1) Build the Distance Idea on the Grid: Δx, Δy, and a Right Triangle
To find the length of a segment between two points on the coordinate plane, start by comparing their coordinates. Suppose the points are A(x1, y1) and B(x2, y2).
- Horizontal change (run):
Δx = x2 − x1 - Vertical change (rise):
Δy = y2 − y1
Now imagine drawing a right triangle whose hypotenuse is the segment AB. One leg is the horizontal move of length |Δx|, and the other leg is the vertical move of length |Δy|. The absolute values are used because lengths are nonnegative, even if the coordinate differences are negative.
A quick way to visualize the triangle is to introduce a “corner” point C that shares the x-coordinate of one point and the y-coordinate of the other, for example C(x2, y1). Then AC is horizontal and CB is vertical, making a right angle at C.
2) Derive the Distance Formula Using the Pythagorean Theorem
In the right triangle, the legs have lengths |Δx| and |Δy|, and the hypotenuse has length d, the distance between the two points. By the Pythagorean Theorem:
d2 = (|Δx|)2 + (|Δy|)2
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Because squaring removes the sign, (|Δx|)2 = (Δx)2 and similarly for Δy. So:
d2 = (Δx)2 + (Δy)2
Taking the positive square root (distance cannot be negative) gives the distance formula:
d = √((x2 − x1)2 + (y2 − y1)2)
| Piece | What it represents on the grid | Why it’s squared |
|---|---|---|
(x2 − x1) | signed horizontal change (Δx) | to use Pythagorean Theorem and remove sign |
(y2 − y1) | signed vertical change (Δy) | to use Pythagorean Theorem and remove sign |
√(…) | turns squared length back into length | distance is the positive root |
3) Worked Examples (from simple to diagonal)
Example 1: Horizontal segment (same y-value)
Find the distance between P(−2, 4) and Q(5, 4).
Δx = 5 − (−2) = 7Δy = 4 − 4 = 0
Distance:
d = √(72 + 02) = √49 = 7
Grid check: the points are on the same horizontal line, so the distance is just the horizontal run: 7 units.
Example 2: Vertical segment (same x-value)
Find the distance between R(3, −1) and S(3, 8).
Δx = 3 − 3 = 0Δy = 8 − (−1) = 9
Distance:
d = √(02 + 92) = √81 = 9
Grid check: same vertical line, so the distance is the vertical rise: 9 units.
Example 3: Diagonal with a 3–4–5 triangle
Find the distance between A(1, 2) and B(4, 6).
Δx = 4 − 1 = 3Δy = 6 − 2 = 4
Distance:
d = √(32 + 42) = √(9 + 16) = √25 = 5
Grid check: moving 3 units right and 4 units up forms a right triangle; the hypotenuse is 5 units.
Example 4: Diagonal with negatives and a non-integer distance
Find the distance between C(−3, 5) and D(2, −1).
Δx = 2 − (−3) = 5Δy = −1 − 5 = −6
Distance (exact):
d = √(52 + (−6)2) = √(25 + 36) = √61
Distance (rounded):
√61 ≈ 7.81 (to the nearest hundredth)
Grid estimation: the legs are about 5 and 6, so the hypotenuse should be a bit less than 5 + 6 = 11 and a bit more than the larger leg 6. Since 7.81 is between 6 and 11 and close to √(49 + 16) = √65 ≈ 8.06, the value is reasonable.
4) Practice: Compute Distances, Then Verify by Estimation
For each pair of points: (1) compute Δx and Δy, (2) use d = √((Δx)2 + (Δy)2), (3) give an exact answer, and (4) give a rounded answer (nearest tenth unless stated). Then do a quick grid estimate: compare the distance to the longer leg and to the sum of legs.
Practice Set A
- Distance between
(−1, 3)and(6, 3). - Distance between
(4, −5)and(4, 2). - Distance between
(0, 0)and(3, 4). - Distance between
(−2, −1)and(5, 3). - Distance between
(−4, 7)and(1, −5). Round to the nearest hundredth.
Answer Key with Estimation Checks
1) (−1, 3) to (6, 3)
Δx = 6 − (−1) = 7,Δy = 3 − 3 = 0d = √(72 + 02) = 7- Exact:
7; Rounded:7.0 - Estimate: horizontal segment, so exactly 7.
2) (4, −5) to (4, 2)
Δx = 0,Δy = 2 − (−5) = 7d = √(0 + 49) = 7- Exact:
7; Rounded:7.0 - Estimate: vertical segment, so exactly 7.
3) (0, 0) to (3, 4)
Δx = 3,Δy = 4d = √(9 + 16) = √25 = 5- Exact:
5; Rounded:5.0 - Estimate: between 4 and 7; 5 fits well.
4) (−2, −1) to (5, 3)
Δx = 5 − (−2) = 7,Δy = 3 − (−1) = 4d = √(72 + 42) = √(49 + 16) = √65- Exact:
√65; Rounded:√65 ≈ 8.1 - Estimate: longer leg is 7, sum is 11, so distance should be between 7 and 11; 8.1 is reasonable.
5) (−4, 7) to (1, −5)
Δx = 1 − (−4) = 5,Δy = −5 − 7 = −12d = √(52 + (−12)2) = √(25 + 144) = √169 = 13- Exact:
13; Rounded:13.00 - Estimate: between 12 and 17; exactly 13 matches a 5–12–13 triangle.
