Leaving Cert Mathematics Higher Paper 2 free practice questions with answers

Question 1 (30 marks) [Coordinate geometry of the line]

(a) The three points A(−1, 2), B(6, 9) and C(2, −5) form a triangle. θ is the acute angle between [AB] and [AC]. (i) Find the slope of the line AB and the slope of the line AC. (ii) Hence, or otherwise, find the value of θ. Give your answer correct to 1 decimal place.

(a) Answer: (i) slope AB = 1, slope AC = • (i) Slope = • Slope of AC = (−5 − 2)/(2 − (−1)) = • (ii) Use tan θ = |

• θ = tan⁻¹(2.5000) = 68.2°.

Angle between two lines: find the two slopes, then use tan θ = |

Formulae and Tables) and take tan⁻¹.

(b) The line ℓ has equation x − y + 4 = 0. The point R has co-ordinates (8, 0). Find the shortest distance from R to the line ℓ. Give your answer in the form p√q, where p, q ∈ ℕ.

(b) Answer: d = 6√2

• The shortest distance is the perpendicular distance: d = |ax₁ + by₁ + c| / √(a² + b²), with a = 1, b = −1, c = 4 and (x₁, y₁) = (8, 0).
• Numerator: |1(8) + (−1)(0) + 4| = |12| = 12.
• Denominator: √(1² + (−1²)) = √2.
• d = 12 / (1√2) = 12√2 / 2 = 6√2.

Distance from a point to a line: use |ax₁ + by₁ + c| / √(a² + b²), then rationalise the denominator to reach the form p√q.

(c) The line ℓ has equation 5x + 3y − 11 = 0. The point P has co-ordinates (−3, −2). Find the equation of the line through P that is perpendicular to ℓ. Give your answer in the form px + qy + r = 0, where p, q, r ∈

ℤ.

(c) Answer: 3x − 5y − 1 = 0

• Slope of ℓ = • The perpendicular slope is the negative reciprocal:
• Use y − y₁ = m(x − x₁) through (−3, −2): y − (−2) = 3/5(x − (−3)).
• Multiply by 5 and rearrange to get 3x − 5y − 1 = 0.

Perpendicular lines: slopes multiply to −1 (negative reciprocal); then use y − y₁ = m(x − x₁) and tidy to integer coefficients.

−7

3 ; (ii) θ ≈ 68.2° y₂ − y₁ x₂ − x₁ . Slope of AB = (9 − 2)/(6 − (−1)) = 1.

−7

3 . m₁ − m₂

1 + m₁m₂ | = |(1 − (−7

3 ))/(1 + (1)(−7

3 ))| = 5

2 . m₁ − m₂

1 + m₁m₂ | (from the −a b = −5

3 .

3

5 .