Grade 12 — Term 3
16 study notes · 223 practice questions · Aligned with NIE syllabus
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This term in the Idasara Academy app has 16 lessons (NIE Units 17, 18, 19, 20). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
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Lessons in This Term
Equation of a straight line
What you must know
- Family of parallel lines: Any line parallel to ax + by + c = 0 can be expressed as ax + by + k = 0; any line perpendicular to it can be expressed as bx - ay + k = 0.
- Distance formula between parallel lines: Perpendicular distance between ax + by + c1 = 0 and ax + by + c2 = 0 is d = |c1 - c2| / √(a^2 + b^2).
- Collinearity test: Points A, B, C are collinear if gradient(AB) = gradient(BC), or if area of triangle ABC = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = 0.
Try one question
What is the equation of the straight line passing through the point (2, 3) with gradient m = -1/2?
- A. 2x + y - 7 = 0
- B. x - 2y + 4 = 0
- C. x + 2y - 8 = 0
- D. x + 2y - 4 = 0
Show the answer
Answer: C. x + 2y - 8 = 0. Using the point-slope form y - y₁ = m(x - x₁): y - 3 = (-1/2)(x - 2) => 2(y - 3) = -(x - 2) => 2y - 6 = -x + 2 => x + 2y - 8 = 0.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Line through the intersection of two lines
What you must know
- Avoid computing intersection coordinates: Using L1 + λ*L2 = 0 avoids the clumsy, tedious step of finding explicit coordinates (x0, y0), drastically reducing algebra and preventing arithmetic errors.
- Handling missing L2 line: If the required line is L2 itself, λ diverges to infinity; use symmetric formulation l*L1 + m*L2 = 0 where both lines are naturally included.
- Concurrency determinant condition: Lines a1*x + b1*y + c1 = 0, a2*x + b2*y + c2 = 0, a3*x + b3*y + c3 = 0 are concurrent iff: |a1 b1 c1; a2 b2 c2; a3 b3 c3| = 0.
Try one question
What is the equation of the straight line passing through the intersection of L₁ ≡ x + y - 4 = 0 and L₂ ≡ 2x - y - 2 = 0 and also passing through the origin (0, 0)?
- A. y = x
- B. y = 2x
- C. y = -x
- D. 2y = x
Show the answer
Answer: A. y = x. Solving simultaneously: Adding L₁ + L₂: 3x - 6 = 0 => x = 2. From L₁: 2 + y - 4 = 0 => y = 2. Intersection point is (2, 2). The straight line passing through (0, 0) and (2, 2) has gradient m = (2 - 0)/(2 - 0) = 1, so its equation is y = x.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Position of two points relative to a line
What you must know
- Ratio of division of line segment by line: The line L: ax + by + c = 0 divides the line segment joining P(x1, y1) and Q(x2, y2) in the ratio m : n = - L(x1, y1) / L(x2, y2); internal division if ratio > 0 (opposite signs), external division if ratio < 0 (same signs).
- Point inside a triangle: Point P(x0, y0) lies strictly INSIDE triangle ABC if P and vertex A lie on the same side of line BC, P and B lie on same side of line AC, and P and C lie on same side of line AB.
- Linear programming feasible regions: Systems of linear inequalities like ax + by + c ≥ 0 define convex polygonal feasible regions bounded by half-planes.
Try one question
What is the relative position of the points P(1, 2) and Q(3, -1) with respect to the straight line L ≡ 2x + 3y - 6 = 0?
- A. P and Q lie on opposite sides of the line
- B. P lies on the line and Q lies off the line
- C. Both P and Q lie exactly on the line
- D. P and Q lie on the same side of the line
Show the answer
Answer: A. P and Q lie on opposite sides of the line. Evaluating L(x, y) = 2x + 3y - 6 at both points: For P(1, 2): L(1, 2) = 2(1) + 3(2) - 6 = 2 + 6 - 6 = +2 (> 0). For Q(3, -1): L(3, -1) = 2(3) + 3(-1) - 6 = 6 - 3 - 6 = -3 (< 0). Because the evaluated values have opposite algebraic signs (+2 and -3), P and Q lie on opposite sides of the line.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Cartesian coordinates and the straight line
What you must know
- Area of triangle from vertex coordinates: Area Δ = (1/2) |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|; evaluated as determinant; if area = 0, the three points are collinear.
