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UNIT 7 Coordinate Geometry LESSON 8 Coordinate Geometry Unit Test
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Do you want a ready-to-print unit test with answer key and scoring rubric? Below is a complete Coordinate Geometry Unit Test (Unit 7 — Lesson 8) with directions, 22 questions of mixed format, a 60-minute suggested time, point values, and a full answer key with brief solutions.
Coordinate Geometry — Unit Test (Unit 7, Lesson 8)
Time: 60 minutes. Allowed: calculator, ruler, graph paper.
Total points: 100
Instructions: Show all work. Answers must be simplified (exact form) unless a decimal is requested.
Multiple choice (6 points — 1 point each)
1. Distance between A(3, −2) and B(−1, 4):
A) 6 B) 2√13 C) √52 D) 10
2. Midpoint of (0, 5) and (8, −3):
A) (4, 1) B) (8, 1) C) (4, −1) D) (−4, 1)
3. Slope of the line through (2, 7) and (5, −2):
A) 3 B) −3 C) −9/3 D) 9
4. Equation (slope-intercept) of the line through (1, 2) with slope 3:
A) y = 3x + 2 B) y = 3x − 1 C) y = x + 1 D) y = 2x + 1
5. Which pair of slopes are perpendicular?
A) 1/2 and 2 B) 1/2 and −2 C) −1/2 and −2 D) 1/2 and 1/2
6. Equation of a circle with center (2, −1) and radius 5:
A) (x − 2)^2 + (y + 1)^2 = 25
B) (x + 2)^2 + (y − 1)^2 = 25
C) (x − 2)^2 + (y − 1)^2 = 25
D) (x + 2)^2 + (y + 1)^2 = 25
Short answer / quick computations (40 points — indicated points)
7. (3 pts) Find the equation of the line through (−3, 4) and (2, −1). Give in slope-intercept form.
8. (2 pts) Find the coordinates of the intersection of y = 2x + 3 and y = −x + 6.
9. (2 pts) What is the distance from the point (3, 4) to the x-axis?
10. (3 pts) For triangle with vertices (0,0), (6,0), (6,8): find the centroid (intersection of medians).
11. (2 pts) Find the slope of a line perpendicular to 4x − 5y = 20.
12. (3 pts) Find the equation in standard form of the line passing through (4,2) and parallel to 3x − 2y = 6.
13. (2 pts) Reflect the point (2, −3) across the x-axis. Give coordinates.
14. (2 pts) What is the image of (−1, 4) under the translation (x, y) → (x + 5, y − 2)?
15. (3 pts) Find the area of the triangle with vertices (0,0), (6,0), (6,8).
16. (2 pts) Are the points (1,2), (3,6), (5,10) collinear? Yes or No — justify with a one-line calculation.
Problems (longer, show work) (54 points)
17. (5 pts) Find the equation of the perpendicular bisector of the segment joining (2, 3) and (8, −1). Give final equation in slope-intercept or standard form.
18. (4 pts) Find the coordinates of the image of point (4, −1) after a 90° counterclockwise rotation about the origin.
19. (5 pts) Find the perpendicular distance from point (3,4) to the line 3x + 4y − 10 = 0. Show formula and result.
20. (6 pts) Find the equation of the tangent line to the circle x^2 + y^2 = 25 at the point (3,4).
21. (8 pts) Triangle A(1,2), B(5,2), C(5,6). Perform a dilation centered at A with scale factor 1/2. Find coordinates A', B', C'. Then compute the area of triangle A'B'C'.
22. (12 pts) Let L1: x + 2y − 3 = 0 and L2: 2x − y + 1 = 0.
a) (4 pts) Find their point of intersection P.
b) (8 pts) Find the equation of the line through P that is perpendicular to L2. Give answer in standard form.
Answer key and brief solutions
Multiple choice
1. B. Distance: sqrt[(3 − (−1))^2 + (−2 − 4)^2] = sqrt[4^2 + (−6)^2] = sqrt(16 + 36) = sqrt52 = 2√13.
2. A. Midpoint = ((0+8)/2, (5+(−3))/2) = (4,1).
3. B. Slope = (−2 − 7)/(5 − 2) = −9/3 = −3.
4. B. y − 2 = 3(x − 1) → y = 3x − 1.
5. B. Slopes m1 = 1/2 and m2 = −2 are negative reciprocals → perpendicular.
6. A. (x − 2)^2 + (y + 1)^2 = 25.
Short answer solutions
7. (3 pts) Points (−3,4) and (2,−1): slope m = (−1 − 4)/(2 − (−3)) = −5/5 = −1.
Use y = mx + b with (2, −1): −1 = −1(2) + b → b = 1. Equation: y = −x + 1.
