Higher June 2023 Paper 2R Q25
25 The straight line with equation \(\;y - 2x = 7\;\) is the perpendicular bisector of the line \(AB\) where \(A\) is the point with coordinates \((j, 7)\) and \(B\) is the point with coordinates \((6, k)\)
Find the coordinates of the midpoint of the line \(AB\)
Show clear algebraic working.
(6)
| Scheme | Marks |
|---|---|
| (gradient of \(AB\) =) \(\text{``}-\dfrac{1}{2}\text{''}\) or \(\text{``}2\text{''}m = -1\) | M1 |
(gradient of \(AB\) =) \(\dfrac{k - 7}{6 - j}\) oe or (midpoint of \(AB\) =) \(\left(\dfrac{j + 6}{2}, \dfrac{k + 7}{2}\right)\) oe | M1 |
\(\dfrac{k - 7}{6 - j} = -\dfrac{1}{2}\) oe or \(2k - j = 8\) oe or \(\left(\dfrac{k + 7}{2}\right) - 2\left(\dfrac{j + 6}{2}\right) = 7\) oe or \(k - 2j = 19\) oe | M1 |
\(\dfrac{k - 7}{6 - j} = -\dfrac{1}{2}\) oe or \(2k - j = 8\) oe and \(\left(\dfrac{k + 7}{2}\right) - 2\left(\dfrac{j + 6}{2}\right) = 7\) oe or \(k - 2j = 19\) oe | A1 |
| \(k = -1\) and \(j = -10\) | A1 |
| Working required Answer: (\(-2\), 3) | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for the use of \(m_1 \times m_2 = -1\) or
for \(\text{``}-\dfrac{1}{2}\text{''}\) embedded in a linear equation
eg \(y = \text{``}-\dfrac{1}{2}\text{''}x + c\)
M1: for a correct expression for the gradient which may be seen in an equation
or
for a correct expression for the midpoint which may be seen in an equation.
M1: for setting up a correct equation for \(AB\) in terms of gradient
or
for setting up a correct equation for the line given and the midpoint
A1: for 2 correct equations
A1: for a correct value of \(k\) and a correct value of \(j\)
A1: dep on previous M1