C1 January 2013 Q5
5. The line \(l_1\) has equation \(y = -2x + 3\)
The line \(l_2\) is perpendicular to \(l_1\) and passes through the point \((5, 6)\).
The line \(l_2\) crosses the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\).
Given that \(O\) is the origin,
| Scheme | Marks |
|---|---|
| Gradient of \(l_2\) is \(\dfrac{1}{2}\) or 0.5 or \(\dfrac{-1}{-2}\) | B1 |
| Either \(y - 6 = \text{"}\tfrac{1}{2}\text{"}(x - 5)\) or \(y = \text{"}\tfrac{1}{2}\text{"}x + c\) and \(6 = \text{"}\tfrac{1}{2}\text{"}(5) + c \Rightarrow c = (\text{"}\tfrac{7}{2}\text{"})\) | M1 |
| \(x - 2y + 7 = 0\) or \(-x + 2y - 7 = 0\) or \(k(x - 2y + 7) = 0\) with k an integer | A1 |
| (3) |
Notes
B1: Must have ½ or 0.5 or \(\dfrac{-1}{-2}\) o.e. stated and stops, or used in their line equation
M1: Full method to obtain an equation of the line through (5,6) with their “\(m\)”. So \(y - 6 = m(x - 5)\) with their gradient or uses \(y = mx + c\) with \((5, 6)\) and their gradient to find \(c\). Allow any numerical gradient here including −2 or −1 but not zero. (Allow (6,5) as a slip if \(y - y_1 = m(x - x_1)\) is quoted first)
A1: Accept any multiple of the correct equation, provided that the coefficients are integers and equation = 0 e.g. \(-x + 2y - 7 = 0\) or \(k(x - 2y + 7) = 0\) or even \(2y - x - 7 = 0\)
Special Case: In (a) and (b): Produces parallel line instead of perpendicular line: So uses \(m = -2\) This is not treated as a misread as it simplifies the question. The marks will usually be B0 M1 A0, M1 A0, M1 A0 i.e. maximum of 3/7
| Scheme | Marks |
|---|---|
| Puts \(x = 0\), or \(y = 0\) in their equation and solves to find appropriate co-ordinate | M1 |
| \(x\)-coordinate of \(A\) is \(-7\) and \(y\)-coordinate of \(B\) is \(\tfrac{7}{2}\). | A1 cao |
| (2) |
Notes
M1: Either one of the \(x\) or \(y\) coordinates using their equation
A1: Needs both correct values. Accept any correct equivalent. Need not be written as co-ordinates. Even just −7 and 3.5 with no indication which is which may be awarded the A1.
| Scheme | Marks |
|---|---|
| Area \(OAB = \dfrac{1}{2}(7)\left(\dfrac{7}{2}\right) = \dfrac{49}{4}\) (units)\(^2\) Applies \(\pm\tfrac{1}{2}\)(base)(height) | M1 |
| \(\dfrac{49}{4}\) | A1cso |
| (2) | |
| (7 marks) |
Notes
M1: Any correct method for area of triangle \(AOB\), with their values for co-ordinates of \(A\) and \(B\) (may include negatives) Method usually half base times height but determinants could be used.
A1: Any exact equivalent to 49/4, e.g. 12.25. (negative final answer is A0 but replacing by positive is A1) Do not need units.
c.s.o. implies if A0 is scored in (b) then A0 is scored in (c) as well. However if candidate has correct line equation in (a) of wrong form may score A0 in (a) and A1 in (b) and (c)
Note: Special cases: \(\dfrac{1}{2}(-7)\left(+\dfrac{7}{2}\right) = -\dfrac{49}{4}\) (units)\(^2\) is M1 A0 but changing sign to area \(= +\dfrac{49}{4}\) gets M1A1 (recovery)
N.B. Candidates making sign errors in (b) and obtaining +7 and \(-\tfrac{7}{2}\) may also get \(\dfrac{49}{4}\) as their answer following previous errors. They should be awarded A0 as this answer is not ft and is for correct solution only