Geometry Coordinate Geometry and Proofs Worksheet Solutions (PDF)

Geometry math worksheet solutions: Coordinate Geometry and Proofs – Maths PDF

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This coordinate geometry answers for Geometry covers coordinate geometry and proofs (Common Core G-GPE.B.4, G-GPE.B.5, G-GPE.B.6, G-GPE.B.7), with a free PDF to print.

Use this answer key to check the 20 problems on coordinate geometry. Every solution shows the formula or theorem used, each step of the work, and the final answer in bold. If an answer differs from yours, compare the steps line by line to find where the work changed.

Geometry · Common Core State Standards: G-GPE.B.4, G-GPE.B.5, G-GPE.B.6, G-GPE.B.7

Coordinate geometry answers: exercises 1 to 10

Exercise 1 – Find a slope

$\frac{10 – 2}{5 – 1} = \frac{8}{4} = 2$

Exercise 2 – Parallel line

Parallel lines have the same slope.

Parallel lines have the same slope, 2.

The y-intercept is 3.

y = 2x + 3

Exercise 3 – Are they parallel?

Slope 1: $\frac{4 – 1}{4 + 2} = \frac{1}{2}$. Slope 2: $\frac{3}{6} = \frac{1}{2}$.

The slopes are equal and the lines are different, so they are parallel.

Yes

Exercise 4 – Perpendicular slope

Take the opposite reciprocal.

-4/3

Exercise 5 – Perpendicular line

Perpendicular slopes are opposite reciprocals.

The perpendicular slope is $-2$.

$y + 1 = -2(x – 4)$

$y = -2x + 7$

y = −2x + 7

Exercise 6 – Are they perpendicular?

$\frac{2}{3} \cdot \left( -\frac{3}{2} \right) = -1$.

Yes

Exercise 7 – Distance

$\sqrt{(5 + 3)^{2} + (8 – 2)^{2}} = \sqrt{64 + 36} = 10$

Exercise 8 – Partition 2 : 3

P splits AB in the ratio 2 : 3 from A.

P is $\frac{2}{5}$ of the way from A to B.

$\left( \frac{2}{5} \cdot 10,\ \frac{2}{5} \cdot 5 \right) = (4, 2)$

P(4, 2)

Exercise 9 – Partition 1 : 2

$\frac{1}{3}$ of the way: $x = -2 + \frac{1}{3}(9) = 1$, $y = 4 + \frac{1}{3}(-9) = 1$.

P(1, 1)

Exercise 10 – Partition 3 : 1

$\frac{3}{4}$ of the way: $x = 1 + \frac{3}{4}(8) = 7$, $y = 2 + \frac{3}{4}(8) = 8$.

P(7, 8)

Exercises 11 and up

Exercise 11 – Perimeter of a triangle

$AB = 4$, $BC = 3$, $AC = \sqrt{4^{2} + 3^{2}} = 5$.

$4 + 3 + 5 = 12$

12 units

Exercise 12 – Area of a rectangle

Width $7 – 1 = 6$, height $5 – 1 = 4$.

$A = 6 \cdot 4 = 24$ and $P = 2(6 + 4) = 20$.

Area 24; perimeter 20

Exercise 13 – Area of a triangle

Base AB = 8 (horizontal). Height = the y-coordinate of C = 6.

$\frac{1}{2} \cdot 8 \cdot 6 = 24$

24 square units

Exercise 14 – Area of a trapezoid

The parallel sides are the bottom (length 6) and the top (length 4), and the height is 4.

$\frac{1}{2}(6 + 4) \cdot 4 = 20$

20 square units

Exercise 15 – Isosceles triangle

$AC = \sqrt{2^{2} + 5^{2}} = \sqrt{29}$ and $BC = \sqrt{(-2)^{2} + 5^{2}} = \sqrt{29}$.

Two sides are equal, so the triangle is isosceles (AB = 4 is the base).

AC = BC = √29

Exercise 16 – Right angle test

Slope of AB: $\frac{2}{6} = \frac{1}{3}$. Slope of AC: $\frac{3}{-1} = -3$.

$\frac{1}{3} \cdot (-3) = -1$, so AB is perpendicular to AC.

The angle at A is a right angle

Exercise 17 – Parallelogram proof

Opposite sides have slopes 1/5 and 2.

Slope of AB: $\frac{1}{5}$. Slope of DC: $\frac{5 – 4}{7 – 2} = \frac{1}{5}$.

Slope of BC: $\frac{5 – 1}{7 – 5} = 2$. Slope of AD: $\frac{4}{2} = 2$.

Both pairs of opposite sides are parallel, so ABCD is a parallelogram.

Parallelogram

Exercise 18 – Meeting point on a map

Meeting point M(8, 9).

$(2 + \frac{3}{4} \cdot 8,\ 3 + \frac{3}{4} \cdot 8) = (8, 9)$.

Distance AB: $\sqrt{8^{2} + 8^{2}} = 8\sqrt{2} = 11.3$. Three quarters of that is $0.75 \cdot 11.31 = 8.5$.

(8, 9); about 8.5 miles

Exercise 19 – New road

Slope of the existing road: $\frac{2 – 4}{5 – 0} = -\frac{2}{5}$.

The perpendicular slope is the opposite reciprocal.

5/2

Exercise 20 – Fence cost

Sides: 12, 9 and $\sqrt{12^{2} + 9^{2}} = 15$.

Perimeter: $12 + 9 + 15 = 36$ feet.

Cost: $36 \cdot \$3.50 = \$126$.

36 feet; $126

Key facts behind this coordinate geometry answers

Key facts

Parallel lines: equal slopes. Perpendicular lines: slopes multiply to -1.
Point P with AP:PB = a:b is a/(a+b) of the way from A to B.
Distance: square root of (x2 – x1) squared plus (y2 – y1) squared.
Perimeter = sum of distances; area uses base times height or decomposition.
Coordinate proofs test slopes (parallel, perpendicular) and distances (equal sides).

How to use this coordinate geometry answers

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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