View a few samples from our curriculum. Each of these shows problems 1 - 10 (of 34) followed by the solutions.
Week 7, Day 2: Coordinate Geometry
1. Point M is located in Quadrant III of the coordinate plane. It is exactly 7 units away from the x-axis and exactly 4 units away from the y-axis. What are the coordinates of Point M?
A. (-7, -4)
B. (4, 7)
C. (-4, -7)
D. (7, 4)
2. What is the distance between points A(0, 0) and B(3, 4) on the coordinate plane?
A. 7
B. 5
C. √7
D. 25
3. Triangle ABC is drawn on the coordinate plane with vertices at A(-2, 1), B(4, 1), and C(4, 9). What is the perimeter of triangle ABC?
A.14
B. 20
C. 24
D. 48
4. A student was asked to find the distance between points A(2, 1) and B(8, 9). The student's complete work is shown here.
Step 1: Find the differences. x-difference: 8 - 2 = 6. y-difference: 9 - 1 = 8.
Step 2: Add the differences. 6 + 8 = 14.
Answer: The distance is 14.
Three students, Maya, Devon, and Elena, propose explanations for the student's error. Which student correctly identifies the mistake?
A. Devon: The student forgot to square the differences before adding and forgot to take the square root at the end.
B. Maya: The student should have used the midpoint formula instead of the distance formula.
C. Elena: The student should have multiplied the differences rather than adding them.
D. The student's work is correct; the distance is 14.
5. Compare the two quantities described here.
Quantity A: The distance between the points P(-2, 1) and Q(1, 5).
Quantity B: The distance between the points R(3, -4) and S(3, 2).
Which of the following statements about Quantity A and Quantity B is true?
A. Quantity A is greater than Quantity B.
B. The two quantities are equal.
C. Quantity B is greater than Quantity A.
D. The relationship cannot be determined from the information given.
6.Triangle ABC has vertices at A(0, 0), B(6, 0), and C(0, 8). What is the area of triangle ABC, in square units?
A. 14
B. 10
C. 48
D. 24
7. Quadrilateral ABCD has vertices at A(1, 1), B(6, 1), C(5, 4), and D(2, 4). What is the area of quadrilateral ABCD, in square units?
A. 8
B. 15
C. 12
D. 24
8.Three vertices of a square are shown at A(1, 1), B(5, 1), and C(5, 5). What are the coordinates of the fourth vertex, D?
A. (5, 1)
B. (1, 5)
C. (1, -5)
D. (3, 3)
9. A triangle has vertices at P(-3, 2), Q(1, 5), and R(5, 2). Which of the following best classifies triangle PQR?
A. Scalene
B. Right
C. Equilateral
D. Isosceles
10. A circle is drawn on a coordinate plane. The center of the circle is located at (2, 1), and the point (7, 13) lies exactly on the outer edge of the circle. What is the exact area of this circle, in square units?
A.26π
B.144π
C.169π
D.13π
Solutions
1. To find the coordinates, translate distances into x and y values and apply the correct signs for the quadrant.
The distance from the x-axis (up or down) determines the y-coordinate. Since it’s 7 units away from the x-axis, the y-value is either 7 or -7. The distance from the y-axis (left or right) determines the x-coordinate. Since it’s 4 units away from the y-axis, the x-value is either 4 or -The problem states Point M is in Quadrant III, where both x and y must be negative. Therefore, the x-coordinate is -4, and the y-coordinate is -7. The point is (-4, -7).
2. Apply the distance formula: d = √((x₂ - x₁)² + (y₂ - y₁)²)
With (x₁, y₁) = (0, 0) and (x₂, y₂) = (3, 4):
d = √((3 - 0)² + (4 - 0)²) = √(9 + 16) = √25 = 5
The distance is 5 units.
3. To find the perimeter, calculate the length of all three sides and then add them together.
Side AB is horizontal (y-coordinates are both 1). Length = |4 - (-2)| = 6
Side BC is vertical (x-coordinates are both 4). Length = |9 - 1| = 8
Side AC is the diagonal.
Use the Pythagorean theorem with legs 6 and 8: 6² + 8² = c²
36 + 64 = c²
100 = c², which means c = 10.
The length of AC is 10.
Perimeter = 6 + 8 + 10 = 24
4. The distance formula is d = √((x₂ - x₁)² + (y₂ - y₁)²). The student found the correct differences 6 and 8 but skipped the squaring step AND the final square-root step, going straight to 6 + 8 = 14. The correct calculation is √(6² + 8²) = √(36 + 64) = √100 = 10. Devon's explanation pinpoints both missing steps. The points (2, 1) and (8, 9) form a 6-8-10 Pythagorean triple.
5. Compute each distance, then compare.
Quantity A: d = √((1 - (-2))² + (5 - 1)²) = √(3² + 4²) = √(9 + 16) = √25 = 5 (a 3-4-5 triangle)
Quantity B: R and S share the same x-coordinate (x = 3), so the segment is vertical. The distance is |2 - (-4)| = 6
Since 6 > 5, Quantity B is greater than Quantity A by 1 unit.
6. The triangle has a right angle at the origin A(0, 0) because AB lies along the x-axis and AC lies along the y-axis. The two legs serve as the base and height:
Base AB = 6 (horizontal distance from x = 0 to x = 6)
Height AC = 8 (vertical distance from y = 0 to y = 8)
Area of a triangle = (1/2)(base)(height) = (1/2)(6)(8) = 24 square units
7. Quadrilateral ABCD is a trapezoid. The two parallel sides are horizontal:
AB lies along y = 1 with length 6 - 1 = 5, and DC lies along y = 4 with length 5 - 2 = 3.
The perpendicular distance between the parallel sides is the vertical gap, 4 - 1 = 3.
Area of a trapezoid = (1/2)(b₁ + b₂)(h) = (1/2)(5 + 3)(3) = (1/2)(8)(3) = 12 square units
8. The square has side-length 4 (from A to B: 5 - 1 = 4) and is axis-aligned. The three given vertices form an L-shape, with the missing vertex D directly above A and directly to the left of C. D must share its x-coordinate with A (x = 1) and its y-coordinate with C (y = 5). So D = (1, 5).
9. To classify the triangle, find the lengths of all three sides using the distance formula or Pythagorean theorem.
Side PQ: Horizontal change = 4 (from -3 to 1), vertical change is 3 (from 2 to 5). 42 + 32 = 25, so PQ = 5.
Side QR: Horizontal change = 4 (from 1 to 5), vertical change is 3 (from 5 down to 2). 42 + 32 = 25, so QR = 5.
Side PR is horizontal (y-coordinates are both 2). Length = |5 - (-3)| = 8.
The side lengths are 5, 5, and 8. Because exactly two sides are equal in length, the triangle is Isosceles.
10. To find the area of the circle, first find its radius. The radius is the straight-line distance from the center (2, 1) to the point on the edge (7, 13). Use the distance formula (or Pythagorean theorem):
Horizontal change = 7 - 2 = 5
Vertical change = 13 - 1 = 12
Square the changes and add them together: 5² + 12² = 25 + 144 = 169
Take the square root: √169 = 13
The radius of the circle is 13.
Use the area formula for a circle:
Area = π * r²
Area = π * 132 = 169π
