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Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: 6.NS.C.6b, 6.NS.C.6c, 6.NS.C.8, 6.G.A.3 · Level: Grade 6
Problem 1 Reading coordinates
Read right or left first, then up or down.
A(2, 3), B(−4, 1), C(−3, −5), D(5, −2).
Problem 2 Naming quadrants
Check the signs. (7, −3): + and −, Quadrant IV. (−2, −8): − and −, III. (−6, 4): − and +, II. (1.5, 9): + and +, I. IV, III, II, I.
Problem 3 Points on the axes
A zero coordinate puts the point on an axis. (0, −4): y-axis, 4 units below the origin. (5, 0): x-axis, 5 units right. (−3, 0): x-axis, 3 units left. (0, 6): y-axis, 6 units above.
Problem 4 Reflecting a point
Across the x-axis the y-sign flips: (8, −3). Across the y-axis the x-sign flips: (−8, 3). (8, −3) and (−8, 3).
Problem 5 Two reflections
- x changes sign: (6, 2).
- y changes sign: (−7, 3).
Problem 6 Horizontal distance
The y-coordinates match. The x-values have different signs, so add absolute values: 9 + 5 = 14 units.
Problem 7 Vertical distance, same signs
The x-coordinates match. Both y-values are negative, so subtract: 9 − 2 = 7 units.
Problem 8 Rational coordinates
Each grid line is 1 unit, and points are at half-units.
E(−1.5, 2.5) in Quadrant II; F(2.5, −0.5) in Quadrant IV; G(−3.5, −2) in Quadrant III.
Problem 9 Two reflections in a row
First: (−3, 4) becomes (−3, −4). Second: the x-sign flips, giving (3, −4). (3, −4).
Problem 10 Finding missing coordinates
- The y-coordinate changes sign: y = 8.
- The x-coordinate changes sign: x = 5.
Problem 11 Rectangle side lengths
AB: x from −4 to 3 at y = −2, so 4 + 3 = 7. BC: y from −2 to 1 at x = 3, so 2 + 1 = 3.
Opposite sides match, so the sides are 7, 3, 7, 3. Perimeter = 7 + 3 + 7 + 3 = 20 units.
Problem 12 Right triangle legs
Vertical side: from (1, 1) to (1, 6), so 6 − 1 = 5. Horizontal side: from (1, 1) to (5, 1), so 5 − 1 = 4. Vertical 5 units, horizontal 4 units.
Problem 13 Trail map
The y-coordinates match, so use the x-values: 5 + 3 = 8. 8 miles.
Problem 14 Reflections across the y-axis
A y-axis reflection keeps y and flips the sign of x. (a) yes. (b) no, y also flips. (c) yes. (a) and (c).
Problem 15 Park walk
Library to pool: same x = −6, so 3 + 4 = 7 blocks. Pool to cafe: same y = −4, so 6 + 5 = 11 blocks. 7 + 11 = 18 blocks.
Problem 16 Rectangle perimeter and area
AB: 5 + 2 = 7. BC: 1 + 4 = 5. The sides are 7 and 5. Perimeter = 2(7 + 5) = 24. Area = 7 × 5 = 35. Sides 7 and 5, perimeter 24 units, area 35 square units.
Problem 17 Square garden
Side: from x = −2 to 4 gives 2 + 4 = 6. The fourth vertex is above (−2, −2) at (−2, 4). Reflect across the y-axis by flipping each x: (2, −2), (−4, −2), (−4, 4), (2, 4). Fourth vertex (−2, 4); side 6; reflection (2, −2), (−4, −2), (−4, 4), (2, 4).
Problem 18 Parallelogram heights
Top: from x = −6 to −1, so 6 − 1 = 5. Bottom: from x = −4 to 1, so 4 + 1 = 5. Vertical gap: y = 2 to y = −2, so 2 + 2 = 4. Horizontal sides 5 and 5; vertical distance 4.
Problem 19 Reflecting a triangle
Flip each x: P′(−1, 2), Q′(−4, 2), R′(−4, 6). PQ is 4 − 1 = 3 and P′Q′ is 4 − 1 = 3; QR and Q′R′ are both 6 − 2 = 4. New vertices (−1, 2), (−4, 2), (−4, 6); side lengths stay the same because a reflection only flips the figure.
Problem 20 Explain the error
The y-coordinate is 0, so the point is on the x-axis, 4 units left of the origin, and axes belong to no quadrant. Quadrant II needs a negative x and a positive y. (−4, 0) is on the x-axis; for example (−2, 7) is in Quadrant II.