8. (2 pts) Solve 2x + 3 = −x + 6 → 3x = 3 → x = 1. y = 5. Intersection: (1,5).
9. (2 pts) Distance to x-axis = |y| = |4| = 4.
10. (3 pts) Centroid = average of coordinates: ((0+6+6)/3, (0+0+8)/3) = (4, 8/3).
11. (2 pts) 4x − 5y = 20 → y = (4/5)x − 4 so slope is 4/5. Perpendicular slope = −5/4.
12. (3 pts) 3x − 2y = 6 → slope = 3/2. Through (4,2): y = (3/2)x + b → 2 = 6 + b → b = −4. So y = (3/2)x − 4.
Standard: 3x − 2y − 8 = 0.
13. (2 pts) Reflection across x-axis: (2, −3) → (2, 3).
14. (2 pts) (−1, 4) → (−1+5, 4−2) = (4, 2).
15. (3 pts) Right triangle base = 6, height = 8 → area = 1/2 * 6 * 8 = 24.
16. (2 pts) Slope from (1,2)→(3,6): (6−2)/(3−1) = 4/2 = 2. From (3,6)→(5,10): (10−6)/(5−3) = 4/2 = 2. Slopes equal ⇒ collinear: Yes.
Problems
17. (5 pts) Endpoints (2,3) and (8,−1).
Midpoint M = ((2+8)/2, (3+(−1))/2) = (5,1).
Slope of segment = (−1 − 3)/(8 − 2) = −4/6 = −2/3.
Perpendicular slope = 3/2.
Equation through M: y − 1 = (3/2)(x − 5) → y = (3/2)x − 15/2 + 1 = (3/2)x − 13/2.
Standard: 3x − 2y − 13 = 0.
18. (4 pts) 90° CCW rotation about origin: (x,y) → (−y, x). So (4, −1) → (1, 4).
19. (5 pts) Distance formula from point (x0,y0) to ax + by + c = 0:
d = |ax0 + by0 + c| / sqrt(a^2 + b^2).
For 3x + 4y − 10 = 0 and point (3,4):
numerator = |3*3 + 4*4 − 10| = |9 + 16 − 10| = |15| = 15.
denominator = sqrt(3^2 + 4^2) = 5.
d = 15/5 = 3.
20. (6 pts) Circle x^2 + y^2 = 25. At point (3,4) the radius vector has slope 4/3. Tangent slope = negative reciprocal = −3/4.
Line through (3,4): y − 4 = (−3/4)(x − 3).
Multiply 4: 4y − 16 = −3x + 9 → 3x + 4y = 25. (Final: 3x + 4y − 25 = 0.)
21. (8 pts) Dilation center A(1,2) with k = 1/2.
A' = A (center) = (1,2).
Vector AB = (5−1, 2−2) = (4, 0) → B' = A + 1/2*(4,0) = (1+2, 2+0) = (3,2).
Vector AC = (5−1, 6−2) = (4,4) → C' = A + 1/2*(4,4) = (1+2, 2+2) = (3,4).
Area original triangle ABC: base AB = 4, height from C = 4 → area = 1/2 * 4 * 4 = 8.
Area scales by k^2 = (1/2)^2 = 1/4 → area A'B'C' = 8 * 1/4 = 2.
22. (12 pts)
a) (4 pts) Solve system:
x + 2y − 3 = 0 → x = 3 − 2y.
Substitute into 2x − y + 1 = 0:
2(3 − 2y) − y + 1 = 0 → 6 − 4y − y + 1 = 0 → 7 − 5y = 0 → y = 7/5.
x = 3 − 2(7/5) = 3 − 14/5 = 15/5 − 14/5 = 1/5.
Intersection P = (1/5, 7/5).
b) (8 pts) L2 is 2x − y + 1 = 0 → rewrite y = 2x + 1, slope m2 = 2.
Perpendicular slope m = −1/2.
Line through P(1/5, 7/5): y − 7/5 = (−1/2)(x − 1/5).
Multiply both sides by 10 to clear fractions:
10(y − 7/5) = −5(x − 1/5) → 10y − 14 = −5x + 1.
Rearrange: 5x + 10y − 15 = 0 → divide by 5: x + 2y − 3 = 0.
(Note: This is interestingly the same equation as L1 — that can happen; check: indeed P lies on L1, and L1 has slope −1/2? Wait L1: x + 2y − 3 = 0 → y = (−1/2)x + 3/2, slope −1/2. So L1 is itself perpendicular to L2. So the requested line is L1.)
Final standard form: x + 2y − 3 = 0.
Scoring suggestion
- Multiple choice: 6 pts (1 each)
- Short-answer (7–16): as indicated sums to 40 pts
- Problems (17–22): as indicated sums to 54 pts
Total = 100 pts.
