October 2021 Paper 2 Q5

OCR ACurrent spec8 marksCo-ordinate Geometry

5 In this question you must show detailed reasoning.

Points \(A\), \(B\) and \(C\) have coordinates \((0, 6)\), \((7, 5)\) and \((6, -2)\) respectively.

(a) Find an equation of the perpendicular bisector of \(AB\). [3]

(b) Hence, or otherwise, find an equation of the circle that passes through points \(A\), \(B\) and \(C\). [5]

Mark scheme (a)

Scheme	Marks
Midpoint \(AB\) is \((3.5, 5.5)\); Gradient \(AB = -\frac{1}{7}\)	B1
Gradient of perpendicular bisector \(-1/\left(-\frac{1}{7}\right)\)	M1
\(y - 5.5 = 7(x - 3.5)\) oe ISW	A1
	[3]

B1: Both. Allow midpoint \(= \left(\frac{0 + 7}{2}, \frac{6 + 5}{2}\right)\) ISW

M1: \((= 7)\)

A1: cao. Correct answer, no working or inadequate working: SC B2

B1: Both
M1: ft their midpt and gradient, NOT \(-\frac{1}{7}\)
A1: cao. Any correct form

Scheme	Marks
\(x^2 + (y - 6)^2 = (x - 7)^2 + (y - 5)^2\)	M1 M1
\(-12y + 36 = -14x - 10y + 49 + 25\) ISW	A1

M1: Attempt expansion
A1: cao. Any correct form eg \(y = 7x - 19\)

Mark scheme (b)

Scheme	Marks
Perpendicular bisector of \(BC\) is \(x + 7y - 17 = 0\) OR of \(CA\) is \(4y = 3x - 1\)	B1
Example method, perp bisectors of \(AB\) & \(BC\): \(x + 7(7x - 19) - 17 = 0\) \((\Rightarrow x = 3)\)	M1
Centre is \((3, 2)\)	B1
eg Radius\(^2 = 3^2 + (6 - 2)^2 = 25\)	M1
Equn of circle is \((x - 3)^2 + (y - 2)^2 = 25\) or \(x^2 - 6x + y^2 - 4y = 12\) oe	A1ft
	[5]

B1: Any correct form for another perp bisector

M1: Attempt solve simultaneously equations of two perpendicular bisectors. Can be implied

B1: cao. NB, if centre \(= (3, 2)\) without clear working, B0M0B1

M1: Correct method for \(r^2\) or \(r\) using their centre & \(A\) or \(B\) or \(C\)

A1ft: ISW. ft their centre & radius, dep both M1 marks

Scheme	Marks
Grad \(BC\) is 7 so \(BC\) & \(AB\) perpendicular	M1
Hence \(AC\) is a diameter	B1