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Coordinate Geometry
NOTE: Clicking an entry in the right column will take you directly to that question!
(BLUE = Paper 1, RED = Paper 2)
| Since 2015, the following subtopic... | Has explicitly appeared in... |
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Area of a Triangle |
Has not appeared |
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Triangle Concurrencies |
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Equation of a Line |
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Features of a Line |
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Divide Line Segment into Ratio |
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Perpendicular distance from a Point to a Line |
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Angle between Two Lines |
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Equation of a Circle |
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Tangent to a Circle |
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Intersection of a Line and a Circle |
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Touching Circles |
2022 Paper 2 Question 2
(a) The points \(A (8,-4)\) and \(B(-1,3)\) are the endpoints of the line segment \([AB]\).
Find the coordinates of the point \(C\), which divides \([AB]\) internally in the ratio \(4:1\).
\(\left(\dfrac{4}{5},\dfrac{8}{5}\right)\)
\begin{align}C&=\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)\\&=\left(\frac{(1)(8)+(4)(-1)}{4+1},\frac{(1)(-4)+(4)(3)}{4+1}\right)\\&=\left(\frac{4}{5},\frac{8}{5}\right)\end{align}
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(b) The line \(l\) has a slope of \(m\) and contains the point \((q,r)\), where \(m,q,r\in\mathbb{R}\) are all positive.
Find the co-ordinates of the point where \(l\) cuts the \(y\)-axis, in terms of \(m\), \(q\) and \(r\).
\((0,r-mq)\)
\begin{align}y=mx+c\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r=mq+c\end{align}
\begin{align}\downarrow\end{align}
\begin{align}c=r-mq\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(0,r-mq)\end{align}
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(c) The line \(k\) has a slope of \(-2\).
The line \(j\) makes an angle of \(30^{\circ}\) with \(k\).
Find one possible value of the slope of the line \(j\).
Give your answer in the form \(d+e\sqrt{f}\), where \(d,e,f\in\mathbb{Z}\).
\(8+5\sqrt{3}\)
\begin{align}\tan\theta=\pm\frac{m_1-m_2}{1+m_1m_2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan30^{\circ}=\frac{-2-m_2}{1+(-2)m_2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{1}{\sqrt{3}}=\frac{-2-m_2}{1-2m_2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1-2m_2=-2\sqrt{3}-\sqrt{3}m_2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_2&=\frac{1+2\sqrt{3}}{2-\sqrt{3}}\\&=\frac{1+2\sqrt{3}}{2-\sqrt{3}}\times\frac{2+\sqrt{3}}{2+\sqrt{3}}\\&=\frac{2+\sqrt{3}+4\sqrt{3}+6}{4+2\sqrt{3}-2\sqrt{3}-3}\\&=8+5\sqrt{3}\end{align}
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2022 Paper 2 Question 3
(a) The circle \(c\) has equation \(x^2+y^2-2x+8y+k=0\), where \(k\in\mathbb{R}\).
The radius of \(c\) is \(5\sqrt{3}\).
Find the value of \(k\).
\(k=-58\)
\begin{align}\sqrt{g^2+f^2-c}=5\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sqrt{(-1)^2+4^2-k}=5\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1+16-k=75\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k=-58\end{align}
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(b) The circle \((x-5)^2+(y+2)^2=20\) has a tangent at the point \((9,-4)\).
Find the slope of this tangent.
\(2\)
The circle has a centre \((5,-2)\). The slope of the line going through this point and \((9,-4)\) is
\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-4-(-2)}{9-5}\\&=-\frac{1}{2}\end{align}
The tangent therefore has a slope of \(2\).
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(c) Two circles each have both the \(x\)-axis and the \(y\)-axis as tangents, and each contains the point \((1,-8)\), as shown in the diagram.
Find the equation of each of these circles.
\((x-5)^2+(y+5)^2=25\) and \((x-13)^2+(y+13)^2=169\)
\begin{align}(x-h)^2+(y-k)^2=r^2\end{align}
Centre: \((h,k)=(r,-r)\)
Point on circumference: \((x,y)=(1,-8)\)
\begin{align}\downarrow\end{align}
\begin{align}(1-r)^2+(-8-(-r))^2=r^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}1-2r+r^2+64-16r+r^2=r^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r^2-18r+65=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(r-5)(r-13)=0\end{align}
\begin{align}\downarrow\end{align}
\(r=5\) or \(r=13\)
\begin{align}\downarrow\end{align}
\begin{align}(x-5)^2+(y+5)^2=25\end{align}
or
\begin{align}(x-13)^2+(y+13)^2=169\end{align}
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2021 Paper 2 Question 2
(a) The line \(3x-6y+2=0\) contains the point \(\left(k,\dfrac{2k+2}{3}\right)\) where \(k\in\mathbb{R}\).
