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3 Coordinate geometry
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Stage 5.1-2

3.03 The distance between two points

Lesson
Worksheet
Practice
Lesson

Distance between two points

Points that lie on a horizontal line share the same y-value. They would have coordinates that look something like this: (2,5) and (-4,5). More generally, two points that lie on a horizontal line could have coordinates (a,b) and (c,b).

If you can recognise that the points lie on a horizontal line then the distance between them is the distance between the x-values: the largest x-value minus the smallest x-value.

Points that lie on a vertical line share the same x-value. They would have coordinates that look something like this: (2,5) and (2, 29). More generally, two points that lie on a vertical line could have coordinates (a, b) and (a, c).

If you can recognise that the points lie on a vertical line then the distance between them is the distance between the y-values: the largest y-value minus the smallest y-value.

What if we want to find the distance between two points that are not on a horizontal or vertical line?

We already learned how to use Pythagoras' theorem to calculate the side lengths in a right triangle. Pythagoras' theorem states: a^2 + b^2 = c^2 where a and b are the shorter side lengths, and c is the hypothenuse.

We can also use Pythagoras' theorem to find the distance between two points on an xy-plane.

The distance between two points (x_1, y_1) and (x_2, y_2) is given by: d=\sqrt{{\text{run}}^2 + {\text{rise}}^2} where the run is the horizontal distance between the two x-coordinates and the rise is the vertical distance between the two y-coordinates.

Examples

Example 1

The points A (-3, -2), B (-3, -4) and C (1, -4) are the vertices of a right angle triangle. as shown on the number plane.

-5
-4
-3
-2
-1
1
2
3
4
5
x
-5
-4
-3
-2
-1
1
2
3
4
5
y
a

Find the length of interval AB.

Worked Solution
Create a strategy

Find the difference between the y-values.

Apply the idea
\displaystyle AB\displaystyle =\displaystyle -2-(-4)Subtract the y-values
\displaystyle =\displaystyle -2+4Multiply signs
\displaystyle =\displaystyle 2 \text{ units}Evaluate
b

Find the length of interval BC.

Worked Solution
Create a strategy

Find the difference between the x-values.

Apply the idea
\displaystyle BC\displaystyle =\displaystyle 1-(-3)Subtract the x-values
\displaystyle =\displaystyle 1+3Multiply signs
\displaystyle =\displaystyle 4 \text{ units}Evaluate
c

Use Pythagoras' theorem to find the length of the interval AC to three decimal places.

Worked Solution
Create a strategy

Use the answers in parts (a) and (b) in Pythagoras' theorem.

Apply the idea
\displaystyle c^2\displaystyle =\displaystyle a^2 + b^2Pythagoras' theorem
\displaystyle AC^2\displaystyle =\displaystyle AB^2 + BC^2Substitute c=AC, a=AB, b=BC
\displaystyle {AC}^2\displaystyle =\displaystyle 2^2 + 4^2Substitute AB=2 and BC=4
\displaystyle {AC}^2\displaystyle =\displaystyle 20Evaluate the right side
\displaystyle AC\displaystyle =\displaystyle \sqrt{20}Take the square root of both sides
\displaystyle =\displaystyle 4.472Evaluate to three decimal places
Idea summary

Two points that lie on a horizontal line have coordinates of the form (a,b) and (c,b). The distance between the points is the largest x-value minus the smallest x-value.

Two points that lie on a vertical line have coordinates of the form (a, b) and (a, c). The distance between the points is the largest y-value minus the smallest y-value.

The distance between two points (x_1, y_1) and (x_2, y_2) is given by:

\displaystyle d=\sqrt{{\text{run}}^2 + {\text{rise}}^2}
\bm{\text{'run'}}
is the horizontal distance between the two x-coordinates
\bm{\text{'rise'}}
is the vertical distance between the two y-coordinates

Outcomes

MA5.1-6NA

determines the midpoint, gradient and length of an interval, and graphs linear relationships

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