C1 January 2011 Q9

EdexcelOld spec11 marksCo-ordinate Geometry

9. The line \(L_1\) has equation \(2y - 3x - k = 0\), where \(k\) is a constant.

Given that the point \(A\,(1, 4)\) lies on \(L_1\), find

(a) the value of \(k\), (1)

(b) the gradient of \(L_1\). (2)

The line \(L_2\) passes through \(A\) and is perpendicular to \(L_1\).

(c) Find an equation of \(L_2\) giving your answer in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (4)

The line \(L_2\) crosses the \(x\)-axis at the point \(B\).

(d) Find the coordinates of \(B\). (2)

(e) Find the exact length of \(AB\). (2)

Scheme	Marks
\((8 - 3 - k = 0)\) so \(\underline{k = 5}\)	B1
	(1)

Mark scheme (b)

Scheme	Marks
\(2y = 3x + k\)	M1
\(y = \dfrac{3}{2}x + \ldots\) and so \(m = \dfrac{3}{2}\) o.e.	A1
	(2)

M1: for an attempt to rearrange to \(y = \ldots\)

A1: for clear statement that gradient is 1.5, can be \(m = 1.5\) o.e.

Mark scheme (c)

Scheme	Marks
Perpendicular gradient \(= -\dfrac{2}{3}\)	B1ft
Equation of line is: \(\quad y - 4 = -\dfrac{2}{3}(x - 1)\)	M1A1ft
\(\underline{3y + 2x - 14 = 0}\) o.e.	A1
	(4)

B1ft: for using the perpendicular gradient rule correctly on their “1.5”

M1: for an attempt at finding the equation of the line through \(A\) using their gradient. Allow a sign slip

1^st A1ft: for a correct equation of the line follow through their changed gradient

2^nd A1: as printed or equivalent with integer coefficients – allow \(\underline{3y + 2x = 14}\) or \(\underline{3y = 14 - 2x}\)

Mark scheme (d)

Scheme	Marks
\(y = 0,\ \Rightarrow\ B(7, 0)\) or \(\underline{x = 7}\) \(x = 7\) or \(-\dfrac{c}{a}\)	M1A1ft
	(2)

M1: for use of \(y = 0\) to find \(x = \ldots\) in their equation

A1ft: for \(x = 7\) or \(-\dfrac{c}{a}\)

Mark scheme (e)

M1: for an attempt to find \(AB\) or \(AB^2\)

A1: for any correct surd form- need not be simplified