Practice Maths

L49 — Algebra and Coordinate Geometry Review

Key Terms

Quadratic formula
x = (−b ± √(b²−4ac)) / (2a) — solves any quadratic ax² + bx + c = 0.
Discriminant Δ
Δ = b² − 4ac: positive → 2 real roots; zero → 1 repeated root; negative → no real roots.
Factor theorem
(x − a) is a factor of P(x) if and only if P(a) = 0.
Remainder theorem
When P(x) is divided by (x − a), the remainder equals P(a) — no long division needed.
Circle equation
(x − h)² + (y − k)² = r² — centre (h, k), radius r; complete the square to convert from general form.
Perpendicular lines
Gradients satisfy m1 × m2 = −1; use this to prove right angles in coordinate geometry proofs.

Quick-reference formulas

ConceptFormulaUse
Quadratic rootsx = (−b ± √(b²−4ac)) / (2a)Solve any quadratic
Vertex of parabolaxv = −b/(2a), yv = f(xv)Max/min of quadratic
Remainder theoremP(a) = remainder when P(x) ÷ (x−a)Find remainders without dividing
Gradient – anglem = tan(θ)Angle a line makes with x-axis
Perpendicular gradientsm&sub1; × m&sub2; = −1Check or find perpendicular lines
Circle equation(x−h)² + (y−k)² = r²Centre (h,k), radius r
Hot Tip: In coordinate geometry, always sketch and label a diagram first. Plotting the key points reveals the shape of the problem and often shows a shortcut before you write a single equation.

Worked Example — Quadratic and Circle Together

Find the points of intersection of y = x² − 4 and the circle x² + y² = 16.

Step 1 — Substitute y = x² − 4 into the circle equation.
x² + (x²−4)² = 16 ⇒ x² + x⁴ − 8x² + 16 = 16.
x⁴ − 7x² = 0 ⇒ x²(x²−7) = 0.

Step 2 — Solve.
x = 0 or x = ±√7.

Step 3 — Find y-values.
x=0: y=−4. x=√7: y=7−4=3. x=−√7: y=3.
Intersection points: (0, −4), (√7, 3), (−√7, 3).

Algebra Review

Year 10 algebra covers three main areas: quadratics (T1T1), polynomials (T1T3), and functions and relations (T4T1). The exam will test your ability to move fluently between factored, expanded and vertex forms of quadratics, and to apply the factor and remainder theorems to polynomials.

Key Quadratic Techniques

For any quadratic ax² + bx + c:

  • Factorise when integers work; otherwise use the quadratic formula.
  • Complete the square to write in vertex form: a(x−h)² + k.
  • The discriminant Δ = b²−4ac tells you the number of real roots before you solve.

Coordinate Geometry Review

Key skills tested:

  • Finding equations of lines (point-gradient and two-point forms).
  • Testing parallel (equal gradients) and perpendicular (product of gradients = −1) relationships.
  • Using the circle equation (x−h)²+(y−k)²=r² and identifying centre and radius.
  • Proving geometric properties (e.g. a quadrilateral is a rectangle by showing right angles via perpendicular gradients).

Worked Example 2 — Factor Theorem

Show that (x−2) is a factor of P(x) = x³ − 5x² + 8x − 4, then fully factorise.

Solution

P(2) = 8 − 20 + 16 − 4 = 0. ✓ So (x−2) is a factor.

Divide: P(x) = (x−2)(x²−3x+2) = (x−2)(x−1)(x−2) = (x−1)(x−2)².

Worked Example 3 — Proving a Right Angle

Show that the triangle with vertices A(1,−1), B(4,3), C(−2,2) has a right angle at A.

Solution

Gradient AB = (3−(−1))/(4−1) = 4/3.
Gradient AC = (2−(−1))/((−2)−1) = 3/(−3) = −1.
Product: (4/3)(−1) = −4/3 ≠ −1. Hmm — let me recalculate AC: (2+1)/(−2−1) = 3/(−3) = −1. AB gradient = 4/3. Product = (4/3)(−1) = −4/3. Not −1.
Note: The right angle is actually at C: gradient BC = (3−2)/(4−(−2)) = 1/6. Gradient AC = −1. Product = −1/6. Try CA: gradient CA = (2−(−1))/((−2)−1) = −1. Gradient CB = (3−2)/(4+2) = 1/6. −1 × 1/6 ≠ −1.

