Coordinate Geometry Proofs: Geometry Math Worksheet Answer Key – PDF Download

Contents

This answer key shows a complete worked solution for every problem on the coordinate geometry worksheet. Compare each step with your own work, check the slope and distance calculations, and note where a figure helps verify a result.

GeometryCommon Core: G-GPE.B.4 · G-GPE.B.5 · G-GPE.B.6 · G-GPE.B.7

Problem 1: Vertical distance

Both points have the same x-coordinate, so the segment is vertical. Its length is $|4-(-6)|=10$.

The length is 10 units.

Problem 2: Distance with a 3-4-5 triangle

$d=\sqrt{(4-1)^{2}+(6-2)^{2}}=\sqrt{9+16}=\sqrt{25}=5$.

The distance is 5 units.

Problem 3: Distance between two points

$d=\sqrt{(4-(-2))^{2}+(7-(-1))^{2}}=\sqrt{36+64}=\sqrt{100}=10$.

The distance is 10 units.

Problem 4: Distance as a radical

$d=\sqrt{(1-(-3))^{2}+(4-2)^{2}}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$. Since $\sqrt{5}\approx 2.236$, $2\sqrt{5}\approx 4.47$.

$2\sqrt{5}\approx 4.47$ units

Problem 5: Midpoint of a segment

$M=(\frac{2+10}{2},\frac{8+(-4)}{2})=(6,2)$.

$M=(6,2)$

Problem 6: Midpoint with fractions

$M=(\frac{-5+2}{2},\frac{3+6}{2})=(-\frac{3}{2},\frac{9}{2})$.

$M=(-\frac{3}{2},\frac{9}{2})=(-1.5,\,4.5)$

Problem 7: Missing endpoint

Let $B=(x,y)$. Then $\frac{-2+x}{2}=3$ gives $x=8$, and $\frac{4+y}{2}=-1$ gives $y=-6$.

$B=(8,-6)$

Problem 8: Slopes of two lines

Slope of $a$: $\frac{8-2}{4-1}=2$. Slope of $b$: $\frac{2-5}{6-0}=-\frac{1}{2}$. The slopes are not equal, but $2\cdot(-\frac{1}{2})=-1$.

The lines are perpendicular.

Line a has slope 2 and line b has slope −1/2; they meet at a right angle.

Problem 9: Parallel line through a point

A parallel line has the same slope, 3. Point-slope form: $y-(-1)=3(x-4)$, so $y+1=3x-12$.

$y=3x-13$

Check: when $x=4$, $y=3(4)-13=-1$.

Problem 10: Perpendicular line through a point

The perpendicular slope is the opposite reciprocal of $-\frac{2}{3}$, which is $\frac{3}{2}$. Then $y-2=\frac{3}{2}(x-6)$, so $y=\frac{3}{2}x-9+2$.

$y=\frac{3}{2}x-7$

Check: when $x=6$, $y=9-7=2$.

Problem 11: Point one third of the way

The fraction is $\frac{1}{1+2}=\frac{1}{3}$. The run is $12-0=12$ and the rise is $14-2=12$. Move $\frac{1}{3}(12)=4$ right and 4 up from $A$.

$P=(4,6)$

P is 1/3 of the way: run 12 ÷ 3 = 4, rise 12 ÷ 3 = 4.

Problem 12: Ratio 3 to 2

The fraction is $\frac{3}{3+2}=\frac{3}{5}$. Run: $4-(-6)=10$, so move $\frac{3}{5}(10)=6$. Rise: $-8-2=-10$, so move $\frac{3}{5}(-10)=-6$. Then $(-6+6,\,2-6)=(0,-4)$.

$(0,-4)$

Problem 13: Perimeter of a triangle

$AB=6$ (horizontal) and $BC=8$ (vertical). $AC=\sqrt{6^{2}+8^{2}}=\sqrt{100}=10$. Perimeter: $6+8+10=24$.

The perimeter is 24 units.

Legs 6 and 8, hypotenuse 10: perimeter 24.

Problem 14: Area of a tilted rectangle

$PQ=\sqrt{4^{2}+2^{2}}=\sqrt{20}=2\sqrt{5}$ and $QR=\sqrt{(-1)^{2}+2^{2}}=\sqrt{5}$. Area $=2\sqrt{5}\cdot\sqrt{5}=2\cdot 5=10$.

The area is 10 square units.

Check: slope of $\overline{PQ}$ is $\frac{1}{2}$ and slope of $\overline{QR}$ is $-2$; their product is $-1$, so the angle at $Q$ is right.

Side lengths 2√5 and √5 give area 10.

Problem 15: Area of a triangle

$\overline{AB}$ is horizontal, so the base is $6-(-2)=8$. The height is the vertical distance from $C$ to the line $y=1$: $7-1=6$. Area $=\frac{1}{2}(8)(6)=24$.

The area is 24 square units.

Problem 16: Classify a triangle by sides

$AB=6$. $AC=\sqrt{3^{2}+5^{2}}=\sqrt{34}$. $BC=\sqrt{(-3)^{2}+5^{2}}=\sqrt{34}$. Two sides are equal, and $\sqrt{34}\approx 5.83\neq 6$.

