June 2023 Paper 1 Q9
9 The points \(P\) and \(Q\) have coordinates \((-6, 15)\) and \((12, 19)\) respectively.
(a)
(i) Find the coordinates of the midpoint of \(PQ\) [1 mark]
(ii) Find the equation of the perpendicular bisector of \(PQ\)
Give your answer in the form \(ax + by = c\) where \(a\), \(b\) and \(c\) are integers. [4 marks]
(b)
(i) A circle passes through the points \(P\) and \(Q\)
The centre of the circle lies on the line with equation \(2x - 5y = -30\)
Find the equation of the circle. [3 marks]
(ii) The circle intersects the coordinate axes at \(n\) points.
State the value of \(n\) [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains \((3, 17)\) Condone position vectors, missing brackets or \(x = 3\) and \(y = 17\) | B1 | 1.1b |
| (1) | ||
| (ii) Obtains gradient of PQ PI correct gradient used in equation of perpendicular bisector. | B1 | 1.1b |
| Forms an equation of a line Either using the negative reciprocal of their gradient or their midpoint | M1 | 3.1a |
| Forms an equation of a line using the negative reciprocal of their gradient and their midpoint | M1 | 1.1a |
| Obtains \(9x + 2y = 61\) OE in the required form. | A1 | 2.1 |
| (4) |
Typical solution
(i)
\[(3, 17)\](ii)
\[m_{PQ} = \frac{19 - 15}{12 - -6} = \frac{4}{18}\]\[y - 17 = -\frac{9}{2}(x - 3)\]\[2y - 34 = -9x + 27\]\[9x + 2y = 61\]| Scheme | Marks | AO |
|---|---|---|
| (i) Solves simultaneously using their \(9x + 2y = 61\) from (a)(ii) with \(2x - 5y = -30\) to obtain the centre of the circle PI by \((5, 8)\) or \(x = 5\), \(y = 8\) | M1 | 3.1a |
| Uses \(P\) or \(Q\) and their centre to find the radius or radius2 | M1 | 3.1a |
| Obtains \((x - 5)^2 + (y - 8)^2 = 170\) ACF Eg \(x^2 - 10x + y^2 - 16y = 81\) | A1 | 1.1b |
| (3) | ||
| (ii) States 4 Must have the correct centre and correct radius or radius2 | R1 | 2.2a |
| (1) | ||
| (9 marks) |
Typical solution
(i)
Centre \((5, 8)\)
\[(x - 5)^2 + (y - 8)^2 = r^2\]\[(12 - 5)^2 + (19 - 8)^2 = 170\]\[(x - 5)^2 + (y - 8)^2 = 170\](ii)
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