C1 January 2014 (IAL) Q6

EdexcelOld spec11 marksCo-ordinate Geometry

Figure 2: lines l1 and l2 meeting at right angles at B; l1 crosses the y-axis at A and l2 crosses the x-axis at C; rectangle ABCD shaded — Figure 2

The straight line \(l_1\) has equation \(2y = 3x + 7\)

The line \(l_1\) crosses the \(y\)-axis at the point \(A\) as shown in Figure 2.

(a)

(i) State the gradient of \(l_1\)

(ii) Write down the coordinates of the point \(A\).

(2)

Another straight line \(l_2\) intersects \(l_1\) at the point \(B\,(1, 5)\) and crosses the \(x\)-axis at the point \(C\), as shown in Figure 2.

Given that \(\angle ABC = 90^\circ\),

(b) find an equation of \(l_2\) in the form \(ax + by + c = 0\), where \(a\), \(b\) and \(c\) are integers. (4)

The rectangle \(ABCD\), shown shaded in Figure 2, has vertices at the points \(A\), \(B\), \(C\) and \(D\).

Mark scheme (a)

Scheme	Marks
(i) \(\dfrac{3}{2}\) or equivalents such as 1.5	B1
(ii) \((0, 3.5)\) Accept \(y = 3\tfrac{1}{2}\)	B1
	(2)

Notes

B1: cao gradient =1.5. Accept equivalences such as \(\dfrac{3}{2}\)

B1: cao intercept =\((0, 3.5)\). Accept 3.5, \(y\)=3.5 and equivalences such as \(\dfrac{7}{2}\)

Mark scheme (b)

Scheme	Marks
Perpendicular gradient \(l_2 = -\dfrac{2}{3}\)	B1ft
Equation of line is: \(y - 5 = -\dfrac{2}{3}(x - 1)\)	M1A1
\(3y + 2x - 17 = 0\)	A1
	(4)

Notes

B1ft: For using the perpendicular gradient rule, \(m_1 = -\dfrac{1}{m_2}\) on their ‘1.5’.
Accept \(-\dfrac{1}{\text{‘}1.5\text{’}}\) or this as part of their equation for \(l_2\) Eg. \(-\dfrac{1}{\text{‘}1.5\text{’}} = \dfrac{y - \ldots}{x - \ldots}\)

M1: For an attempt at finding the equation of \(l_2\) using (1,5) and their adapted gradient.
Condone for this mark a gradient of \(\dfrac{3}{2}\) going to \(\dfrac{2}{3}\). Eg. Allow for \(\dfrac{y - 5}{x - 1} = \dfrac{2}{3}\)
If the form \(y = mx + c\) is used it must be a full method to find \(c\) with (1,5) and an adapted gradient.

A1: For an a correct unsimplified equation of the line through (1,5) with the correct gradient.
Allow \(\dfrac{y - 5}{x - 1} = -\dfrac{2}{3}\) and \(5 = -\dfrac{2}{3}\times 1 + c \Rightarrow c = \dfrac{17}{3}\)

A1: cso \(\pm(3y + 2x - 17) = 0\)

An example of B1ftM0A0A0 would be \(-\dfrac{1}{3} = \dfrac{y - 5}{x + 1}\) following a gradient of ‘3’ in part (a)
An example of B1ftM1A0A0 would be \(-\dfrac{1}{3} = \dfrac{y - 5}{x - 1}\) following a gradient of ‘3’ in part (a)
An example of B0ftM1A0A0 would be \(\dfrac{1}{3} = \dfrac{y - 5}{x - 1}\) following a gradient of ‘3’ in part (a)

Mark scheme (c)

Scheme	Marks
Point C: \(y = 0 \Rightarrow 2x = 17 \Rightarrow x = 8.5\) oe	M1, A1
\(AB = \sqrt{(1 - 0)^2 + (5 - 3.5)^2} = \left(\dfrac{\sqrt{13}}{2}\right)\) \(BC = \sqrt{(8.5 - 1)^2 + (5 - 0)^2} = \left(\dfrac{\sqrt{325}}{2}\right)\)	M1 (either)
Area rectangle = \(AB\times BC = \dfrac{\sqrt{13}}{2}\times\dfrac{\sqrt{325}}{2} = \dfrac{\sqrt{13}}{2}\times\dfrac{\sqrt{13}\sqrt{25}}{2} = \dfrac{5\times 13}{4} = 16.25\) oe	dM1A1
	(5)
	(11 marks)

Notes

M1: An attempt to use their equation found in part b to find the \(x\) coordinate of \(C\)
They must either use the equation of \(l_2\) and set \(y = 0 \Rightarrow x = \ldots\) or use its gradient \(\dfrac{17.5}{x} = \dfrac{3}{2} \Rightarrow x = ..\)

A1: \(C = (8.5, 0)\). Allow equivalents such as \(x\) =8.5 at \(C\)

M1: An attempt to use \(\sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) for \(AB\) or \(BC\). There is no need to ‘calculate’ these.
Evidence of an attempt would be \(AB^2 = 1^2 + 1.5^2 \Rightarrow AB = ..\)

dM1: Multiplying together their values of \(AB\) and \(BC\) to find area \(ABCD\)
It is dependent upon both M’s having been scored.

A1: cao 16.25 or equivalents such as \(\dfrac{65}{4}\).