Question
The area of an equilateral triangle is \(1\) cm2. Determine the area of:
The points A, B have coordinates (\(1\), \(0\)), (\(0\), \(1\)) respectively. The point P(\(x\), \(y\)) moves in such a way that \({\rm{AP}} = k{\rm{BP}}\) where \(k \in {\mathbb{R}^ + }\) .
the circumscribed circle.
the inscribed circle.
When \(k = 1\) , show that the locus of P is a straight line.
When \(k \ne 1\) , the locus of P is a circle.
(i) Find, in terms of \(k\) , the coordinates of C, the centre of this circle.
(ii) Find the equation of the locus of C as \(k\) varies.
Answer/Explanation
Markscheme
consider the above diagram – [AD] and [BE] are the medians and O is therefore both the incentre and the circumcentre (R1)
let \({\rm{AB}} = d\) and let \(R\) denote the radius of the circumcircle
then,
\(R = {\rm{AO}} = {\rm{AE}}\sec {30^ \circ }\) M1
\( = \frac{d}{2} \times \frac{2}{{\sqrt 3 }} = \frac{d}{{\sqrt 3 }}\) (A1)
area of circumcircle \( = \pi {R^2} = \frac{{\pi {d^2}}}{3}\) A1
area of triangle \({\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{AB}}{\rm{.AC}}\sin {\rm{BAC}}\) M1
\( = \frac{{\sqrt 3 {d^2}}}{4}\) (A1)
\(\frac{{\sqrt 3 {d^2}}}{4} = 1 \Rightarrow {d^2} = \frac{4}{{\sqrt 3 }}\) A1
area of circumcircle \( = \frac{{4\pi }}{{3\sqrt 3 }}\) (\(2.42\)) A1
[8 marks]
let \(r\) denote the radius of the incircle
then
\(r = {\rm{OE}} = {\rm{AE}}\tan {30^ \circ }\) M1
\( = \frac{d}{{2\sqrt 3 }}\) (A1)
area of incircle \( = \pi {r^2} = \frac{{\pi {d^2}}}{{12}}\)
\( = \frac{\pi }{{3\sqrt 3 }}\) (\(0.605\)) A1
[3 marks]
\({\rm{A}}{{\rm{P}}^2} = {(x – 1)^2} + {y^2}\) and \({\rm{B}}{{\rm{P}}^2} = {x^2} + {(y – 1)^2}\) A1
\({x^2} – 2x + 1 + {y^2} = {x^2} + {y^2} – 2y + 1\) M1
\(y = x\) which is the equation of a straight line A1
[3 marks]
(i) \({x^2} – 2x + 1 + {y^2} = {k^2}({x^2} + {y^2} – 2y + 1)\) M1
\(({k^2} – 1){x^2} + ({k^2} – 1){y^2} + 2x – 2{k^2}y + {k^2} – 1 = 0\) A1
\({x^2} + {y^2} + \frac{{2x}}{{{k^2} – 1}} – \frac{{2{k^2}y}}{{{k^2} – 1}} + 1 = 0\) A1
by completing the squares or quoting the standard result, M1
coordinates of C are
\(\left( { – \frac{1}{{{k^2} – 1}},\frac{{{k^2}}}{{{k^2} – 1}}} \right)\) A1
(ii) let (\(x\), \(y\)) be the coordinates of C
attempting to find \(k\) or \({{k^2}}\) , (M1)
\({k^2} = 1 – \frac{1}{x}\) (A1)
\(y = \frac{{1 – \frac{1}{x}}}{{ – \frac{1}{x}}}\) (M1)
\(y = 1 – x\) A1
[9 marks]