Would you like this converted into a printable PDF, a version with only questions (no answer key), or a version adapted to a particular grade level (e.g., Grades 8, 9 or Honors/IB/AP)?
Coordinate Geometry — Unit Test (Unit 7, Lesson 8)
Time: 60 minutes. Allowed: calculator, ruler, graph paper.
Total points: 100
Instructions: Show all work. Answers must be simplified (exact form) unless a decimal is requested.
Multiple choice (6 points — 1 point each)
1. Distance between A(3, −2) and B(−1, 4):
A) 6 B) 2√13 C) √52 D) 10
2. Midpoint of (0, 5) and (8, −3):
A) (4, 1) B) (8, 1) C) (4, −1) D) (−4, 1)
3. Slope of the line through (2, 7) and (5, −2):
A) 3 B) −3 C) −9/3 D) 9
4. Equation (slope-intercept) of the line through (1, 2) with slope 3:
A) y = 3x + 2 B) y = 3x − 1 C) y = x + 1 D) y = 2x + 1
5. Which pair of slopes are perpendicular?
A) 1/2 and 2 B) 1/2 and −2 C) −1/2 and −2 D) 1/2 and 1/2
6. Equation of a circle with center (2, −1) and radius 5:
A) (x − 2)^2 + (y + 1)^2 = 25
B) (x + 2)^2 + (y − 1)^2 = 25
C) (x − 2)^2 + (y − 1)^2 = 25
D) (x + 2)^2 + (y + 1)^2 = 25
Short answer / quick computations (40 points — indicated points)
7. (3 pts) Find the equation of the line through (−3, 4) and (2, −1). Give in slope-intercept form.
8. (2 pts) Find the coordinates of the intersection of y = 2x + 3 and y = −x + 6.
9. (2 pts) What is the distance from the point (3, 4) to the x-axis?
10. (3 pts) For triangle with vertices (0,0), (6,0), (6,8): find the centroid (intersection of medians).
11. (2 pts) Find the slope of a line perpendicular to 4x − 5y = 20.
12. (3 pts) Find the equation in standard form of the line passing through (4,2) and parallel to 3x − 2y = 6.
13. (2 pts) Reflect the point (2, −3) across the x-axis. Give coordinates.
14. (2 pts) What is the image of (−1, 4) under the translation (x, y) → (x + 5, y − 2)?
15. (3 pts) Find the area of the triangle with vertices (0,0), (6,0), (6,8).
16. (2 pts) Are the points (1,2), (3,6), (5,10) collinear? Yes or No — justify with a one-line calculation.
Problems (longer, show work) (54 points)
17. (5 pts) Find the equation of the perpendicular bisector of the segment joining (2, 3) and (8, −1). Give final equation in slope-intercept or standard form.
18. (4 pts) Find the coordinates of the image of point (4, −1) after a 90° counterclockwise rotation about the origin.
19. (5 pts) Find the perpendicular distance from point (3,4) to the line 3x + 4y − 10 = 0. Show formula and result.
20. (6 pts) Find the equation of the tangent line to the circle x^2 + y^2 = 25 at the point (3,4).
21. (8 pts) Triangle A(1,2), B(5,2), C(5,6). Perform a dilation centered at A with scale factor 1/2. Find coordinates A', B', C'. Then compute the area of triangle A'B'C'.
22. (12 pts) Let L1: x + 2y − 3 = 0 and L2: 2x − y + 1 = 0.
a) (4 pts) Find their point of intersection P.
b) (8 pts) Find the equation of the line through P that is perpendicular to L2. Give answer in standard form.
Answer key and brief solutions
Multiple choice
1. B. Distance: sqrt[(3 − (−1))^2 + (−2 − 4)^2] = sqrt[4^2 + (−6)^2] = sqrt(16 + 36) = sqrt52 = 2√13.
2. A. Midpoint = ((0+8)/2, (5+(−3))/2) = (4,1).
3. B. Slope = (−2 − 7)/(5 − 2) = −9/3 = −3.
4. B. y − 2 = 3(x − 1) → y = 3x − 1.
5. B. Slopes m1 = 1/2 and m2 = −2 are negative reciprocals → perpendicular.
6. A. (x − 2)^2 + (y + 1)^2 = 25.
Short answer solutions
7. (3 pts) Points (−3,4) and (2,−1): slope m = (−1 − 4)/(2 − (−3)) = −5/5 = −1.
Use y = mx + b with (2, −1): −1 = −1(2) + b → b = 1. Equation: y = −x + 1.
8. (2 pts) Solve 2x + 3 = −x + 6 → 3x = 3 → x = 1. y = 5. Intersection: (1,5).
9. (2 pts) Distance to x-axis = |y| = |4| = 4.