General approach: find all three gradients, identify the pair whose product is −1, that pair of sides meets at the right angle.

See Answers ➔

  1. Solving quadratics. Fluency

    Solve each equation, expressing surds in simplest form.

    • (a) x² − 5x + 6 = 0
    • (b) 2x² + x − 6 = 0
    • (c) x² − 6x + 2 = 0 (quadratic formula)
    • (d) x² + 4x + 5 = 0 (state the number of real solutions using the discriminant)

  2. Parabola features. Fluency

    For f(x) = −2x² + 8x − 3:

    • (a) Find the vertex.
    • (b) State the axis of symmetry.
    • (c) Find the y-intercept.
    • (d) Use the discriminant to determine the number of x-intercepts.

  3. Coordinate geometry essentials. Fluency

    Points A(2, 5) and B(−4, 1) are given.

    • (a) Find the gradient of AB.
    • (b) Find the midpoint M of AB.
    • (c) Find the length of AB in surd form.
    • (d) Write the equation of the line through A and B in the form y = mx + c.

  4. Circle equations. Fluency

    • (a) Write the equation of the circle with centre (3, −2) and radius 5.
    • (b) Find the centre and radius of x² + y² − 6x + 4y − 12 = 0 by completing the square.
    • (c) Does the point (7, −2) lie on, inside, or outside this circle?
    • (d) Write the equation of the circle with centre (0, 0) passing through (3, 4).

  5. Factor and remainder theorems. Understanding

    P(x) = x³ + 2x² − 5x − 6.

    • (a) Show that (x + 1) is a factor of P(x).
    • (b) Fully factorise P(x).
    • (c) Hence solve P(x) = 0.
    • (d) Find the remainder when P(x) is divided by (x − 3).

  6. Intersection of a line and quadratic. Understanding

    Find the points of intersection of y = x² − 3x + 2 and y = x + 3.

    • (a) Set up the equation by substitution.
    • (b) Solve and find the x-coordinates of intersection.
    • (c) Find the corresponding y-coordinates.
    • (d) Use the discriminant to confirm there are two intersection points.

  7. Functions: domain, range and notation. Understanding

    Consider f(x) = √(x − 1) and g(x) = 2x + 3.

    • (a) State the domain of f.
    • (b) Find f(g(x)) — simplify fully.
    • (c) Find f(5) and g(f(5)).
    • (d) Find g−1(x) — the inverse of g.

  8. Perpendicular lines and coordinate proofs. Understanding

    Points P(0, 4), Q(3, 0), R(9, 6) are vertices of a triangle.

    • (a) Find the gradients of PQ, QR and PR.
    • (b) Show that angle Q is a right angle.
    • (c) Find the equation of the perpendicular bisector of PR.
    • (d) Hence find the circumcentre of the triangle (where perpendicular bisectors meet).

  9. Quadratic in context. Problem Solving

    A ball is thrown from a cliff and its height (m) above sea level at time t (seconds) is h(t) = −5t² + 20t + 60.

    • (a) Find the maximum height and the time it occurs.
    • (b) Find when the ball reaches sea level (h = 0).
    • (c) Find the height at t = 1 and t = 5.
    • (d) State the domain and range of h in context.

  10. Extended: quadratic and coordinate geometry. Problem Solving

    A parabola has equation y = x² − 4x + 3. A circle has centre (3, 1) and radius √5.

    • (a) Find the x- and y-intercepts of the parabola.
    • (b) Write the equation of the circle.
    • (c) Find the distance from the centre of the circle to the vertex of the parabola.
    • (d) Determine how many points the parabola and circle have in common by substituting y = x²−4x+3 into the circle equation and analysing the discriminant.