← Back to the Coordinate Geometry Proofs worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-GPE.B.4, G-GPE.B.5, G-GPE.B.6, G-GPE.B.7 · Level: Grade 10 Geometry
Problem 1 Distance Between Two Points
d = √((4 − (−3))² + (6 − 1)²) = √(7² + 5²) = √74.
√74 ≈ 8.60.
d = √74 ≈ 8.60
Problem 2 Midpoint of a Segment
x: (−7 + 5) ÷ 2 = −1. y: (4 + −10) ÷ 2 = −3.
(−1, −3)
Problem 3 Missing Endpoint
3 = (8 + x) ÷ 2, so x = −2. −2 = (1 + y) ÷ 2, so y = −5.
K = (−2, −5)
Problem 4 Parallel or Not?
Slope of p = 8 − 24 − 1 = 63 = 2.
Slope of q = 5 − (−3)2 − (−2) = 84 = 2.
Equal slopes, so the lines are parallel.
Problem 5 Perpendicular or Not?
Slope of m = 34. Slope of n = −86 = −43.
Product = 34 · (−43) = −1.
Yes, the lines are perpendicular.
Problem 6 Parallel Line Through a Point
Parallel lines share the slope −4. Use y = −4x + b with (2, −3): −3 = −8 + b, so b = 5.
y = −4x + 5
Problem 7 Perpendicular Line Through a Point
Solve for y: y = 32x − 4, slope 32. The perpendicular slope is −23.
y − 1 = −23(x + 6), so y = −23x − 4 + 1 = −23x − 3.
y = −23x − 3
Problem 8 Dividing a Segment 1 : 3
The parts total 1 + 3 = 4, so the point is 14 of the way from A to B. The changes are 16 in x and −12 in y.
x: −6 + 14(16) = −2. y: 2 + 14(−12) = −1.
P = (−2, −1)
Problem 9 Dividing a Segment 3 : 2
P is 35 of the way from A to B. The changes are 15 in x and −10 in y.
x: −5 + 35(15) = 4. y: 2 + 35(−10) = −4.
P = (4, −4)
Problem 10 Perimeter and Area of a Right Triangle
AB = 7 (horizontal) and BC = 5 (vertical), so the triangle has a right angle at B. AC = √(7² + 5²) = √74 ≈ 8.6.
Perimeter = 12 + √74 ≈ 20.6. Area = ½ · 7 · 5 = 17.5.
Perimeter = 12 + √74 ≈ 20.6; area = 17.5 square units
Problem 11 Area by Enclosing Rectangle
The enclosing rectangle runs from x = 1 to 7 and y = 1 to 8: 6 by 7, area 42.
Corner triangles: ½ · 6 · 2 = 6 (bottom), ½ · 3 · 5 = 7.5 (upper right), ½ · 3 · 7 = 10.5 (left).
Area = 42 − 6 − 7.5 − 10.5 = 18.
18 square units
Problem 12 Classify by Side Lengths
DE = √(5² + 4²) = √41. EF = √(5² + (−4)²) = √41. DF = √(10² + 0²) = 10.
Two sides are equal (√41 ≈ 6.40) and the third is 10.
Isosceles
Problem 13 Prove a Right Triangle
Slope PQ = 53. Slope PR = −610 = −35.
Product = 53 · (−35) = −1, so PQ ⊥ PR and the right angle is at P.
PQ = √(9 + 25) = √34. PR = √(100 + 36) = √136 = 2√34. Area = ½ · √34 · 2√34 = 34.
Right angle at P; area = 34 square units
Problem 14 Prove a Parallelogram
Slope AB = −1 − 02 − (−3) = −15. Slope DC = 3 − 44 − (−1) = −15.
Slope AD = 42 = 2. Slope BC = 42 = 2.
Both pairs of opposite sides are parallel, so ABCD is a parallelogram.
Problem 15 Prove a Rectangle
Slope WX = −25, slope ZY = 2 − 44 − (−1) = −25. Slope XY = 52, slope WZ = 52. Opposite sides are parallel, so it is a parallelogram.
Slope WX · slope XY = −25 · 52 = −1, so there is a right angle at X.
Diagonals: WY = √(7² + 3²) = √58 and XZ = √((−3)² + 7²) = √58, equal as expected for a rectangle.
A parallelogram with a right angle is a rectangle.
Problem 16 Hikers Meet Halfway
Midpoint = (−4 + 82, 2 + −62) = (2, −2).
Distance from Imani = √(6² + (−4)²) = √52 ≈ 7.21. Corbin walks the same distance.
Meet at (2, −2); each walks ≈ 7.2 mi
Problem 17 Picnic Table on the Trail
x: 23(18) = 12. y: 23(12) = 8. So the table is at (12, 8).
Distance = √(12² + 8²) = √208 ≈ 14.42 units = 1,442 yards.
(12, 8); about 1,442 yards
Problem 18 Fencing a Trapezoid Field
(a) AB and DC are horizontal, so they are parallel; the other sides have slopes 52 and −52, not parallel. AD = √(2² + 5²) = √29 and BC = √((−2)² + 5²) = √29, so the legs are equal: isosceles trapezoid.
(b) Perimeter = 10 + 6 + 2√29 = 16 + 2√29 ≈ 26.77 units = 268 yards. Area = ½(10 + 6)(5) = 40 square units = 4,000 square yards.
(b) Fence ≈ 268 yd; area = 4,000 yd²
Problem 19 Triangular Park
Enclosing rectangle: x from 1 to 9 and y from 2 to 10, so 8 by 8 = 64.
Corner triangles: ½ · 8 · 2 = 8, ½ · 4 · 6 = 12, ½ · 4 · 8 = 16. Area = 64 − 8 − 12 − 16 = 28.
Sides: √(64 + 4) = √68 ≈ 8.25, √(16 + 36) = √52 ≈ 7.21, √(16 + 64) = √80 ≈ 8.94.
Area = 28 square blocks; perimeter ≈ 24.4 blocks
Problem 20 Midpoints Form a Parallelogram
Midpoints: AB → (4, 1); BC → (9, 5.5); CD → (6, 8); DA → (1, 3.5).
Slope (4, 1) to (9, 5.5) = 4.55 = 0.9; slope (1, 3.5) to (6, 8) = 4.55 = 0.9.
Slope (9, 5.5) to (6, 8) = 2.5−3; slope (4, 1) to (1, 3.5) = 2.5−3.
Opposite sides have equal slopes, so the midpoint quadrilateral is a parallelogram.