Key takeaway: This geometry coordinate test answer key grades nine questions worth 20 points. Also, solutions show all steps, and each note says where a point is gained. So allow half marks when the method is sound but a number slips. Because slopes or lengths both make a valid proof, accept either route.
Use this answer key to score the coordinate geometry test out of 20 points. Each solution is fully worked, and the scoring note after it shows how the points are awarded, including partial credit for correct methods with small arithmetic slips.
Geometry Coordinate Test Answer Key: questions 1–5
1. Distance and midpoint (2 points)
- \(AB=\sqrt{(2-(-4))^{2}+(-5-3)^{2}}=\sqrt{36+64}=10\). \(AB=10\)
- \(M=(\frac{-4+2}{2},\frac{3+(-5)}{2})\). \(M=(-1,-1)\)
Scoring: 1 point for the correct distance (substitution and result), 1 point for the correct midpoint.
2. Parallel or perpendicular (2 points)
Slope of line 1: \(\frac{4-(-2)}{4-1}=2\). Slope of line 2: \(\frac{3-5}{3-(-1)}=-\frac{1}{2}\). Slope of line 3: \(\frac{4-0}{2-0}=2\).
Lines 1 and 3 are parallel (equal slopes). Lines 1 and 2 are perpendicular and lines 2 and 3 are perpendicular (\(2\cdot(-\frac{1}{2})=-1\)).
Scoring: 1 point for the three correct slopes; 1 point for naming the parallel pair and both perpendicular pairs with the product-of-slopes reason.
3. Perpendicular line (2 points)
The perpendicular slope is the opposite reciprocal of \(-\frac{3}{4}\): \(\frac{4}{3}\). Then \(y-5=\frac{4}{3}(x-3)\), so \(y=\frac{4}{3}x-4+5\).
\(y=\frac{4}{3}x+1\) Check: \(\frac{4}{3}(3)+1=5\).
Scoring: 1 point for the slope \(\frac{4}{3}\); 1 point for the correct equation.
4. Partition a segment (2 points)
The run is \(5-(-7)=12\) and the rise is \(-2-4=-6\).
- Fraction \(\frac{1}{3}\): \((-7+4,\,4-2)\). \((-3,2)\)
- Fraction \(\frac{3}{4}\): \((-7+9,\,4-4.5)\). \((2,-0.5)\)
Scoring: 1 point for each correct point (a correct fraction with an arithmetic slip earns no point).
5. Triangle perimeter and area (3 points)
\(PQ=8\) (horizontal). \(QR=\sqrt{(2-6)^{2}+(5-(-1))^{2}}=\sqrt{52}=2\sqrt{13}\). \(PR=\sqrt{(2-(-2))^{2}+(5-(-1))^{2}}=\sqrt{52}=2\sqrt{13}\).
- Perimeter \(=8+4\sqrt{13}\approx 8+14.42\). \(8+4\sqrt{13}\approx 22.42\) units
- Base \(PQ=8\), height \(5-(-1)=6\). Area \(=\frac{1}{2}(8)(6)=24\) square units
Scoring: Part (a): 1 point for the three side lengths, 1 point for the exact and rounded perimeter. Part (b): 1 point for the correct area.
Geometry Coordinate Test Answer Key: questions 6–9
6. Missing endpoint (2 points)
- \(\frac{-5+x}{2}=1\) gives \(x=7\); \(\frac{4+y}{2}=-2\) gives \(y=-8\). \(B=(7,-8)\)
- \(AB=\sqrt{(7-(-5))^{2}+(-8-4)^{2}}=\sqrt{144+144}=\sqrt{288}\). \(AB=12\sqrt{2}\)
Scoring: 1 point for each part.
