C1 January 2008 Q4
4. The point \(A\,(-6, 4)\) and the point \(B\,(8, -3)\) lie on the line \(L\).
| Scheme | Marks |
|---|---|
| \(m = \dfrac{4 - (-3)}{-6 - 8}\) or \(\dfrac{-3 - 4}{8 - (-6)}, \qquad = \dfrac{7}{-14}\) or \(\dfrac{-7}{14} \qquad \left(= -\dfrac{1}{2}\right)\) | M1, A1 |
| Equation: \(y - 4 = -\dfrac{1}{2}\left(x - (-6)\right)\) or \(y - (-3) = -\dfrac{1}{2}(x - 8)\) | M1 |
| \(x + 2y - 2 = 0\) (or equiv. with integer coefficients… must have ‘= 0’) (e.g. \(14y + 7x - 14 = 0\) and \(14 - 7x - 14y = 0\) are acceptable) | A1 |
| (4) |
Notes
1st M: Attempt to use \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\) (may be implicit in an equation of \(L\)).
2nd M: Attempting straight line equation in any form, e.g \(y - y_1 = m\left(x - x_1\right)\), \(\dfrac{y - y_1}{x - x_1} = m\), with any value of \(m\) (except 0 or \(\infty\)) and either (–6, 4) or (8, –3).
N.B.: It is also possible to use a different point which lies on the line, such as the midpoint of \(AB\) (1, 0.5).
Alternatively, the 2nd M may be scored by using \(y = mx + c\) with a numerical gradient and substituting (–6, 4) or (8, –3) to find the value of \(c\).
Having coords the wrong way round, e.g. \(y - (-6) = -\dfrac{1}{2}(x - 4)\), loses the 2nd M mark unless a correct general formula is seen, e.g. \(y - y_1 = m\left(x - x_1\right)\).
| Scheme | Marks |
|---|---|
| \((-6 - 8)^2 + \left(4 - (-3)\right)^2\) | M1 |
| \(14^2 + 7^2\) or \((-14)^2 + 7^2\) or \(14^2 + (-7)^2\) (M1 A1 may be implied by 245) | A1 |
| \(AB = \sqrt{14^2 + 7^2}\) or \(\sqrt{7^2\left(2^2 + 1^2\right)}\) or \(\sqrt{245}\) | |
| \(7\sqrt{5}\) | A1cso |
| (3) | |
| (7 marks) |
Notes
M: Attempting to use \(\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2\).
Missing bracket, e.g. \(-14^2 + 7^2\) implies M1 if no earlier version is seen.\(-14^2 + 7^2\) with no further work would be M1 A0.
\(-14^2 + 7^2\) followed by ‘recovery’ can score full marks.