Higher June 2022 Paper 2 Q22

EdexcelHigherCurrent spec7 marksLinear Graphs & Gradients Simultaneous Equations

22 \(ABCD\) is a kite, with diagonals \(AC\) and \(BD\), drawn on a centimetre square grid, with a scale of 1 cm for 1 unit on each axis.

\(A\) is the point with coordinates \((-3, 4)\)

The diagonals of the kite intersect at the point \(M\) with coordinates \((0, 2)\)

Given that \(AB = AD = 6.5\) cm and the \(x\) coordinate of \(B\) is positive,

find the coordinates of the points \(B\) and \(D\).

(7)

Mark scheme

Scheme	Marks
(gradient \(AM\) =) \(\dfrac{4 - 2}{-3 - 0}\) oe \(\left(= -\dfrac{2}{3}\right)\)	M1
\(y = \dfrac{3}{2}x + 2\) or eg \(\dfrac{y - 2}{x} = \dfrac{3}{2}\) oe	M1
\((x - -3)^2 + (y - 4)^2 = 6.5^2\) or \((x - 0)^2 + (y - 2)^2 = 6.5^2 - \left[(-3 - 0)^2 + (4 - 2)^2\right]\) oe eg \(x^2 + (y - 2)^2 = 29.25\)	M1
eg \(x^2 + 6x + 9 + y^2 - 8y + 16 - 42.25 = 0\) oe or \(x^2 + y^2 - 4y + 4 - 29.25 = 0\) oe	M1
eg \(x^2 + 6x + 9 + \left(\dfrac{3}{2}x + 2\right)^2 - 8\left(\dfrac{3}{2}x + 2\right) + 16 - 42.25 = 0\) \(\left(\dfrac{2y - 4}{3}\right)^2 + y^2 - 4y + 4 - 29.25 = 0\) oe	M1
eg \(\dfrac{13}{4}x^2 = \dfrac{117}{4}\) or oe \(13y^2 - 52y - 211.25 = 0\)	M1
(3, 6.5) (−3, −2.5)	A1
	(7)
	(7 marks)

Notes

M1: A correct method to find gradient of \(AM\)

M1: For the correct equation of the line passing through \(BD\) or for a correct expression involving the \(x\) and \(y\) coordinates of point \(B\) or point \(D\)

M1: A correct equation in \(x\) and \(y\) to find the coordinates of \(B\) and \(D\)

M1: Brackets expanded

M1: For a correct substitution into a correct equation to get an equation in either \(x\) only or \(y\) only

M1: A fully correct simplified equation in \(x\) or in \(y\) – all brackets expanded and like terms grouped.

A1: correct coordinates
SCB3 for one pair of correct coordinates or both \(x\) values correct or both \(y\) values correct

Sketch: A at height 2 above L, LM = 3 with M at (0, 2), AM = root 13, AB = 6.5, BM = 3 root 13 over 2, and B above N on the extension of LM — Mark scheme (Alt 1)

Scheme	Marks
\((AM =)\;\sqrt{3^2 + 2^2}\;\left(= \sqrt{13} = 3.605\ldots\right)\) or \((AM^2 =)\;3^2 + 2^2\;(= 13)\)	M1
\((BM =)\;\sqrt{6.5^2 - \text{``}{\sqrt{13}}\text{''}^2}\;\left(= \sqrt{29.25} = \dfrac{3\sqrt{13}}{2}\;5.4083\ldots\right)\)	M1
(SF =) \(\dfrac{\sqrt{29.25}}{\sqrt{13}} = \dfrac{3}{2}\) oe or \(MN = x\), \(BN = 1.5x\) (see diag) or \((LAM =)\;\sin^{-1}\dfrac{3}{\sqrt{13}}\;(= 56.3\ldots)\) oe or \((LMA =)\;\cos^{-1}\dfrac{3}{\sqrt{13}}\;(= 33.6\ldots)\) or	M1
eg \(\overrightarrow{MB}_x = \dfrac{3}{2} \times 2\) or \(\overrightarrow{MB}_y = \dfrac{3}{2} \times 3\) or \(\overrightarrow{MD}_x = -\dfrac{3}{2} \times 2\) or \(\overrightarrow{MD}_y = -\dfrac{3}{2} \times 3\) oe or \(x^2 + (1.5x)^2 = \sqrt{29.25}^2\) or \(MN = \sqrt{29.25}\cos 56.3\ldots\;(= 3)\) oe or \(BN = \sqrt{29.25}\sin 56.3\ldots\;(= 4.5)\) oe	M1
\(\overrightarrow{MB}_x = \dfrac{3}{2} \times 2\) and \(\overrightarrow{MB}_y = \dfrac{3}{2} \times 3\) or \(\overrightarrow{MD}_x = -\dfrac{3}{2} \times 2\) and \(\overrightarrow{MD}_y = -\dfrac{3}{2} \times 3\) oe or \(x^2 + 2.25x^2 = 29.25\) or \(MN = \dfrac{3\sqrt{13}}{2}\cos 56.309\ldots\;(= 3)\) and \(BN = \dfrac{3\sqrt{13}}{2}\sin 56.309\ldots\;(= 4.5)\) oe	M1
eg (0, 2) is translated \(\begin{pmatrix} 3 \\ 4.5 \end{pmatrix}\) or \((0 + 3, 2 + 4.5)\;(= (3, 6.5))\) or (0, 2) is translated \(\begin{pmatrix} -3 \\ -4.5 \end{pmatrix}\) or \((0 - 3, 2 - 4.5)\;(= (-3, -2.5))\) oe or \(3.25x^2 = 29.25\)	M1
(3, 6.5) (−3, −2.5)	A1