7. Prove a rectangle (3 points)
| Statements | Reasons |
|---|---|
| \(\text{slope }JK=\frac{-2-0}{1-(-3)}=-\frac{1}{2}\); \(\text{slope }ML=\frac{4-6}{4-0}=-\frac{1}{2}\) | Slope formula |
| \(\text{slope }KL=\frac{4-(-2)}{4-1}=2\); \(\text{slope }JM=\frac{6-0}{0-(-3)}=2\) | Slope formula |
| JKLM is a parallelogram | Both pairs of opposite sides have equal slopes, so they are parallel |
| \(-\frac{1}{2}\cdot 2=-1\), so \(\overline{JK}\perp\overline{KL}\) | Slopes with product \(-1\) are perpendicular |
| JKLM is a rectangle | A parallelogram with a right angle |
\(JK=\sqrt{4^{2}+(-2)^{2}}=\sqrt{20}=2\sqrt{5}\) and \(KL=\sqrt{3^{2}+6^{2}}=\sqrt{45}=3\sqrt{5}\). Area \(=2\sqrt{5}\cdot 3\sqrt{5}=30\).
JKLM is a rectangle with area 30 square units.
Scoring: 1 point for showing both pairs of opposite sides parallel; 1 point for showing a right angle; 1 point for the correct area.
8. Campus map (2 points)
- \(GC=\sqrt{8^{2}+6^{2}}=10\) units, and \(10\times 50=500\). 500 feet
- \((1+\frac{3}{4}(8),\,2+\frac{3}{4}(6))=(7,\,6.5)\). \((7,6.5)\)
Scoring: 1 point for each part.
9. Classify a triangle (2 points)
\(AB=\sqrt{3^{2}+4^{2}}=5\), \(BC=\sqrt{4^{2}+(-3)^{2}}=5\), \(AC=\sqrt{7^{2}+1^{2}}=\sqrt{50}\). Two sides are equal and the third is different.
Slope of \(\overline{AB}\) is \(\frac{4}{3}\) and slope of \(\overline{BC}\) is \(-\frac{3}{4}\); the product is \(-1\). (Also \(5^{2}+5^{2}=50=(\sqrt{50})^{2}\).)
The triangle is isosceles and right, with the right angle at \(B\).
Scoring: 1 point for the three lengths and the classification; 1 point for a valid right-angle justification.
Scoring summary
20 points: add the points of every question of this geometry coordinate test answer key.
How do you score the coordinate test?
In question 1, \(AB=10\) earns one point, so the midpoint \((-1,-1)\) earns the other. Then question 2 gives slopes 2, \(-\frac{1}{2}\) and 2, so lines 1 and 3 are parallel. Also, line 2 is perpendicular to both, because \(2\cdot(-\frac{1}{2})=-1\). Meanwhile, question 3 gives \(y=\frac{4}{3}x+1\), since the perpendicular slope flips and changes sign. Likewise, question 6 gives \(B(7,-8)\). Therefore, award a point for each coordinate. Finally, question 5 gives a perimeter of \(8+4\sqrt{13}\) and an area of 24, because \(PQ=8\) and \(PR=QR=\sqrt{52}\).
Where does partial credit apply?
For the rectangle in question 7, then, award one point for the slopes, one for showing a right angle, and one for the conclusion. However, if a student proves only a parallelogram, give two points. Instead, ask for the product of adjacent slopes. Likewise, in question 8, a correct distance in units earns one point, and the conversion to feet earns another. Since the gym and cafeteria are 10 units apart, the answer is 500 feet. Still, a bare 10 earns only one point. In addition, question 9 earns full credit for two equal sides of length 5, so the label is isosceles. As a result, a student who also notes the right angle shows deeper insight.
What should you reteach?
If question 6 is missed, students likely averaged the wrong numbers. Therefore, write the midpoint equation \(\frac{x_A+x_B}{2}=1\) and solve for the unknown. Also, repeat it for the y-coordinate. If question 3 is missed, students probably kept the original slope. In that case, practice flipping the fraction first. Then change the sign, and finally use the point. Also, check the new line against the given point. As a result, a sign slip shows up at once. Likewise, ask students to graph the line, because a picture exposes a wrong slope. Therefore, the check takes only a minute.
Where do the results point?
Begin by grouping the errors as formula, algebra or proof wording. Then give each group two similar practice items. Also, let students rewrite one lost proof in complete sentences. Finally, a short quiz later shows who is ready. Also, keep each student’s lost points in a list. Then compare the list with the next unit’s quiz. As a result, patterns appear quickly.
See every chapter on the Geometry math page.