- Incenter coordinates formula: Triangle ABC with side lengths a, b, c opposite to vertices A(x1, y1), B(x2, y2), C(x3, y3) has Incenter I = ((a*x1 + b*x2 + c*x3)/(a + b + c), (a*y1 + b*y2 + c*y3)/(a + b + c)).
- Orthocenter and Circumcenter relationship (Euler line): In any non-equilateral triangle, Orthocenter H, Centroid G, and Circumcenter O are collinear on the Euler Line, with G dividing HO in ratio HG : GO = 2 : 1.
Try one question
What is the distance between the two points A(-1, 4) and B(5, -4) in the Cartesian plane?
- A. 8 units
- B. 12 units
- C. 14 units
- D. 10 units
Show the answer
Answer: D. 10 units. Using the distance formula d = √[(x₂ - x₁)² + (y₂ - y₁)²]: d = √[(5 - (-1))² + (-4 - 4)²] = √[6² + (-8)²] = √[36 + 64] = √100 = 10 units.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Perpendicular distance from a point to a line
What you must know
- Distinguishing Acute from Obtuse angle bisector: Make constant terms c1, c2 positive; compute (a1*a2 + b1*b2): if a1*a2 + b1*b2 > 0, the '+' sign yields obtuse angle bisector and '-' yields acute angle bisector; if a1*a2 + b1*b2 < 0, '+' yields acute and '-' yields obtuse.
- Angle bisector containing origin: When c1, c2 > 0, the bisector corresponding to the '+' sign contains the origin.
- Perpendicular distance from Origin: Distance from O(0, 0) to ax + by + c = 0 simplifies directly to d = |c| / √(a^2 + b^2).
Try one question
What is the perpendicular distance from the point (1, -2) to the straight line 3x - 4y + 9 = 0?
- A. 5 units
- B. 2 units
- C. 4 units
- D. 3 units
Show the answer
Answer: C. 4 units. Using the perpendicular distance formula d = |a x₀ + b y₀ + c| / √(a² + b²): d = |3(1) - 4(-2) + 9| / √(3² + (-4)²) = |3 + 8 + 9| / √(9 + 16) = |20| / √25 = 20 / 5 = 4 units.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Cartesian equation of a circle
What you must know
- Circle through three non-collinear points: Substitute (x1, y1), (x2, y2), (x3, y3) into x^2 + y^2 + 2gx + 2fy + c = 0 to obtain three linear equations in parameters g, f, c, and solve simultaneously.
- Point position relative to a circle: For point P(x1, y1) and S ≡ x^2 + y^2 + 2gx + 2fy + c = 0: P lies outside the circle if S1 > 0; P lies on the circle if S1 = 0; P lies inside the circle if S1 < 0 (where S1 = x1^2 + y1^2 + 2gx1 + 2fy1 + c).
- Intercepts made on coordinate axes: Circle x^2 + y^2 + 2gx + 2fy + c = 0 intercepts length 2√(g^2 - c) on the x-axis, and length 2√(f^2 - c) on the y-axis; tangent to x-axis if g^2 = c, tangent to y-axis if f^2 = c.
Try one question
What are the center and radius of the circle defined by the general Cartesian equation x² + y² - 6x + 8y - 11 = 0?
- A. Center = (3, -4), Radius = 36
- B. Center = (-3, 4), Radius = 6
- C. Center = (-6, 8), Radius = 11
- D. Center = (3, -4), Radius = 6
Show the answer
Answer: D. Center = (3, -4), Radius = 6. Comparing with x² + y² + 2gx + 2fy + c = 0: 2g = -6 => g = -3; 2f = 8 => f = 4; c = -11. Center = (-g, -f) = (3, -4). Radius r = √(g² + f² - c) = √((-3)² + 4² - (-11)) = √(9 + 16 + 11) = √36 = 6.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Position of a straight line relative to a circle
What you must know
- Equation of tangent at point P(x1, y1) on circle S = 0: For x^2 + y^2 = a^2, tangent is x*x1 + y*y1 = a^2; for general circle x^2 + y^2 + 2gx + 2fy + c = 0, tangent is x*x1 + y*y1 + g(x + x1) + f(y + y1) + c = 0 (T = 0).
