Length of a Line Segment

How long is a slanted segment on a grid? You can’t just count squares, but you can turn the segment into the hypotenuse of a right triangle and use the Pythagorean theorem. That idea gives the length formula (also called the distance formula), which you’ll use to find perimeters, compare side lengths, and later to build the equation of a circle.

Key ideas

Horizontal and vertical segments

If two points are on the same horizontal line, the length is just the difference in $x$-coordinates. From $(-2, 3)$ to $(5, 3)$ the length is $5 - (-2) = 7$. For a vertical segment, use the difference in $y$-coordinates. Lengths are never negative, so subtract the smaller from the larger.

Slanted segments: build a right triangle

For a slanted segment $AB$, draw a horizontal leg from $A$ and a vertical leg up (or down) to $B$. They meet at a right angle, so $AB$ is the hypotenuse.

The horizontal leg is $6$, the vertical leg is $8$, so $AB = \sqrt{6^2 + 8^2} = 10$.

The horizontal leg is the change in $x$, and the vertical leg is the change in $y$. By the Pythagorean theorem:

$$AB^2 = (\text{change in } x)^2 + (\text{change in } y)^2$$

The length formula

The length of the segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ is

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

Squaring makes every term positive, so it doesn’t matter which point you call $A$, and a negative change in $x$ or $y$ is fine.

Exact and decimal answers

Often the answer is a square root that isn’t a whole number, like $\sqrt{50}$. That’s the exact answer. You can also give a decimal approximation, like $\sqrt{50} \approx 7.07$. Keep the exact form while you’re still working, and round only at the end. If you’ve learned to simplify radicals, you can write $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$.

Classifying triangles by side lengths

Find all three side lengths, then compare:

Type	Side lengths
Equilateral	all three sides equal
Isosceles	exactly two sides equal
Scalene	no sides equal

To check for a right triangle, test the Pythagorean theorem backwards: if the squares of the two shorter sides add up to the square of the longest side, the triangle has a right angle. (You can also use slopes, as on verifying geometric properties.)

Worked examples

Example 1: A whole-number length

Find the length of the segment joining $A(-2, -1)$ and $B(4, 7)$.

Solution.

$$\begin{aligned} AB &= \sqrt{(4 - (-2))^2 + (7 - (-1))^2} \\ &= \sqrt{6^2 + 8^2} \\ &= \sqrt{36 + 64} \\ &= \sqrt{100} \\ &= 10 \end{aligned}$$

This is the segment in the figure above: legs of $6$ and $8$, hypotenuse $10$.

Example 2: Exact and decimal answers

Find the length of $CD$, with $C(-3, 4)$ and $D(2, -1)$. Give the exact answer and a decimal to two decimal places.

Solution.

$$\begin{aligned} CD &= \sqrt{(2 - (-3))^2 + (-1 - 4)^2} \\ &= \sqrt{5^2 + (-5)^2} \\ &= \sqrt{25 + 25} \\ &= \sqrt{50} \end{aligned}$$

The exact length is $\sqrt{50}$ (or $5\sqrt{2}$), which is about $7.07$.

Notice that $(-5)^2 = 25$. Squaring a negative change gives a positive number.

Example 3: Classifying a triangle

A triangle has vertices $A(1, 1)$, $B(5, 3)$ and $C(3, 7)$. Classify it by its side lengths, and decide whether it’s a right triangle.

Solution. Find all three lengths:

$$\begin{aligned} AB &= \sqrt{(5 - 1)^2 + (3 - 1)^2} = \sqrt{16 + 4} = \sqrt{20} \\ BC &= \sqrt{(3 - 5)^2 + (7 - 3)^2} = \sqrt{4 + 16} = \sqrt{20} \\ AC &= \sqrt{(3 - 1)^2 + (7 - 1)^2} = \sqrt{4 + 36} = \sqrt{40} \end{aligned}$$

$AB = BC$, so the triangle is isosceles.

Now test for a right angle. The longest side is $AC$. Compare squares:

$$AB^2 + BC^2 = 20 + 20 = 40 = AC^2$$

The squares match, so the triangle is also a right triangle, with the right angle at $B$ (the vertex opposite the longest side). It’s an isosceles right triangle.

Squared lengths are easy here: $AB^2 = 20$, so you never need to round.

