C1 January 2005 Q8
8.
The points \(A(1, 7)\), \(B(20, 7)\) and \(C(p, q)\) form the vertices of a triangle \(ABC\), as shown in Figure 2. The point \(D(8, 2)\) is the mid-point of \(AC\).
(a) Find the value of \(p\) and the value of \(q\). (2)
The line \(l\), which passes through \(D\) and is perpendicular to \(AC\), intersects \(AB\) at \(E\).
(b) Find an equation for \(l\), in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (5)
(c) Find the exact \(x\)-coordinate of \(E\). (2)
| Scheme | Marks |
|---|---|
| \(p = 15,\ q = -3\) | B1 B1 |
| (2) |
Notes
(a) Special case:
If B0 B0 from main scheme, allow M1 for a correct method, e.g. \(8 = \dfrac{1 + p}{2}\).
| Scheme | Marks |
|---|---|
| Grad. of line \(ADC\): \(m = -\dfrac{5}{7}\), Grad. of perp. line \(= -\dfrac{1}{m}\ \left(= \dfrac{7}{5}\right)\) | B1, M1 |
| Equation of \(l\): \(y - 2 = \dfrac{7}{5}(x - 8)\) | M1 A1ft |
| \(7x - 5y - 46 = 0\) (Allow rearrangements, e.g. \(5y = 7x - 46\)) | A1 |
| (5) |
Notes
(b) Finding eqn. of \(ADC\) instead of \(l\) scores M1 A0 A0.
| Scheme | Marks |
|---|---|
| Substitute \(y = 7\) into equation of \(l\) and find \(x = \ldots\) | M1 |
| \(\dfrac{81}{7}\) or \(11\dfrac{4}{7}\) (or exact equiv.) | A1 |
| (2) | |
| (9 marks) |