Find the value of \(k\).
\(k=-2\)
\begin{align}3x-6y+2=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3k-6\left(\frac{2k+2}{3}\right)+2=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3k-4k-4+2=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-k-2=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k=-2\end{align}
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(b) The point \(P(s,t)\) is on the line \(x-2y-8=0\).
The point \(P\) is also a distance of \(1\) unit from the line \(4x+3y+6=0\).
Find a value of \(s\) and the corresponding value of \(t\).
\(s=2\) and \(t=-3\) or \(s=\dfrac{2}{11}\) or \(t=-\dfrac{43}{11}\)
\begin{align}x-2y-8=0\end{align}
\begin{align}(x,y)=(s,t)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}s-2t-8=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}s=2t+8\end{align}
and
\begin{align}\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{|4s+3t+6|}{\sqrt{4^2+3^2}}=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{|4(2t+8)+3t+6|}{5}=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|8t+32+3t+6|=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|11t+38|=5\end{align}
Choose positive option:
\begin{align}11t+38=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=-\frac{43}{11}\end{align}
and
\begin{align}s&=2t+8\\&=2\left(\frac{-43}{11}\right)+8\\&=\frac{2}{11}\end{align}
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(c) The points \(A(4,2)\) and \(C(16,11)\) are vertices of the triangle \(ABC\) shown below.
\(D\) and \(E\) are points on \([CA]\) and \([CB]\) respectively.
The ratio \(|AD|:|DC|\) is \(2:1\).
(i) Find \(|AD|\).
(ii) \([AB]\) and \([DE]\) are horizontal line segments.
\(|AB|=33\) units.
Find the coordinates of \(B\) and of \(E\).
(i) \(10\)
(ii) \((23,8)\)
(i)
\begin{align}|AC|&=\sqrt{(16-4)^2+(11-2)^2}\\&=\sqrt{12^2+9^2}\\&=\sqrt{225}\\&=15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|AD|&=\frac{2}{3}(15)\\&=10\end{align}
(ii)
\begin{align}B&=(4+33,2)\\&=(37,2)\end{align}
and
\begin{align}E&=\left(16+\frac{1}{3}(21),11-\frac{1}{3}(9)\right)\\&=(23,8)\end{align}
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2021 Paper 2 Question 3
(a) The circle \(k\) has centre \(C(1,-2)\) and chord \([AB]\) where \(|AB|=4\sqrt{3}\).
The point \(D(3,2)\) is the midpoint of the chord \([AB]\), as shown below.
Find the radius of \(k\). Give your answer in the form \(a\sqrt{b}\), where \(a,b\in\mathbb{N}\).
\(4\sqrt{2}\)
\begin{align}|AD|&=\frac{1}{2}(4\sqrt{3})\\&=2\sqrt{3}\end{align}
and
\begin{align}|CD|&=\sqrt{(3-1)^2+(2-(-2))^2}\\&=\sqrt{2^2+4^2}\\&=\sqrt{20}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}r&=|CA|\\&=\sqrt{|AD|^2+|CD|^2}\\&=\sqrt{(2\sqrt{3})^2+(\sqrt{20})^2}\\&=4\sqrt{2}\end{align}
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(b)
(i) Show that the circles \(c:x^2+y^2+4x-2y-95=0\) and \(s:(x-7)^2+(y-13)^2=25\) touch externally.
(ii) There are an infinite number of circles which touch circle \(c\) externally at the same point that \(s\) touches \(c\).
Find the coordinates of the centre of one of these circles, apart from circle \(s\).
(i) The answer is already in the question!
(ii) e.g. \((16,25)\)
(i)
Centre of \(c\) is \((-2,1)\) and
\begin{align}r_c&=\sqrt{g^2+f^2-c}\\&=\sqrt{2^2+(-1)^2-(-95)}\\&=\sqrt{100}\\&=10\end{align}
and
Centre of \(s\) is \((7,13)\) and \(r_s=5\).
\begin{align}\downarrow\end{align}
\begin{align}r_c+r_s=15\end{align}
The distance \(d\) between their centres is
\begin{align}d&=\sqrt{(-2-7)^2+(1-13)^2}\\&=\sqrt{81+144}\\&=\sqrt{225}\\&=15\end{align}
Therefore, they touch externally.
(ii)
\begin{align}(7+9,13+12)=(16,25)\end{align}
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2020 Paper 2 Question 1
(a) The coordinates of three points are \(A(2,-6)\), \(B(6,-12)\) and \(C(-4,3)\).
Find the perpendicular distance from \(A\) to \(BC\).
