always positive

Distance Between Two Points (Leaving Cert Mathematics): Revision Notes