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ESO Y4A · Mathematics
coordinate geometry · Coordinate geometry

Distance between two points

An ant walking in a straight line from one point to another travels exactly the magnitude of the vector joining them. That is why the distance between two points is not a new formula: it is the same magnitude formula you already know, applied to the vector $\vec{AB}$.

Before you start, make sure you know: Pythagoras' theorem (2nd ESO)
Before you start, make sure you know: Operations with integers (2nd ESO)

The distance formula

Right-angled triangle with hypotenuse d(A,B) and horizontal and vertical legs parallel to the coordinate axes. A B leg x leg y d(A,B)
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In a test, students are asked to find the distance between points $A(-2, 1)$ and $B(1, 5)$. A student's response is…

In a test, students are asked to find the distance between points $A(-2, 1)$ and $B(1, 5)$. A student's response is $d(A, B) = \sqrt{(1 - (-2))^2 - (5 - 1)^2} = \sqrt{9 - 16} = \sqrt{-7}$, concluding that the distance does not exist. Identify the sign error in the formula used and calculate the correct distance.

Solve
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Calculate the exact distance between points $A(3, -2)$ and $B(-1, 1)$ in the Cartesian plane.

Calculate the exact distance between points $A(3, -2)$ and $B(-1, 1)$ in the Cartesian plane.

Solve

Perimeters and triangles

By adding the distances between consecutive vertices you can calculate the perimeter of any polygon. And by comparing the three side lengths of a triangle (using the converse of Pythagoras' theorem) you can tell whether it is right-angled, or whether it is isosceles or equilateral.

Triangle in the Cartesian plane with vertices at A, B and C for calculating its sides and perimeter. A B C
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Calculate the perimetre of the triangle whose vertices are located at coordinates $A(0, 0)$, $B(6, 0)$ and $C(3, 4)$.

Calculate the perimetre of the triangle whose vertices are located at coordinates $A(0, 0)$, $B(6, 0)$ and $C(3, 4)$.

Solve
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Prove whether the triangle with vertices $A(-1, 2)$, $B(2, -2)$ and $C(6, 1)$ is a right-angled triangle by calculating the lengths of its three sides and applying the…

Prove whether the triangle with vertices $A(-1, 2)$, $B(2, -2)$ and $C(6, 1)$ is a right-angled triangle by calculating the lengths of its three sides and applying the converse of Pythagoras' theorem.

Solve

Equidistant points

A point is equidistant from two others when it is the same distance from both. Many problems (aerials that must cover several villages equally, sensors that must be placed the same distance from two machines) are solved by equating two distances and solving for the unknown.

Point P equidistant from points M and N, showing the equal-length segments joining them. M N P
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Check mathematically whether point $P(2, 3)$ is equidistant from points $M(5, -1)$ and $N(-1, -1)$.

Check mathematically whether point $P(2, 3)$ is equidistant from points $M(5, -1)$ and $N(-1, -1)$.

Solve
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Find the coordinates of the point on the x-axis that is equidistant from the points $A(1, 4)$ and $B(5, -2)$.

Find the coordinates of the point on the x-axis that is equidistant from the points $A(1, 4)$ and $B(5, -2)$.

Solve

Distance from a point to a line (a preview)

Some problems ask for the shortest distance from a point to an entire line, not to another specific point. That minimum distance is always the one along the perpendicular to the line through the point. Once you know the general equation $Ax + By + C = 0$ of a line (next section), you will be able to calculate it directly:

$$d(P, r) = \frac{|A p_1 + B p_2 + C|}{\sqrt{A^2+B^2}}$$ Distance from a point to a line measured along the perpendicular segment between them. P r minimum d
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A comet is at position $C(10, 15)$ and a space probe follows a linear trajectory given by $2x - y - 1 = 0$. A novice scientist chooses a random point on the trajectory,…

A comet is at position $C(10, 15)$ and a space probe follows a linear trajectory given by $2x - y - 1 = 0$. A novice scientist chooses a random point on the trajectory, for example $P(1, 1)$, and calculates the distance from the comet to the trajectory as $d(C, P) = \sqrt{(10-1)^2 + (15-1)^2} = \sqrt{81 + 196} = \sqrt{277}$. Explain why this method does not guarantee finding the actual distance to the line. Apply the correct analytical formula and find the exact value of this distance.

r C P Incorrect distance Actual distance
Solve

Summary

See all the exercises in this section