Apply the Pythagorean Theorem to find the distance between two points in a coordinate system

Learning goal: Find the distance between two points in a coordinate system by drawing the right triangle whose hypotenuse joins them, reading the horizontal and vertical legs from the coordinates, and applying the Pythagorean Theorem; give exact and approximate distances and use them to find perimeters and to classify figures drawn on a grid.

Why a coordinate plane makes right triangles. In a coordinate system, horizontal lines (constant y) and vertical lines (constant x) are perpendicular. Take any two points A and B that are not on the same horizontal or vertical line. Draw a horizontal segment from A and a vertical segment from B; they meet at a third point C, and the angle at C is a right angle. The segment AB is therefore the hypotenuse of right triangle ACB, and its length is the distance between A and B.

Reading the legs from the coordinates. The horizontal leg has length equal to the difference of the x-coordinates, and the vertical leg has length equal to the difference of the y-coordinates. Length is never negative, so take the difference as a positive number: horizontal leg = |x₂ − x₁| and vertical leg = |y₂ − y₁|. Then the distance is d = √((x₂ − x₁)² + (y₂ − y₁)²). The order of subtraction does not matter, because a number and its opposite have the same square: (5 − (−3))² and (−3 − 5)² are both 64.

A fully worked model. Let A = (−3, 7) and B = (5, 1). The corner point C shares the x-coordinate of B and the y-coordinate of A, so C = (5, 7). Horizontal leg AC: from x = −3 to x = 5 is 5 − (−3) = 8 units. Vertical leg CB: from y = 7 down to y = 1 is 7 − 1 = 6 units. Then AB² = 8² + 6² = 64 + 36 = 100, so AB = 10 units. The diagram shows this triangle. Notice the step that is easy to skip: the horizontal leg crosses the y-axis, and its length is 8, not 5 − 3 = 2; subtracting a negative number adds.

When the answer is irrational. Let P = (2, 3) and Q = (7, 1). Horizontal leg 7 − 2 = 5, vertical leg 3 − 1 = 2. PQ² = 25 + 4 = 29, so PQ = √29. This is irrational; 25 < 29 < 36 places it between 5 and 6, and 5.4² = 29.16 is slightly too large. Refining, 5.38² = 28.9444 < 29 < 29.0521 = 5.39², and the midpoint test 5.385² = 28.998225 < 29 puts it above 5.385, so PQ ≈ 5.39 to the nearest hundredth. Give the exact form √29 unless the problem asks for a decimal, and state the precision when you round.

Points on the same line. If two points share a y-coordinate, such as (−2, 3) and (3, 3), the distance is simply the horizontal difference, 3 − (−2) = 5. The formula still works, because the vertical leg is 0 and √(25 + 0) = 5, but there is no triangle to draw.

Grid units mean something. When a map uses a coordinate system with 1 unit = 1 km, a computed distance of 10 units is 10 km, in a straight line. This is the straight-line distance, not the distance along streets; a street route that follows the legs would be 8 + 6 = 14 km in the model above.

Using distances to describe figures. Distances on the grid let you find the perimeter of a polygon by adding its side lengths, and they let you classify triangles. A triangle with two equal side lengths is isosceles. A triangle is right when its three side lengths satisfy a² + b² = c² with c the longest side, by the converse of the Pythagorean Theorem. Since the squared distances come straight from the formula, you can test the converse without taking any square roots: compare the sums of squared lengths.

Misconceptions, and why they are wrong. Counting diagonal grid squares as if each were one unit long is the most common error. From (0, 0) to (8, 8) a student may count eight diagonal steps and report 8. Each diagonal step crosses a unit square from corner to corner, and that diagonal is √(1 + 1) = √2 ≈ 1.41, so the true distance is 8√2 ≈ 11.31. Mixing coordinates, subtracting an x from a y, produces a leg that does not exist in the figure. Forgetting the square root reports the squared distance; a distance of 100 between (−3, 7) and (5, 1) is impossible because the two legs together are only 14. Finally, the distance between two points is the same in both directions; if your value from A to B differs from your value from B to A, one subtraction went wrong.

Connecting to the theorem. Everything here is the Pythagorean Theorem applied in a coordinate system; there is no separate rule to memorize. When you can draw the triangle, read the legs and take the root, you can find the distance between any two points on the plane.

Worked distance with negative coordinates: Find the distance from (−4, −2) to (1, 4). Horizontal leg: 1 − (−4) = 5. Vertical leg: 4 − (−2) = 6. Distance squared: 25 + 36 = 61, so the distance is √61. Since 49 < 61 < 64, it lies between 7 and 8; 7.8² = 60.84 and 7.9² = 62.41, so the distance is about 7.8, and a calculator gives 7.81. Both subtractions involve a negative coordinate, so subtracting produces an addition; the leg is 5, not 3.

Worked perimeter and classification: Triangle with vertices (1, 1), (4, 5) and (4, 1). Side from (1, 1) to (4, 1): horizontal, 3 units. Side from (4, 1) to (4, 5): vertical, 4 units. Side from (1, 1) to (4, 5): legs 3 and 4, so √(9 + 16) = 5 units. Perimeter: 3 + 4 + 5 = 12 units. The triangle is right at (4, 1), where a horizontal and a vertical side meet, and the converse agrees: 3² + 4² = 25 = 5².

Worked isosceles test: Triangle P = (0, 0), Q = (6, 2), R = (2, 6). PQ² = 36 + 4 = 40. PR² = 4 + 36 = 40. QR² = (6 − 2)² + (2 − 6)² = 16 + 16 = 32. PQ and PR are both √40 ≈ 6.32, so the triangle is isosceles. Is it right? The longest sides are the two of length √40; check 32 + 40 = 72, which is not 40, so no pair of sides satisfies a² + b² = c² and the triangle is not right. The squared lengths were compared directly, without taking roots.

Plot-and-measure record: On grid paper plot A = (−3, 7) and B = (5, 1), draw the corner point C = (5, 7) and the right triangle, and label the legs 8 and 6 and the hypotenuse 10. Then plot (2, 3) and (7, 1), draw the triangle, label legs 5 and 2, and write the hypotenuse as √29 ≈ 5.39. For each, measure the hypotenuse with a ruler in grid units and compare with the computed value, recording the difference.

Independent performance: Choose four points in different quadrants that form a convex quadrilateral (every corner points outward; no dent), and find all four side lengths and the perimeter. Success criteria: each side has its right triangle drawn or its legs stated from the coordinates; every subtraction with a negative coordinate is shown; each length is given in exact form and, if irrational, as a decimal to two places; the perimeter is stated with units; and both diagonals are computed and used, with the converse of the Pythagorean Theorem, to decide which corners of the quadrilateral are right angles: each diagonal splits the quadrilateral into two triangles and tests the two corners it does not pass through, so one diagonal settles two corners and the second diagonal settles the other two; show the squared lengths that were compared.