- Equation of Normal at P(x1, y1): Normal passes through center (-g, -f); for x^2 + y^2 = a^2, normal is x*y1 - y*x1 = 0 (y = (y1/x1)x).
- Midpoint of chord: The chord of circle S = 0 whose midpoint is (x1, y1) has equation T = S1.
Try one question
What is the condition on the constant c for the straight line y = 2x + c to be a tangent to the circle x² + y² = 5?
- A. c = ±√5
- B. c = ±10
- C. c = ±25
- D. c = ±5
Show the answer
Answer: D. c = ±5. The circle x² + y² = r² has radius r = √5 and center at (0, 0). The line is 2x - y + c = 0. For tangency, the perpendicular distance from center (0, 0) to line must equal radius r: |c| / √(2² + (-1)²) = √5 => |c| / √5 = √5 => |c| = 5 => c = ±5.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Tangents to a circle from an external point
What you must know
- Method 1 vs Method 2 for finding individual tangents: Method 1 uses point-gradient line y - y1 = m(x - x1) and equates perpendicular distance from center to radius r to solve for the two values of m; Method 2 solves the quadratic joint equation S*S1 = T^2 into linear factors.
- Vertical tangent awareness: If the quadratic in m yields only one finite solution, the second tangent is vertical (equation x = x1 with infinite gradient).
- Director Circle: The locus of the point of intersection of two perpendicular tangents to circle x^2 + y^2 = a^2 is the concentric circle x^2 + y^2 = 2a^2 (radius = a√2).
Try one question
What is the length of the tangent segment drawn from the external point P(4, 5) to the circle S ≡ x² + y² - 4x - 2y - 6 = 0?
- A. 3 units
- B. 4 units
- C. 9 units
- D. √17 units
Show the answer
Answer: A. 3 units. The length of the tangent from a point (x₁, y₁) to the circle S = 0 is L = √(S₁₁) = √(x₁² + y₁² - 4x₁ - 2y₁ - 6). Substituting (4, 5): L = √(4² + 5² - 4(4) - 2(5) - 6) = √(41 - 32) = √9 = 3 units.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Circle through the intersection of two circles
What you must know
- Determining λ from auxiliary condition: Substitute a given external point, or use condition of passing through origin, or orthogonality with a third circle to determine parameter λ.
- Smallest circle through intersection of circle and line: The circle in the family S + λ*L = 0 having minimum radius is the one where the intercepted chord acts as its DIAMETER (center lies on line L = 0).
- Radical axis exists even for non-intersecting circles: S1 - S2 = 0 is the geometric radical axis of equal tangent lengths regardless of whether the circles intersect, touch, or are completely disjoint.
Try one question
What is the general Cartesian form representing any circle passing through the points of intersection of a given circle S ≡ x² + y² + 2gx + 2fy + c = 0 and a straight line L ≡ lx + my + n = 0, where λ is a real parameter?
- A. S · L = λ
- B. S / L = λ
- C. S² + λ L² = 0
- D. S + λ L = 0
Show the answer
Answer: D. S + λ L = 0. The linear combination S + λ L = 0 represents a second-degree curve with identical coefficients for x² and y² and no xy cross-term, which is satisfied by all points where both S = 0 and L = 0 simultaneously. Hence, S + λ L = 0 represents the coaxal family of circles through the intersection points.
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Position of two circles
What you must know
- Intersection of Direct Common Tangents: Meet at external center of similitude P dividing C1C2 externally in ratio r1 : r2: P = (r1*C2 - r2*C1) / (r1 - r2).
- Intersection of Transverse Common Tangents: Meet at internal center of similitude Q dividing C1C2 internally in ratio r1 : r2: Q = (r1*C2 + r2*C1) / (r1 + r2).
- Lengths of common tangents: Direct common tangent length = √[d^2 - (r1 - r2)^2]; Transverse common tangent length = √[d^2 - (r1 + r2)^2].
Try one question
Two circles have centers C₁ and C₂ and radii r₁ and r₂, respectively. What is the geometric relationship between the two circles when the distance between their centers d = C₁C₂ satisfies d = r₁ + r₂?
