Higher June 2024 Paper 3 Q15
15 A straight line passes through points \(A\,(-5, 9)\), \(B\) and \(C\,(3, -7)\).
Not drawn accurately
(a) \(AB : BC = 1 : 3\)
Work out the coordinates of point \(B\). [3 marks]
(b) Work out the equation of the line perpendicular to \(AC\) that passes through \(C\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Correct method for finding the difference between the \(x\) or \(y\) coordinates for line \(AC\) | M1 | may be on diagram eg \(9 - -7\) or 16 or \(3 - -5\) or 8 |
| Correct method for finding the difference between the \(x\) or \(y\) coordinates for line \(AB\) or line \(BC\) | M1dep | may be on diagram eg \(16 \div (1 + 3)\) or 4 or \(8 \div (1 + 3)\) or 2 or \(16 \times \dfrac{3}{(1 + 3)}\) or 12 or \(8 \times \dfrac{3}{(1 + 3)}\) or 6 |
| \((-3, 5)\) | A1 |
Additional guidance
| Up to M2 may be awarded for correct work, with no or incorrect answer, even if this is seen amongst multiple attempts | |
| Condone any missing minus signs if absolute values for the differences are correct | |
| \((-3, \ldots)\) or \((\ldots, 5)\) | M1M1A0 |
| Answer | Mark | Comments |
|---|---|---|
| \((m_1 =)\ \dfrac{-7 - 9}{3 - -5}\) or \((m_1 =)\ \dfrac{9 - -7}{-5 - 3}\) or \(-2\) | M1 | gradient of \(AC\) |
| \(-1 \div\) their \(-2\) or \(\dfrac{1}{2}\) | M1 | gradient of line perpendicular to \(AC\) their \(-2\) must be identified as a gradient \(\dfrac{1}{2}\) implies M1M1 |
| \(-7 =\) their \(\dfrac{1}{2} \times 3 + c\) or \((c =)\ -8.5\) or \(y - -7 =\) their \(\dfrac{1}{2}(x - 3)\) | M1dep | oe condone any letter for \(c\) dep on 2nd M1 |
| \(y = \dfrac{1}{2}x - 8.5\) | A1 | oe eg \(2y = x - 17\) |
Additional guidance
Check part (a) for working for part (b)