C1 January 2012 Q6

EdexcelOld spec8 marksCo-ordinate Geometry

Figure 1: line l1 through A on the negative x-axis and B on the positive y-axis, and line l2 through B and C on the positive x-axis — Figure 1

The line \(l_1\) has equation \(2x - 3y + 12 = 0\)

(a) Find the gradient of \(l_1\). (1)

The line \(l_1\) crosses the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\), as shown in Figure 1.

The line \(l_2\) is perpendicular to \(l_1\) and passes through \(B\).

(b) Find an equation of \(l_2\). (3)

The line \(l_2\) crosses the \(x\)-axis at the point \(C\).

Scheme	Marks
\((m =)\dfrac{2}{3}\) (or exact equivalent)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(B\): \((0, 4)\) [award when first seen – may be in (c)]	B1
Gradient: \(\dfrac{-1}{m} = -\dfrac{3}{2}\)	M1
\(y - 4 = -\dfrac{3x}{2}\) or equiv. e.g. \(\left(y = -\dfrac{3x}{2} + 4,\quad 3x + 2y - 8 = 0\right)\)	A1
	(3)

Notes

B1: for coordinates of \(B\). Accept 4 marked on \(y\)-axis (clearly labelled)

M1: for use of perpendicular gradient rule. Follow through their value for \(m\)

A1: for a correct equation (any form, need not be simplified). Answer only 3/3

Mark scheme (c)

Scheme	Marks
\(A\): \((-6, 0)\) [award when first seen – may be in (b)]	B1
\(C\): \(\dfrac{3x}{2} = 4 \ \Rightarrow\ x = \dfrac{8}{3}\) [award when first seen – may be in (b)]	B1ft
Area: Using \(\dfrac{1}{2}(x_C - x_A)y_B\)	M1
\(= \dfrac{1}{2}\left(\dfrac{8}{3} + 6\right)4 = \dfrac{52}{3}\ \left(= 17\dfrac{1}{3}\right)\)	A1 cso
	(4)
	(8 marks)

Notes

ALT

Scheme	Marks
\(BC = \dfrac{4}{6}\sqrt{52}\) (from similar triangles) (or possibly using \(C\))	2^nd B1ft
Area: Using \(\dfrac{1}{2}(AB\times BC)\) N.B. \(AB = \sqrt{6^2 + 4^2} = \sqrt{52}\)	M1
\(= \dfrac{1}{2}\times\sqrt{52}\times\left(\dfrac{2}{3}\sqrt{52}\right) = \dfrac{52}{3}\ \left(= 17\dfrac{1}{3}\right)\)	A1

Notes

1^st B1: for the coordinates of \(A\) (clearly labelled). Accept \(-6\) marked on \(x\)-axis

2^nd B1ft: for the coordinates of \(C\) (clearly labelled) or \(AC = \tfrac{26}{3}\).
Accept \(x = \dfrac{8}{3}\) marked on \(x\)-axis. Follow through from \(l_2\) if \(>0\)

M1: for an expression for the area of the triangle (all lengths \(> 0\)). Ft their 4, \(-6\) and \(\dfrac{8}{3}\)

A1 cso: for \(\dfrac{52}{3}\) or exact equivalent seen but must be a single fraction or \(17\tfrac{1}{3}\) or \(17\tfrac{2}{6}\) etc
\(17\tfrac{1}{3}\) on its own can only score full marks if \(A\), \(B\) and \(C\) are all correct.

ALT 2^nd B1ft If they use this approach award this mark for \(C\) (if seen) or \(BC\)Use of Det 2^nd M1 must get as far as: \(\tfrac{1}{2}\left|x_A\times y_B - x_C\times y_B\right|\)