- A. The circles touch each other internally
- B. The circles touch each other externally at exactly one point
- C. The circles intersect each other at two distinct points
- D. One circle lies entirely inside the other without touching
Show the answer
Answer: B. The circles touch each other externally at exactly one point. When the distance between the centers equals the sum of the radii (d = r₁ + r₂), the two circles meet at a single common point on the line of centers between them, touching each other externally.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Condition for two circles to intersect orthogonally
What you must know
- Standard derivation of 2(g1 g2 + f1 f2) = c1 + c2: Distance squared between centers C1(-g1, -f1) and C2(-g2, -f2) is (g1 - g2)^2 + (f1 - f2)^2; equating to r1^2 + r2^2 = (g1^2 + f1^2 - c1) + (g2^2 + f2^2 - c2) expands and simplifies directly to 2(g1 g2 + f1 f2) = c1 + c2.
- Circle orthogonal to three given circles: Its center is the Radical Center of the three circles, and its radius is the length of tangent from the radical center to any of the three circles.
- Self-orthogonal circle impossibility: A real circle cannot be orthogonal to itself since 2(g^2 + f^2) = 2c => g^2 + f^2 - c = 0 (point circle, radius 0).
Try one question
What is the necessary and sufficient algebraic condition for the two circles x² + y² + 2g₁x + 2f₁y + c₁ = 0 and x² + y² + 2g₂x + 2f₂y + c₂ = 0 to intersect orthogonally (at right angles)?
- A. 2g₁g₂ + 2f₁f₂ = c₁ + c₂
- B. g₁g₂ + f₁f₂ = c₁ + c₂
- C. g₁/g₂ = f₁/f₂ = c₁/c₂
- D. 2g₁g₂ - 2f₁f₂ = c₁ - c₂
Show the answer
Answer: A. 2g₁g₂ + 2f₁f₂ = c₁ + c₂. Two circles intersect orthogonally if the tangents at their intersection point are perpendicular, which means the radii to the intersection point form a right triangle with the line of centers: d² = r₁² + r₂². Substituting d² = (g₁ - g₂)² + (f₁ - f₂)², r₁² = g₁² + f₁² - c₁, and r₂² = g₂² + f₂² - c₂ simplifies directly to 2g₁g₂ + 2f₁f₂ = c₁ + c₂.
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Mathematical induction
What you must know
- Divisibility proofs via induction: To prove f(n) is divisible by d, evaluate f(k+1) - f(k) or f(k+1) - m*f(k) to extract d as an explicit factor using inductive hypothesis f(k) = d*M.
- Inequality proofs via induction: When proving f(n) > g(n), start from f(k+1), express it using f(k), substitute f(k) > g(k), and demonstrate that the resulting expression exceeds g(k+1).
- Non-standard base case: If proposition holds for n ≥ 4 (e.g., 2^n > n^2), the base case MUST be verified at n = 4, not n = 1.
Try one question
In proving by the Principle of Mathematical Induction that a proposition P(n) is true for all positive integers n ≥ 1, what are the two mandatory deductive steps required?
- A. Differentiate P(n) with respect to n and set the derivative to zero
- B. Assume P(n) is true for all n and prove P(0) = 0
- C. Base step: Verify P(1) is true; Inductive step: Prove that if P(k) is true, then P(k + 1) is necessarily true
- D. Compute numerical values for n = 1, 2, 3, 4, 5 and extrapolate without proof
Show the answer
Answer: C. Base step: Verify P(1) is true; Inductive step: Prove that if P(k) is true, then P(k + 1) is necessarily true. Mathematical induction consists of: (1) Basis Step: verifying that P(1) holds; and (2) Inductive Step: showing that for any arbitrary integer k ≥ 1, the assumption that P(k) is true (inductive hypothesis) logically guarantees that P(k + 1) is true. By the induction axiom, P(n) is true for all n ∈ ℤ⁺.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Finite series
What you must know
- Arithmetic Mean (AM) and Geometric Mean (GM): For numbers a and b: AM = (a + b)/2; GM = √(ab); for positive distinct numbers, AM > GM.
- Harmonic Progression (H.P.): A sequence whose reciprocals form an A.P.; harmonic mean HM = 2ab / (a + b); connection: AM * HM = (GM)^2.
