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MPM2D Analytic Geometry
Grade 10 Unit 2 — practice questions with full solutions

Length and midpoint of a line segment, equation of a circle, and using coordinates to prove what type of shape you have.

Free MPM2D analytic geometry practice for Grade 10 students in Ontario. These 6 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MPM2D unit test, catching up on a lesson, or reviewing before the final exam.
MPM2D · Grade 10 Analytic Geometry 5 core 1 challenge
1Core
-1123456789-1123456789xyA(1, 1)B(7, 3)C(3, 7)
For triangle ABC shown on the grid: (a) find the length of each side in exact form, (b) classify the triangle, and (c) find the midpoint of BC.
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  1. (a) Use the length formula d = √((x₂−x₁)² + (y₂−y₁)²).
  2. AB: √((7−1)² + (3−1)²) = √(36 + 4) = √40.
  3. AC: √((3−1)² + (7−1)²) = √(4 + 36) = √40.
  4. BC: √((3−7)² + (7−3)²) = √(16 + 16) = √32.
  5. (b) Two sides are equal (AB = AC = √40), so the triangle is isosceles.
  6. (c) Midpoint of BC = ((7+3)÷2, (3+7)÷2) = (5, 5).
Final answer(a) AB = AC = √40, BC = √32   (b) isosceles   (c) (5, 5)
2Core
Find the distance between (−3, 4) and (5, −2), giving an exact answer and a decimal to two places.
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  1. Apply the distance formula: √((5 − (−3))² + ((−2) − 4)²).
  2. Simplify inside: √(8² + (−6)²) = √(64 + 36).
  3. That gives √100 = 10 exactly.
Final answer10 units
3Core
Write the equation of a circle centred at the origin with radius 6, and determine whether the point (4, −5) lies inside, on, or outside it.
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  1. A circle centred at the origin has equation x² + y² = r².
  2. With r = 6: x² + y² = 36.
  3. Test the point: 4² + (−5)² = 16 + 25 = 41.
  4. Since 41 > 36, the point is farther from the origin than the radius.
Final answerx² + y² = 36; the point lies outside
4Core
A triangle has vertices P(−2, 1), Q(4, 3) and R(2, −5). Find the equation of the median from P to QR.
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  1. A median joins a vertex to the midpoint of the opposite side.
  2. Midpoint of QR = ((4+2)÷2, (3+(−5))÷2) = (3, −1).
  3. Slope from P(−2, 1) to (3, −1): (−1 − 1) ÷ (3 − (−2)) = −2 ÷ 5.
  4. Use y = mx + b with P: 1 = (−2⁄5)(−2) + b, so 1 = 4⁄5 + b and b = 1⁄5.
Final answery = −2⁄5x + 1⁄5
5Core
Verify that the quadrilateral with vertices A(0, 0), B(4, 2), C(6, 6) and D(2, 4) is a parallelogram.
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  1. A parallelogram has both pairs of opposite sides parallel — so compare slopes.
  2. Slope AB = (2−0)÷(4−0) = 1⁄2. Slope DC = (6−4)÷(6−2) = 1⁄2. Equal, so AB ∥ DC.
  3. Slope AD = (4−0)÷(2−0) = 2. Slope BC = (6−2)÷(6−4) = 2. Equal, so AD ∥ BC.
  4. Both pairs of opposite sides are parallel, which is the definition of a parallelogram.
Final answerBoth pairs of opposite sides have equal slopes (½ and 2), so it is a parallelogram
6Challenge
Find the equation of the perpendicular bisector of the segment joining A(−1, 2) and B(5, 6).
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  1. The perpendicular bisector passes through the midpoint and is perpendicular to AB.
  2. Midpoint: ((−1+5)÷2, (2+6)÷2) = (2, 4).
  3. Slope AB = (6−2)÷(5−(−1)) = 4÷6 = 2⁄3.
  4. Perpendicular slope is the negative reciprocal: −3⁄2.
  5. Through (2, 4): 4 = −3⁄2(2) + b, so 4 = −3 + b and b = 7.
Final answery = −3⁄2x + 7

MPM2D Analytic Geometry — common questions

Short answers to the things students ask most about this unit.

What is the equation of a circle centred at the origin?

It is x² + y² = r², where r is the radius. A point is inside if x² + y² is less than r², and outside if it is greater.

How do you prove a shape is a parallelogram using coordinates?

Show both pairs of opposite sides have equal slopes. For a rectangle you also show adjacent slopes are negative reciprocals.

More MPM2D practice — grade 10 academic

MPM2D Linear SystemsMPM2D Factoring QuadraticsMPM2D Quadratic Relations & GraphingMPM2D Solving Quadratic EquationsAll MPM2D units
← Unit 1: Linear SystemsUnit 3: Factoring Quadratics →
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