10. (3 pts) Centroid = average of coordinates: ((0+6+6)/3, (0+0+8)/3) = (4, 8/3).
11. (2 pts) 4x − 5y = 20 → y = (4/5)x − 4 so slope is 4/5. Perpendicular slope = −5/4.
12. (3 pts) 3x − 2y = 6 → slope = 3/2. Through (4,2): y = (3/2)x + b → 2 = 6 + b → b = −4. So y = (3/2)x − 4.
Standard: 3x − 2y − 8 = 0.
13. (2 pts) Reflection across x-axis: (2, −3) → (2, 3).
14. (2 pts) (−1, 4) → (−1+5, 4−2) = (4, 2).
15. (3 pts) Right triangle base = 6, height = 8 → area = 1/2 * 6 * 8 = 24.
16. (2 pts) Slope from (1,2)→(3,6): (6−2)/(3−1) = 4/2 = 2. From (3,6)→(5,10): (10−6)/(5−3) = 4/2 = 2. Slopes equal ⇒ collinear: Yes.
Problems
17. (5 pts) Endpoints (2,3) and (8,−1).
Midpoint M = ((2+8)/2, (3+(−1))/2) = (5,1).
Slope of segment = (−1 − 3)/(8 − 2) = −4/6 = −2/3.
Perpendicular slope = 3/2.
Equation through M: y − 1 = (3/2)(x − 5) → y = (3/2)x − 15/2 + 1 = (3/2)x − 13/2.
Standard: 3x − 2y − 13 = 0.
18. (4 pts) 90° CCW rotation about origin: (x,y) → (−y, x). So (4, −1) → (1, 4).
19. (5 pts) Distance formula from point (x0,y0) to ax + by + c = 0:
d = |ax0 + by0 + c| / sqrt(a^2 + b^2).
For 3x + 4y − 10 = 0 and point (3,4):
numerator = |3*3 + 4*4 − 10| = |9 + 16 − 10| = |15| = 15.
denominator = sqrt(3^2 + 4^2) = 5.
d = 15/5 = 3.
20. (6 pts) Circle x^2 + y^2 = 25. At point (3,4) the radius vector has slope 4/3. Tangent slope = negative reciprocal = −3/4.
Line through (3,4): y − 4 = (−3/4)(x − 3).
Multiply 4: 4y − 16 = −3x + 9 → 3x + 4y = 25. (Final: 3x + 4y − 25 = 0.)
21. (8 pts) Dilation center A(1,2) with k = 1/2.
A' = A (center) = (1,2).
Vector AB = (5−1, 2−2) = (4, 0) → B' = A + 1/2*(4,0) = (1+2, 2+0) = (3,2).
Vector AC = (5−1, 6−2) = (4,4) → C' = A + 1/2*(4,4) = (1+2, 2+2) = (3,4).
Area original triangle ABC: base AB = 4, height from C = 4 → area = 1/2 * 4 * 4 = 8.
Area scales by k^2 = (1/2)^2 = 1/4 → area A'B'C' = 8 * 1/4 = 2.
22. (12 pts)
a) (4 pts) Solve system:
x + 2y − 3 = 0 → x = 3 − 2y.
Substitute into 2x − y + 1 = 0:
2(3 − 2y) − y + 1 = 0 → 6 − 4y − y + 1 = 0 → 7 − 5y = 0 → y = 7/5.
x = 3 − 2(7/5) = 3 − 14/5 = 15/5 − 14/5 = 1/5.
Intersection P = (1/5, 7/5).
b) (8 pts) L2 is 2x − y + 1 = 0 → rewrite y = 2x + 1, slope m2 = 2.
Perpendicular slope m = −1/2.
Line through P(1/5, 7/5): y − 7/5 = (−1/2)(x − 1/5).
Multiply both sides by 10 to clear fractions:
10(y − 7/5) = −5(x − 1/5) → 10y − 14 = −5x + 1.
Rearrange: 5x + 10y − 15 = 0 → divide by 5: x + 2y − 3 = 0.
(Note: This is interestingly the same equation as L1 — that can happen; check: indeed P lies on L1, and L1 has slope −1/2? Wait L1: x + 2y − 3 = 0 → y = (−1/2)x + 3/2, slope −1/2. So L1 is itself perpendicular to L2. So the requested line is L1.)
Final standard form: x + 2y − 3 = 0.
Scoring suggestion
- Multiple choice: 6 pts (1 each)
- Short-answer (7–16): as indicated sums to 40 pts
- Problems (17–22): as indicated sums to 54 pts
Total = 100 pts.
Would you like this converted into a printable PDF, a version with only questions (no answer key), or a version adapted to a particular grade level (e.g., Grades 8, 9 or Honors/IB/AP)?
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