- Sigma linearity limitation: Note that ∑ (ur * vr) is NOT equal to (∑ ur) * (∑ vr); sigma does not distribute over products or quotients.
Try one question
In an Arithmetic Progression (AP), the first term is a = 3 and the common difference is d = 4. What is the sum of the first 20 terms (S₂₀)?
- A. 1640
- B. 790
- C. 850
- D. 820
Show the answer
Answer: D. 820. Using the sum formula S_n = (n / 2) [2a + (n - 1)d] with n = 20, a = 3, d = 4: S₂₀ = (20 / 2) [2(3) + (19)(4)] = 10 [6 + 76] = 10 × 82 = 820.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Sums of elementary series
What you must know
- Sum of odd numbers: 1 + 3 + 5 + ... + (2n - 1) = ∑_{r=1}^n (2r - 1) = 2[n(n+1)/2] - n = n^2.
- Sum of even numbers: 2 + 4 + 6 + ... + 2n = 2 ∑_{r=1}^n r = n(n + 1).
- Factorial polynomial representation: Express r(r + 1) or r(r + 1)(r + 2) directly to use telescoping or standard sums: ∑_{r=1}^n r(r + 1) = n(n + 1)(n + 2) / 3.
Try one question
What is the closed-form algebraic formula for the sum of the squares of the first n natural numbers, ∑_{r=1}^n r²?
- A. [n(n + 1)] / 2
- B. [n(n + 1)(2n + 1)] / 6
- C. [n(n + 1) / 2]²
- D. [n(2n + 1)(2n - 1)] / 6
Show the answer
Answer: B. [n(n + 1)(2n + 1)] / 6. The sum of the squares of the first n positive integers is given by the standard identity ∑_{r=1}^n r² = [n(n + 1)(2n + 1)] / 6.
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Summation of series
What you must know
- Identifying f(r) in reciprocal products: For ur = 1 / [(ar + b)(a(r+1) + b)], let f(r) = 1 / (ar + b); then f(r) - f(r + 1) = a / [(ar + b)(a(r+1) + b)] = a * ur, so ur = (1/a)[f(r) - f(r + 1)].
- Split cancellation over two steps: If ur = f(r) - f(r + 2), the sum retains TWO initial terms and TWO terminal terms: Sn = f(1) + f(2) - f(n + 1) - f(n + 2).
- Infinite sum from telescoping: If lim_{n->∞} f(n + 1) = 0, the sum to infinity exists and equals S_∞ = f(1).
Try one question
Using the method of differences (telescoping series), what is the sum to n terms S_n = ∑_{r=1}^n [1 / (r(r + 1))]?
- A. 1 / (n + 1)
- B. n / (n + 1)
- C. (n + 1) / n
- D. n / (2n + 1)
Show the answer
Answer: B. n / (n + 1). Expressing in partial fractions: 1 / [r(r + 1)] = 1/r - 1/(r + 1). Summing from r = 1 to n: S_n = (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n + 1)). Intermediate terms cancel telescopically, leaving S_n = 1 - 1/(n + 1) = n / (n + 1).
A model question written by Idasara on the exam pattern. It is not from a past paper.
Convergence and divergence of series
What you must know
- lim un = 0 does NOT guarantee convergence: The converse of the n-th term test is FALSE; e.g., harmonic series ∑ (1/n) has lim(1/n) = 0, but diverges to infinity.
- Sum to infinity of difference series: If Sn = f(1) - f(n + 1) and lim_{n->∞} f(n + 1) = L, then S_∞ = f(1) - L.
- p-series test: The infinite p-series ∑_{n=1}^∞ 1/n^p converges if and only if p > 1; diverges if p ≤ 1.
Try one question
Under what condition on the common ratio r does an infinite geometric series a + ar + ar² + ... (with a ≠ 0) converge to a finite sum?
- A. r = 1
- B. |r| < 1 (-1 < r < 1)
- C. r > 1
- D. |r| ≤ 1
Show the answer
Answer: B. |r| < 1 (-1 < r < 1). The n-th partial sum of a geometric series is S_n = a(1 - r^n) / (1 - r). As n → ∞, r^n → 0 if and only if |r| < 1. In this case, the series converges to the finite sum S_∞ = a / (1 - r). If |r| ≥ 1, the series diverges.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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