Based on your answer, what can you conclude about the relationship between the points \(A\), \(B\) and \(C\)?
As the distance is zero, all three points are collinear.
\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-(-12)}{-4-6}\\&=-\frac{15}{10}\\&=-\frac{3}{2}\end{align}
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-(-12)=-\frac{3}{2}(x-6)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y+12=-\frac{3}{2}x+9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2y+24=-3x+18\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x+2y+6=0\end{align}
The perpendicular is therefore
\begin{align}d&=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\&=\frac{3(2)+2(-6)+6}{\sqrt{3^2+2^2}}\\&=0\end{align}
and therefore all three points are collinear.
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(b) The diagram below shows two lines \(a\) and \(b\). The equation of \(a\) is \(x-2y+1=0\).
The acute angle between \(a\) and \(b\) is \(\theta\). Line \(b\) makes an angle of \(60^{\circ}\) with the positive sense of the \(x\)-axis, as shown in the diagram.
Find the value of \(\theta\), in degrees, correct to \(3\) decimal places.
\(33.435^{\circ}\)
\begin{align}x-2y+1=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2y=x+1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=\frac{1}{2}x+1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_a=\frac{1}{2}\end{align}
and
\begin{align}m_b&=\tan60^{\circ}\\&=\sqrt{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan\theta&=\pm\frac{m_a-m_b}{1+m_am_b}\\&=\pm\frac{\frac{1}{2}-\sqrt{3}}{1+\left(\frac{1}{2}\right)(\sqrt{3})}\\&=\pm\frac{1-2\sqrt{3}}{2+\sqrt{3}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\theta&=\tan^{-1}\left(\frac{1-2\sqrt{3}}{2+\sqrt{3}}\right)\\&\approx33.435^{\circ}\end{align}
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2020 Paper 2 Question 2
(a) The circle \(c\) has equation \(x^2+y^2-4x+2y-4=0\).
The point \(A\) is the centre of the circle.
The line \(l\) is a tangent to \(c\) at the point \(T\), as shown in the diagram.
The point \(B(5,8)\) is on \(l\).
Find \(|BT|\).
\(|BT|=9\)
\begin{align}|AT|&=\sqrt{g^2+f^2-c}\\&=\sqrt{2^2+(-1)^2-(-4)}\\&=\sqrt{9}\\&=3\end{align}
and
\begin{align}|AB|&=\sqrt{(2-5)^2+(-1-8)^2}\\&=\sqrt{9+81}\\&=\sqrt{90}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|BT|&=\sqrt{|AB|^2-|AT|^2}\\&=\sqrt{(\sqrt{90})^2-3^2}\\&=\sqrt{81}\\&=9\end{align}
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(b) Two circles, \(c_1\) and \(c_2\), have their centres on the \(x\)-axis. Each circle has a radius of \(5\) units.
The point \((1,4)\) lies on each circle. Find the equation of \(c_1\) and the equation of \(c_2\).
\((x-4)^2+y^2=25\) and \((x+2)^2+y^2=25\)
The centre of both circles is \((-g,0)\). Their radii therefore satisfy:
\begin{align}\sqrt{(-g-1)^2+(0-4)^2}=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\sqrt{g^2+2g+1+16}=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g^2+2g+17=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g^2+2g-8=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(g+4)(g-2)=0\end{align}
\begin{align}\downarrow\end{align}
\(g=-4\) and \(g=2\)
\begin{align}\downarrow\end{align}
\begin{align}(x-4)^2+y^2=25\end{align}
and
\begin{align}(x+2)^2+y^2=25\end{align}
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2019 Paper 2 Question 2
(a) The line \(p\) makes an intercept on the \(x\)-axis at \((a,0)\) and on the \(y\)-axis at \((0,b)\), where \(a,b\neq0\).
Show that the equation of \(p\) can be written as \(\dfrac{x}{a}+\dfrac{y}{b}=1\).
The answer is already in the question!
\begin{align}m&=\frac{b-0}{0-a}\\&=-\frac{b}{a}\end{align}
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-0=-\frac{b}{a}(x-a)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}ay=-bx+ab\end{align}
\begin{align}\downarrow\end{align}
\begin{align}bx+ay=ab\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{x}{a}+\frac{y}{b}=1\end{align}
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(b) The line \(l\) has a slope \(m\), and contains the point \(A(6,0)\).
(i) Write the equation of the line \(l\) in term of \(m\).
(ii) The line \(l\) cuts the line \(k:4x+3y=25\) at \(P\).
Find the co-ordinates of \(P\) in terms of \(m\).
Give each co-ordinate as a fraction in its simplest form.
