C1 June 2009 Q8

EdexcelOld spec8 marksCo-ordinate Geometry

Figure 1: line l through A (6, 7) perpendicular to AB, with B (8, 2), C on the y-axis, and triangle OCB shown dashed — Figure 1

The points \(A\) and \(B\) have coordinates \((6, 7)\) and \((8, 2)\) respectively.

The line \(l\) passes through the point \(A\) and is perpendicular to the line \(AB\), as shown in Figure 1.

(a) Find an equation for \(l\) in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (4)

Given that \(l\) intersects the \(y\)-axis at the point \(C\), find

(b) the coordinates of \(C\), (2)

Mark scheme (a)

Scheme	Marks
\(AB\colon\ m = \dfrac{2 - 7}{8 - 6}, \ \left(= -\dfrac{5}{2}\right)\)	B1
Using \(m_1m_2 = -1\colon\ m_2 = \dfrac{2}{5}\)	M1
\(y - 7 = \dfrac{2}{5}(x - 6), \qquad 2x - 5y + 23 = 0 \quad\) (o.e. with integer coefficients)	M1, A1
	(4)

Notes

B1: for an expression for the gradient of \(AB\). Does not need the \(= -2.5\)

1st M1: for use of the perpendicular gradient rule. Follow through their \(m\)

2nd M1: for the use of \((6, 7)\) and their changed gradient to form an equation for \(l\).
Can be awarded for \(\dfrac{y - 7}{x - 6} = \dfrac{2}{5}\) o.e.
Alternative is to use \((6, 7)\) in \(y = mx + c\) to find a value for \(c\). Score when \(c = \ldots\) is reached.

A1: for a correct equation in the required form and must have “\(= 0\)” and integer coefficients

Mark scheme (b)

Scheme	Marks
Using \(x = 0\) in the answer to (a), \(y = \dfrac{23}{5}\) or 4.6	M1, A1ft
	(2)

Notes

M1: for using \(x = 0\) in their answer to part (a) e.g. \(-5y + 23 = 0\)

A1ft: for \(y = \dfrac{23}{5}\) provided that \(x = 0\) clearly seen or \(C\,(0, 4.6)\). Follow through their equation in (a)
If \(x = 0,\ y = 4.6\) are clearly seen but \(C\) is given as \((4.6, 0)\) apply ISW and award the mark.
This A mark requires a simplified fraction or an exact decimal
Accept their 4.6 marked on diagram next to \(C\) for M1A1ft

Mark scheme (c)

Scheme	Marks
Area of triangle \(= \dfrac{1}{2}\times 8\times\dfrac{23}{5} = \dfrac{92}{5}\) (o.e) e.g. \(\left(18\dfrac{2}{5},\ 18.4,\ \dfrac{184}{10}\right)\)	M1 A1
	(2)
	(8 marks)

Notes

M1: for \(\tfrac{1}{2}\times 8\times y_C\) so can follow through their \(y\) coordinate of \(C\).

A1: for 18.4 (o.e.) but their \(y\) coordinate of \(C\) must be positive

Use of 2 triangles or trapezium and triangle
Award M1 when an expression for area of \(OCB\) only is seen

Determinant approach
Award M1 when an expression containing \(\tfrac{1}{2}\times 8\times y_C\) is seen