Example 4: Perimeter of a garden

A garden plan uses a grid in metres. The garden has corners $A(-4, 0)$, $B(0, 3)$, $C(5, 3)$ and $D(2, -1)$. Fencing costs $18 per metre. Find the perimeter, and the cost to fence the garden.

Solution. Find each side, going around the shape in order:

$$\begin{aligned} AB &= \sqrt{(0 - (-4))^2 + (3 - 0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \\ BC &= 5 - 0 = 5 && \text{horizontal} \\ CD &= \sqrt{(2 - 5)^2 + (-1 - 3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \\ DA &= \sqrt{(-4 - 2)^2 + (0 - (-1))^2} = \sqrt{36 + 1} = \sqrt{37} \end{aligned}$$ $$\text{Perimeter} = 5 + 5 + 5 + \sqrt{37} = 15 + \sqrt{37} \approx 15 + 6.083 = 21.083$$

The perimeter is about $21.08$ m. The cost is $21.083 \times 18 \approx 379.49$, so it costs about $379.49 to fence the garden.

Keep the extra decimal places until the last step, so your rounding doesn’t change the cost.

Common mistakes

Adding the changes instead of squaring them. The length from $(-2, -1)$ to $(4, 7)$ is not $6 + 8 = 14$. Walking along the two legs is longer than cutting straight across. Square, add, then take the square root.

Taking the square root of each term separately. $\sqrt{36 + 64}$ is $\sqrt{100} = 10$, not $\sqrt{36} + \sqrt{64} = 14$. Add first, then take the root.

Sign errors with negative coordinates. $4 - (-2) = 6$, not $2$. Use brackets when you substitute a negative coordinate. Once you square, a negative change becomes positive anyway, but only if you subtracted correctly first.

Mixing coordinates from different points. Subtract $x$ from $x$ and $y$ from $y$, using the same two points. Label the points $(x_1, y_1)$ and $(x_2, y_2)$ before you substitute.

Rounding too early. Rounding each side before adding can change the perimeter. Keep exact values (or at least three decimal places) until the final answer.

Testing the wrong sides for a right angle. The longest side must be on its own: check whether the two shorter sides’ squares add to the longest side’s square.

Practice

1. (Warm-up) Find the length of each segment.

• (a) $(1, 2)$ and $(4, 6)$
• (b) $(-3, 7)$ and $(5, 7)$

Solution

(a) $\sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$

(b) The segment is horizontal: $5 - (-3) = 8$.

2. (Warm-up) Find the length of each segment. Give exact answers, and a decimal to two decimal places where needed.

• (a) $(0, 0)$ and $(-5, 12)$
• (b) $(-2, 3)$ and $(4, -1)$

Solution

(a) $\sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

(b) $\sqrt{(4 - (-2))^2 + (-1 - 3)^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21$ (exact: $\sqrt{52} = 2\sqrt{13}$)

3. (Core) Find the length of the segment from $A(-4, -3)$ to $B(3, 1)$, exactly and to two decimal places.

Solution

$$AB = \sqrt{(3 - (-4))^2 + (1 - (-3))^2} = \sqrt{7^2 + 4^2} = \sqrt{49 + 16} = \sqrt{65} \approx 8.06$$

4. (Core) Classify the triangle with vertices $J(-3, 2)$, $K(3, 4)$ and $L(1, -2)$ by its side lengths. Is it a right triangle?

Solution

$$\begin{aligned} JK &= \sqrt{(3 - (-3))^2 + (4 - 2)^2} = \sqrt{36 + 4} = \sqrt{40} \\ KL &= \sqrt{(1 - 3)^2 + (-2 - 4)^2} = \sqrt{4 + 36} = \sqrt{40} \\ JL &= \sqrt{(1 - (-3))^2 + (-2 - 2)^2} = \sqrt{16 + 16} = \sqrt{32} \end{aligned}$$

$JK = KL$, so the triangle is isosceles.

The longest sides are $JK$ and $KL$ (both $\sqrt{40}$). Test: $JL^2 + KL^2 = 32 + 40 = 72$, which is not $JK^2 = 40$. No pair works, so it is not a right triangle.

5. (Core) Find the perimeter of the triangle with vertices $P(-3, 1)$, $Q(3, 4)$ and $R(1, -2)$, to one decimal place.

