Higher January 2022 Paper 2R Q22
22 \(ABC\) is a triangle in which angle \(ABC = 90^\circ\)
\(p\) and \(q\) are integers such that
the coordinates of \(A\) are \((p, 10)\)
the coordinates of \(B\) are \((-1, -5)\)
the coordinates of \(C\) are \((8, q)\)
Given that the gradient of \(AC\) is \(-\dfrac{6}{7}\)
work out the value of \(p\) and the value of \(q\)
(5)
| Scheme | Marks |
|---|---|
\((\text{gradient } AB =)\ \dfrac{10 - -5}{p - -1}\left(= \dfrac{10 + 5}{p + 1} = \dfrac{15}{p + 1}\right)\) oe or \((\text{gradient } BC =)\ \dfrac{q - -5}{8 - -1}\left(= \dfrac{q + 5}{8 + 1} = \dfrac{q + 5}{9}\right)\) oe or \((\text{gradient } AC =)\ \dfrac{10 - q}{p - 8}\) oe or \(\sqrt{(p - -1)^2 + (10 - -5)^2}\) or \((p - -1)^2 + (10 - -5)^2\) or \(\sqrt{(8 - -1)^2 + (q - -5)^2}\) or \((8 - -1)^2 + (q - -5)^2\) or \(\sqrt{(8 - p)^2 + (q - 10)^2}\) or \((8 - p)^2 + (q - 10)^2\) oe | M1 |
• \(\text{‘}{\dfrac{15}{p + 1}}\text{’} \times \text{‘}{\dfrac{q + 5}{9}}\text{’} = -1\) or \(\text{‘}{\dfrac{15}{p + 1}}\text{’} = -\text{‘}{\dfrac{9}{q + 5}}\text{’}\) or \(9p + 15q = -84\) oe • \(\text{‘}{\dfrac{10 - q}{p - 8}}\text{’} = -\dfrac{6}{7}\) or \(6p - 7q = -22\) oe • \((p - -1)^2 + (10 - -5)^2 + (8 - -1)^2 + (q - -5)^2 = (8 - p)^2 + (q - 10)^2\) or \(18p + 30q = -168\) Alternative for the second point • \(\dfrac{6}{7}p + 10 = -8 \times -\dfrac{6}{7} + q\) oe | M2 |
Elimination E.g. \(54p + 90q = -504\) \(54p - 63q = -198\) With subtraction or \(153q = -306\) or \(63p + 105q = -588\) \(90p - 105q = -330\) With the operation of addition or \(153p = -918\) or Substitution E.g. \(6\left(\dfrac{-84 - 15q}{9}\right) = -22\) or \(6p - 7\left(\dfrac{-84 - 9p}{15}\right) = -22\) or \(9\left(\dfrac{-22 + 7q}{6}\right) + 15q = -84\) or \(9p + 15\left(\dfrac{6p + 22}{7}\right) = -84\) | M1 |
| \(p = -6\) and \(q = -2\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for finding the gradient of \(AB\) or \(BC\) or \(AC\)
This may be seen embedded in \(m_1 \times m_2 = -1\)
or
for finding the length of \(AB\) or \(BC\) or \(AC\) (or \(AB^2\) etc)
M2: for two out of the three of:
• using \(m_1 \times m_2 = -1\)
• using the gradient of \(AC\) to form an equation.
• using Pythagoras theorem
If not M2, then M1 for one of the equations.
Alternative for the second point
obtaining this equation by using \(y = mx + c\) with coordinates of \(A\) and \(C\) separately, and then eliminating c)
M1: (dep M3) for correct method to eliminate one variable – multiplying one or both equations so the coefficient of \(x\) or \(y\) is the same in both, with the correct operation to eliminate one variable (condone one arithmetic error)
or
isolating \(p\) or \(q\) in one equation and substituting into the other (condone one arithmetic error).
A1: for \(p = -6\) and \(q = -2\)
Must be clearly identified