The triangle is isosceles but not equilateral.

Problem 17: Right angle with slopes

Slope of $\overline{PQ}$: $\frac{3-1}{5-1}=\frac{1}{2}$. Slope of $\overline{QR}$: $\frac{7-3}{3-5}=-2$. Product: $\frac{1}{2}\cdot(-2)=-1$, so $\overline{PQ}\perp\overline{QR}$.

$PQ=\sqrt{20}$ and $QR=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20}$. Area $=\frac{1}{2}\sqrt{20}\cdot\sqrt{20}=10$.

The right angle is at $Q$ and the area is 10 square units.

Slopes 1/2 and −2 multiply to −1: right angle at Q.

Problem 18: Prove a parallelogram

Midpoint of $\overline{AC}$: $(\frac{1+8}{2},\frac{1+6}{2})=(4.5,\,3.5)$. Midpoint of $\overline{BD}$: $(\frac{6+3}{2},\frac{2+5}{2})=(4.5,\,3.5)$.

The diagonals share a midpoint, so they bisect each other and ABCD is a parallelogram.

Check: from $A$ to $B$ and from $D$ to $C$ the move is 5 right and 1 up.

Both diagonals have midpoint (4.5, 3.5), so ABCD is a parallelogram.

Problem 19: Prove a rhombus

Each side has a 3 by 4 change: $AB=\sqrt{3^{2}+4^{2}}=5$, $BC=5$, $CD=5$, $DA=5$. Four equal sides make a rhombus.

The diagonals lie on the x-axis and the y-axis, so they are perpendicular and cut the figure into four right triangles with legs 3 and 4. Each has area $\frac{1}{2}(3)(4)=6$, so the area is $4\cdot 6=24$.

It is a rhombus with area 24 square units.

All four sides of ABCD measure 5 and the diagonals are perpendicular.

Problem 20: Walk to the library

(a) $SL=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13$ blocks, and $13\times 0.1=1.3$ miles.

(b) $(\frac{2+14}{2},\frac{3+8}{2})=(8,\,5.5)$.

(a) 1.3 miles. (b) The bench is at $(8,5.5)$.

Problem 21: Ranger station on a trail

Move $\frac{2}{5}(15)=6$ right and $\frac{2}{5}(20)=8$ up: $R=(6,8)$.

Whole trail: $\sqrt{15^{2}+20^{2}}=\sqrt{625}=25$ units. The remaining $\frac{3}{5}$ is $15$ units, or $15\times 0.1=1.5$ miles. Check: $RC=\sqrt{9^{2}+12^{2}}=15$.

$R=(6,8)$; the station is 1.5 miles from the campsite.

Problem 22: Garden fence cost

$AB=8$ ft and $BC=6$ ft.

(a) $2(8+6)=28$ ft.

(b) $28\times 4.50=126$.

(d) $AC=\sqrt{8^{2}+6^{2}}=10$ ft.

(a) 28 ft. (b) $126.00. (c) 48 square feet. (d) 10 ft.

Problem 23: Isosceles trapezoid park

Slopes: $AB$ is 0 and $DC$ is 0, so $AB\parallel DC$. $\text{Slope }AD=\frac{4}{2}=2$ and $\text{slope }BC=\frac{4}{-2}=-2$, so the other pair is not parallel. Exactly one pair of parallel sides makes a trapezoid.

Legs: $AD=\sqrt{2^{2}+4^{2}}=\sqrt{20}$ and $BC=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20}$. Equal legs make it isosceles.

Area: a 4 by 4 rectangle (16) plus two right triangles with legs 2 and 4 (each 4) is 24 square units. Each square unit is $10\times 10=100$ square yards, so the area is $24\times 100$ = 2,400 square yards.

It is an isosceles trapezoid with area 2,400 square yards.

Split into a 4 by 4 rectangle and two triangles with legs 2 and 4: 16 + 4 + 4 = 24.

Problem 24: Split a triangle

(a) Fraction $\frac{1}{3}$ of the run 12 is 4, so $D=(4,0)$.

(b) $\overline{AB}$ lies on the x-axis (horizontal) and $\overline{CD}$ lies on the line $x=4$ (vertical), so they are perpendicular.

(d) $AD=4$, $CD=9$, $AC=\sqrt{4^{2}+9^{2}}=\sqrt{97}\approx 9.85$. Perimeter $4+9+9.85=22.85$.

(a) $D=(4,0)$. (c) 54 square units. (d) about 22.85 units.

D = (4, 0) puts CD on the vertical line x = 4, perpendicular to AB.

Problem 25: Drone flight

(a) $PL=\sqrt{(8-(-4))^{2}+(14-(-2))^{2}}=\sqrt{144+256}=\sqrt{400}=20$ units, and $20\times 10=200$ yards.

(b) $(\frac{-4+8}{2},\frac{-2+14}{2})=(2,\,6)$.

(a) 200 yards. (b) The photo point is $(2,6)$.

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