MPM2D Grade 10 Math Exam Review: Practice Test & Formulas [Ontario]

The MPM2D Grade 10 Principles of Mathematics examination is the critical gateway to senior high school university-preparation courses. Mastery of quadratic equations, analytic geometry, and trigonometry in Grade 10 determines whether a student can transition smoothly into Grade 11 Functions (MCR3U).

Under Ontario's Growing Success framework, the final 30% evaluation tests comprehensive procedural skill and multi-step problem solving. This guide provides the complete Ontario formula cheat sheet, K/T/C/A grading criteria, and representative practice questions with worked solutions.

The Complete MPM2D Exam Formula Cheat Sheet

• Linear Systems: Slope-intercept form y = mx + b; Standard form Ax + By + C = 0; Methods of solving: Substitution, Elimination, and Graphing point of intersection.
• Analytic Geometry Formulas: Midpoint M = ((x₁ + x₂)/2, (y₁ + y₂)/2); Distance between points d = √[(x₂ - x₁)² + (y₂ - y₁)²]; Slope m = (y₂ - y₁) / (x₂ - x₁); Parallel lines m₁ = m₂; Perpendicular lines m₁ · m₂ = -1 (negative reciprocals).
• Equation of a Circle centered at origin (0, 0): x² + y² = r², where r is the radius.
• Geometric Definitions: Median joins vertex to opposite midpoint; Altitude is perpendicular line from vertex to opposite side; Perpendicular bisector passes through midpoint of a segment at a 90° angle.
• Quadratic Relations Forms: Standard form y = ax² + bx + c; Factored form y = a(x - r)(x - s) with zeros r and s; Vertex form y = a(x - h)² + k with vertex (h, k) and axis of symmetry x = h.
• Quadratic Formula: x = [-b ± √(b² - 4ac)] / (2a); Discriminant D = b² - 4ac (D > 0: two real roots; D = 0: one repeated root; D < 0: zero real roots).
• Completing the Square: y = ax² + bx + c → factor a from first two terms → add and subtract (b/2a)² inside bracket → group into perfect square a(x - h)² + k.
• Right-Angle Trigonometry (SOH CAH TOA): sin(θ) = Opp/Hyp; cos(θ) = Adj/Hyp; tan(θ) = Opp/Adj; Pythagorean Theorem a² + b² = c².
• Oblique Triangle Trigonometry: Sine Law: a/sin(A) = b/sin(B) = c/sin(C); Cosine Law: c² = a² + b² - 2ab·cos(C) or cos(C) = (a² + b² - c²) / (2ab).

Unit 1 & 2: Linear Systems & Analytic Geometry Practice

Practice Question 1 (Application): Triangle ABC has vertices A(2, 5), B(6, 1), and C(-2, -3). Determine the exact equation of the median from vertex A to side BC in slope-intercept form.

Worked Solution 1:

• Step 1: Understand definition of a median. A median runs from vertex A to the midpoint of opposite side BC. Let M be the midpoint of BC.
• Step 2: Calculate midpoint M of BC using B(6, 1) and C(-2, -3): x_m = (6 + (-2))/2 = 4/2 = 2; y_m = (1 + (-3))/2 = -2/2 = -1. Therefore, M(2, -1).
• Step 3: Find slope between vertex A(2, 5) and midpoint M(2, -1): Notice x₁ = 2 and x₂ = 2! Slope m = (-1 - 5) / (2 - 2) = -6 / 0, which is undefined.
• Step 4: Formulate equation: Because the line has undefined slope and constant x-coordinates, it is a vertical line.
• Final Equation: x = 2. (Note: A classic exam question testing whether students recognize vertical lines!).

Unit 3 & 4: Quadratic Equations & Optimization Practice

Practice Question 2 (Thinking & Application): A soccer ball's trajectory is modeled by h(t) = -5t² + 20t + 1, where h is height in meters and t is time in seconds. A) By completing the square, find the maximum height and when it occurs. B) Determine the exact time the ball hits the ground using the quadratic formula.

Worked Solution 2:

• Part A Completing the Square: Factor -5 from the t-terms: h(t) = -5(t² - 4t) + 1.
• Half of -4 is -2; (-2)² = 4. Add and subtract 4 inside: h(t) = -5(t² - 4t + 4 - 4) + 1 = -5(t² - 4t + 4) + (-5)(-4) + 1 = -5(t - 2)² + 20 + 1 = -5(t - 2)² + 21.
• Vertex coordinates: (2, 21). Maximum height is 21 meters, occurring at exactly t = 2 seconds.
• Part B Ground Intercept: Set h(t) = 0: -5t² + 20t + 1 = 0. Identify coefficients: a = -5, b = 20, c = 1.
• Quadratic Formula: t = [-20 ± √(20² - 4(-5)(1))] / [2(-5)] = [-20 ± √(400 + 20)] / (-10) = (-20 ± √420) / (-10).
• Evaluate roots: √420 ≈ 20.4939. t₁ = (-20 + 20.4939)/(-10) = 0.4939 / -10 ≈ -0.05 s (Inadmissible: time before launch).
• t₂ = (-20 - 20.4939)/(-10) = -40.4939 / -10 ≈ 4.05 s.
• Concluding Statement: The soccer ball reaches a maximum height of 21 m at 2 s, and hits the ground after approximately 4.05 seconds.

Unit 5: Trigonometry & Oblique Triangles Practice

Practice Question 3 (Application): In triangle PQR, p = 8.0 cm, q = 11.0 cm, and angle R = 52°. Determine the length of side r to one decimal place.

Worked Solution 3:

• Step 1: Analyze given information. Two sides and the contained angle are known (SAS). Use the Cosine Law.
• Step 2: State formula: r² = p² + q² - 2pq·cos(R).
• Step 3: Substitute values: r² = (8.0)² + (11.0)² - 2(8.0)(11.0)·cos(52°) = 64 + 121 - 176·cos(52°).
• Step 4: Compute: cos(52°) ≈ 0.61566. 176 · 0.61566 ≈ 108.356. r² = 185 - 108.356 = 76.644.
• Step 5: Take square root: r = √76.644 ≈ 8.75 cm.
• Concluding Statement: Side r is approximately 8.8 cm.

Top 5 MPM2D Exam Traps

• Trap 1: Forgetting to distribute the negative when completing the square: -5(t² - 4t + 4 - 4) becomes -5(t - 2)² + 20, NOT -20.
• Trap 2: Discarding negative quadratic answers without explaining why (e.g. 'inadmissible since time/distance cannot be negative').
• Trap 3: Confusing Median with Altitude and Perpendicular Bisector in analytic geometry.
• Trap 4: Forgetting that the equation of a circle is x² + y² = r² where r is the radius, not the diameter.
• Trap 5: Calculator in wrong angle mode or misidentifying whether a triangle requires Sine Law vs Cosine Law.