Higher June 2023 Paper 1 Q24

24 Points \(P\), \(Q\) and \(R\) (8, 22) form a triangle.

Not drawn accurately

\(PQ\) is a horizontal line, with \(P\) on the \(y\)-axis.

Angle \(PRQ\) is a right angle.

The gradient of \(PR\) is 2

Work out the coordinates of \(Q\). [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 – using the equations of the lines
\(\dfrac{22 - y}{8 - 0} = 2\) or \(22 = 2 \times 8 + c\) or \((c =)\ 22 - 2 \times 8\) or \(c = 6\) or \(P\) is at (0, 6) or \((PR =)\ y = 2x + 6\) or \(y\)-coordinate of \(P\) is 6 or \(y\)-coordinate of \(Q\) is 6	M1	oe equation using any letter \(y\) is the \(y\)-coordinate of \(P\) ignore missing brackets may be seen on diagram may be seen on diagram
\(2m = -1\) or \((m =)\ -\dfrac{1}{2}\)	M1	oe gradient of \(RQ\)
\(22 =\) their \(-\dfrac{1}{2} \times 8 + c\) or \(22 = -4 + c\) or \(c = 26\) or \((RQ =)\ y = -\dfrac{1}{2}x + 26\)	M1dep	oe equation in \(c\) dep on previous mark oe equation of \(RQ\)
their \(\left(-\dfrac{1}{2}x + 26\right) =\) their 6 or \(x\)-coordinate of \(Q\) is 40	M1dep	oe equation in \(x\) where \(x\) is the \(x\)-coordinate of \(Q\) dep on M3 \(-\dfrac{1}{2} = \dfrac{22 - \text{their } 6}{8 - x}\) implies M4 if their 6 is correct or from correct working
(40, 6)	A1
Alternative method 2 – using similar triangles
Drops a perpendicular from \(R\) to point \(S\) on \(PQ\) and uses \(RS = 2PS = 16\) to work out that \(P\) is at (0, 6)	M1	any or no letter eg \(22 - 2 \times 8\)
\(2m = -1\) or \((m =)\ -\dfrac{1}{2}\) or \(\dfrac{RS}{SQ} = \dfrac{1}{2}\)	M1	oe gradient of \(RQ\)
\(16 \times 2\) or 32	M1dep	length of \(SQ\) may be seen on diagram dep on previous mark
8 + their 32 or \(x\)-coordinate of \(Q\) is 40	M1dep
(40, 6)	A1

Note that 40 (for the \(x\)-coordinate of \(Q\)) implies M3 (on alt 2) and implies M4 if 6 is also seen (on alt 1)