Solution

$$\begin{aligned} PQ &= \sqrt{(3 - (-3))^2 + (4 - 1)^2} = \sqrt{36 + 9} = \sqrt{45} \approx 6.708 \\ QR &= \sqrt{(1 - 3)^2 + (-2 - 4)^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.325 \\ PR &= \sqrt{(1 - (-3))^2 + (-2 - 1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \end{aligned}$$$$\text{Perimeter} \approx 6.708 + 6.325 + 5 = 18.033$$

The perimeter is about $18.0$ units.

6. (Core) A delivery drone flies in a straight line from $(-300, 200)$ to $(500, -400)$ on a grid measured in metres.

• (a) How far does it fly?
• (b) At $8$ m/s, how long does the flight take?

Solution

(a)

$$\sqrt{(500 - (-300))^2 + (-400 - 200)^2} = \sqrt{800^2 + (-600)^2} = \sqrt{640\,000 + 360\,000} = \sqrt{1\,000\,000} = 1000$$

The drone flies $1000$ m.

(b) $\text{time} = \dfrac{1000}{8} = 125$ seconds.

7. (Core) The point $(k, 3)$ is $5$ units from $(1, -1)$. Find all possible values of $k$.

Solution

Set the length equal to $5$ and square both sides:

$$\begin{aligned} (k - 1)^2 + (3 - (-1))^2 &= 25 \\ (k - 1)^2 + 16 &= 25 \\ (k - 1)^2 &= 9 \\ k - 1 &= 3 \quad \text{or} \quad k - 1 = -3 \\ k &= 4 \quad \text{or} \quad k = -2 \end{aligned}$$

Check $k = 4$: $\sqrt{3^2 + 4^2} = 5$. ✓ Check $k = -2$: $\sqrt{(-3)^2 + 4^2} = 5$. ✓

8. (Challenge) Find the point on the $y$-axis that is the same distance from $A(-2, 1)$ as from $B(4, 5)$.

Solution

A point on the $y$-axis has the form $(0, y)$. Set the squared distances equal (squaring avoids the square roots):

$$\begin{aligned} (0 - (-2))^2 + (y - 1)^2 &= (0 - 4)^2 + (y - 5)^2 \\ 4 + y^2 - 2y + 1 &= 16 + y^2 - 10y + 25 \\ -2y + 5 &= -10y + 41 \\ 8y &= 36 \\ y &= 4.5 \end{aligned}$$

The point is $(0, 4.5)$.

Check: distance squared to $A$ is $4 + 3.5^2 = 16.25$, and to $B$ is $16 + (-0.5)^2 = 16.25$. ✓

9. (Challenge) A triangle has vertices $A(-4, 2)$, $B(6, 4)$ and $C(2, -6)$. Join the midpoints of the three sides to make a smaller triangle. Find the side lengths of the smaller triangle, and compare them with the sides of triangle $ABC$. What do you notice?

Solution

Midpoints: $AB$ gives $D(1, 3)$, $BC$ gives $E(4, -1)$, and $AC$ gives $F(-1, -2)$.

Smaller triangle:

$$\begin{aligned} DE &= \sqrt{(4 - 1)^2 + (-1 - 3)^2} = \sqrt{9 + 16} = 5 \\ EF &= \sqrt{(-1 - 4)^2 + (-2 - (-1))^2} = \sqrt{25 + 1} = \sqrt{26} \\ DF &= \sqrt{(-1 - 1)^2 + (-2 - 3)^2} = \sqrt{4 + 25} = \sqrt{29} \end{aligned}$$

Triangle $ABC$:

$$\begin{aligned} AC &= \sqrt{(2 - (-4))^2 + (-6 - 2)^2} = \sqrt{36 + 64} = 10 \\ AB &= \sqrt{(6 - (-4))^2 + (4 - 2)^2} = \sqrt{104} = 2\sqrt{26} \\ BC &= \sqrt{(2 - 6)^2 + (-6 - 4)^2} = \sqrt{116} = 2\sqrt{29} \end{aligned}$$

Each side of the smaller triangle is half the side of $ABC$ that it doesn’t touch: $DE = 5$ is half of $AC = 10$, $EF = \sqrt{26}$ is half of $AB = 2\sqrt{26}$, and $DF = \sqrt{29}$ is half of $BC = 2\sqrt{29}$. (You can check $\sqrt{104} = \sqrt{4 \times 26} = 2\sqrt{26}$.) This is the midsegment property, which you’ll prove on verifying geometric properties.