C1 January 2006 Q3

EdexcelOld spec5 marksCo-ordinate Geometry

3. The line \(L\) has equation \(y = 5 - 2x\).

(a) Show that the point \(P\,(3, -1)\) lies on \(L\). (1)

(b) Find an equation of the line perpendicular to \(L\), which passes through \(P\). Give your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (4)

Mark scheme (a)

Scheme	Marks
\(y = 5 - (2\times 3) = -1\) (or equivalent verification) (*)	B1
	(1)

Notes

(a) \(y - (-1) = -2(x - 3) \Rightarrow y = 5 - 2x\) is fine for B1.
Just a table of values including \(x = 3,\ y = -1\) is insufficient.

Mark scheme (b)

Scheme	Marks
Gradient of perpendicular to \(L\) is \(\dfrac{1}{2}\)	B1
\(y - (-1) = \dfrac{1}{2}(x - 3)\) (ft from a changed gradient)	M1 A1ft
\(x - 2y - 5 = 0\) (or equiv. with integer coefficients)	A1
	(4)
	(5 marks)

Notes

(corrected from the printed mark scheme: the first line is printed as “Gradient of \(L\) is \(\frac{1}{2}\)”; the gradient of \(L\) is \(-2\), and \(\frac{1}{2}\) is the gradient of the perpendicular.)

(b) M1: eqn of a line through \((3, -1)\), with any numerical gradient (except 0 or \(\infty\)).

For the M1 A1ft, the equation may be in any form, e.g. \(\dfrac{y - (-1)}{x - 3} = \dfrac{1}{2}\).

Alternatively, the M1 may be scored by using \(y = mx + c\) with a numerical gradient and substituting \((3, -1)\) to find the value of \(c\), with A1ft if the value of \(c\) follows through correctly from a changed gradient.

Allow \(x - 2y = 5\) or equiv., but must be integer coefficients.

The “= 0” can be implied if correct working precedes.