7.4 Problem Solving on the Coordinate Plane
Key Takeaways
- The distance formula d = √((x₂ − x₁)² + (y₂ − y₁)²) is the Pythagorean theorem applied to the horizontal and vertical legs between two points.
- The midpoint formula averages the coordinates: M = ((x₁ + x₂)/2, (y₁ + y₂)/2).
- Horizontal and vertical segment lengths can be found by subtracting the differing coordinate, with no radical required.
- Slope comparison classifies figures on the plane: equal slopes indicate parallel sides and slopes with product −1 indicate perpendicular sides.
- The area of a polygon on the plane is found by decomposition, by subtraction from a bounding rectangle, or for a triangle by using a horizontal or vertical side as the base.
7.4 Problem Solving on the Coordinate Plane
Skill 5 of Competency 3 sits between algebra and geometry: it uses coordinates and algebraic formulas to answer geometric questions. The two workhorse formulas must be memorized, since no reference sheet is supplied.
Distance and midpoint
Distance: d = √((x₂ − x₁)² + (y₂ − y₁)²) Midpoint: M = ((x₁ + x₂)/2, (y₁ + y₂)/2)
The distance formula is the Pythagorean theorem. Between (x₁, y₁) and (x₂, y₂), the horizontal leg is |x₂ − x₁| and the vertical leg is |y₂ − y₁|, so the segment is the hypotenuse of a right triangle. Understanding this means you never have to recall the formula in isolation.
Distance from (−3, 2) to (5, 8): Horizontal leg = 5 − (−3) = 8; vertical leg = 8 − 2 = 6 d = √(64 + 36) = √100 = 10
Midpoint of (−3, 2) and (5, 8): ((−3 + 5)/2, (2 + 8)/2) = (1, 5)
The two formulas are frequently confused. Distance subtracts and squares; midpoint adds and halves. A quick sanity check: the midpoint must lie between the two points, so its coordinates fall between theirs.
For horizontal or vertical segments, skip the formula entirely. From (2, 7) to (2, −4), the length is |7 − (−4)| = 11, no radical needed.
Finding an endpoint from a midpoint
Given one endpoint and the midpoint, solve for the other.
A has coordinates (1, −6) and the midpoint of AB is (4, 2). Find B. (1 + x)/2 = 4 → x = 7, and (−6 + y)/2 = 2 → y = 10. So B = (7, 10).
The shortcut: the midpoint is the same displacement from each endpoint. From (1, −6) to (4, 2) is +3 right and +8 up, so continue the same amount to reach (7, 10).
Classifying figures from coordinates
Compute side lengths with the distance formula and slopes with the slope formula, then apply the definitions from section 7.1.
| Goal | Test |
|---|---|
| Parallelogram | Both pairs of opposite sides have equal slopes |
| Rectangle | Parallelogram plus adjacent slopes with product −1 |
| Rhombus | Parallelogram plus all four sides equal length |
| Square | All four sides equal and adjacent sides perpendicular |
| Right triangle | Two sides have slopes with product −1, or side lengths satisfy a² + b² = c² |
| Isosceles triangle | Two sides have equal length |
Are A(0, 0), B(4, 3), C(7, −1), D(3, −4) the vertices of a square? AB = √(16 + 9) = 5; BC = √(9 + 16) = 5; CD = √(16 + 9) = 5; DA = √(9 + 16) = 5. All sides equal. Slope AB = 3/4; slope BC = (−1 − 3)/(7 − 4) = −4/3. The product is (3/4)(−4/3) = −1, so adjacent sides are perpendicular. Four equal sides and right angles: yes, a square.
Area of a polygon on the plane
Triangle with a horizontal or vertical side. Use that side as the base and the perpendicular distance as the height.
Triangle with vertices (2, 1), (9, 1), (5, 7). Base is horizontal from x = 2 to x = 9, length 7. Height is the vertical distance from y = 1 to y = 7, which is 6. Area = ½(7)(6) = 21 square units.
No convenient side? Enclose the figure in the smallest bounding rectangle and subtract the corner triangles.
Triangle with vertices (1, 1), (6, 3), (3, 7). Bounding rectangle: x from 1 to 6, y from 1 to 7 → area 5 × 6 = 30. Corner triangle 1: legs 5 and 2 → area 5. Corner triangle 2: legs 3 and 4 → area 6. Corner triangle 3: legs 2 and 6 → area 6. Triangle area = 30 − 5 − 6 − 6 = 13 square units.
Verify with the shoelace formula: ½|1(3 − 7) + 6(7 − 1) + 3(1 − 3)| = ½|−4 + 36 − 6| = ½(26) = 13 ✓
Partitioning a segment
An extension of the midpoint idea: to find the point that divides a segment in ratio m : n from A to B, move m/(m + n) of the way.
Find the point one-third of the way from A(−2, 3) to B(7, 12). Change is (9, 9); one third is (3, 3). Point = (−2 + 3, 3 + 3) = (1, 6).
Applied coordinate contexts
Items frequently frame the plane as a map or grid.
A school sits at (2, 5) and a library at (10, 11), with each unit representing one mile. A bus stop is planned at the midpoint. Where is it, and how far is it from the school? Midpoint: (6, 8). Distance from school: √(16 + 9) = 5 miles.
Watch for scale factors in these contexts. If each grid unit is 0.5 miles rather than 1, a computed distance of 10 units means 5 miles. And distinguish straight-line distance — what the distance formula gives — from grid distance along streets, which is |Δx| + |Δy|. For the school and library above, the straight-line distance is 10 units but the along-the-streets distance is 8 + 6 = 14 units. Items specify which one they want, and both values appear among the options.
What is the distance between the points (−6, 4) and (2, −11)?
The midpoint of segment PQ is (−1, 6). If P is (4, −2), what are the coordinates of Q?
A triangle has vertices at (−2, 3), (6, 3), and (4, 9). What is its area?