(i) \(y=mx-6m\)
(ii) \(P:\left(\dfrac{18m+25}{3m+4},\dfrac{m}{3m+4}\right)\)
(i)
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-0=m(x-6)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=mx-6m\end{align}
(ii)
\begin{align}y=mx-6m\end{align}
\begin{align}4x+3y=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}mx-y=6m\end{align}
\begin{align}4x+3y=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3mx-3y=18m\end{align}
\begin{align}4x+3y=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(3m+4)x=18m+25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=\frac{18m+25}{3m+4}\end{align}
and
\begin{align}y&=mx-6m\\&=m\left(\frac{18m+25}{3m+4}\right)-6m\\&=\frac{18m^2+25m-18m^2-24m}{3m+4}\\&=\frac{m}{3m+4}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}P:\left(\frac{18m+25}{3m+4},\frac{m}{3m+4}\right)\end{align}
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2019 Paper 2 Question 3
(a) The point \((-2,k)\) is on the circle \((x-2)^2+(y-3)^2=65\).
Find the two possible values of \(k\), where \(k\in\mathbb{Z}\).
\(k=-4\) or \(k=10\)
\begin{align}(x-2)^2+(y-3)^2=65\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(-2-2)^2+(k-3)^2=65\end{align}
\begin{align}\downarrow\end{align}
\begin{align}16+k^2-6k+9=65\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k^2-6k-40=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(k+4)(k-10)=0\end{align}
\begin{align}\downarrow\end{align}
\(k=-4\) or \(k=10\)
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(b) The circle \(s\) is in the first quadrant. It touches both the \(x\)-axis and the \(y\)-axis.
The line \(t:3x-4y+6=0\) is a tangent to \(s\) as shown. Find the equation of \(s\).
\((x-1)^2+(y-1)^2=1\)
The centre of the circle is \((r,r)\) and the radius is \(r\).
\begin{align}\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=r\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{|3r-4r+6||}{\sqrt{3^2+4^2}}=r\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{|-r+6|}{5}=r\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|-r+6|=5r\end{align}
\(-r+6=5r\) or \(-r+6=-5r\)
\begin{align}\downarrow\end{align}
\(r=1\) or \(r=-\dfrac{3}{2}\)
\begin{align}\downarrow\end{align}
\begin{align}r=1\end{align}
(as we are in the first quadrant).
\begin{align}(x-1)^2+(y-1)^2=1\end{align}
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2018 Paper 2 Question 5
The line ݉\(m:2x+3y+1=0\) is parallel to the line ݊\(n:2x+3y-51=0\).
(a) Verify that \(A(-2,1)\) is on \(m\).
The answer is already in the question!
\begin{align}2x+3y+1&=2(-2)+3(1)+1\\&=-4+3+1\\&=0\end{align}
as required.
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(b) Find the coordinates of \(B\), the point on the line ݊\(n\) closest to \(A\), as shown below.
\((6,13)\)
\begin{align}2x+3y-51=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=-\frac{2}{3}x+17\end{align}
Therefore, the slope of both parallel lines is \(-\dfrac{2}{3}\).
The slope of \(AB\) is therefore \(\dfrac{3}{2}\).
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-1=\frac{3}{2}(x-(-2))\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2y-2=3x+6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x-2y+8=0\end{align}
We will now solve the following two equations:
\begin{align}2x+3y-51=0\end{align}
\begin{align}3x-2y+8=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}6x+9y-153=0\end{align}
\begin{align}6x-4y+16=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}13y-169=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=13\end{align}
and
\begin{align}3x-2y+8=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=\frac{2y-8}{3}\\&=\frac{2(13)-8}{3}\\&=6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}B:(6,13)\end{align}
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(c) Two touching circles, \(s\) and \(t\), are shown in the diagram. ݉\(m\) is a tangent to \(s\) at \(A\) and ݊\(n\) is a tangent to \(t\) at \(B\). The ratio of the radius of \(s\) to the radius of \(t\) is \(1:3\).
Find the equation of s.
\((x+1)^2+(y-2.5)^2=3.25\)
The point at which both circles intersect is:
\begin{align}\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)&=\left(\frac{3(-2)+1(6)}{3+1},\frac{3(1)+1(13)}{3+1}\right)\\&=(0,4)\end{align}
Circle \(s\) therefore has a centre of
\begin{align}\left(\frac{0-2}{2},\frac{4+1}{2} \right)=(-1,2.5)\end{align}
and a radius of
\begin{align}r&=\sqrt{(-1-(-2))^2+(2.5-1)^2}\\&=\sqrt{3.25}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x+1)^2+(y-2.5)^2=3.25\end{align}
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2017 Paper 2 Question 3
\(ABC\) is a triangle where the co-ordinates of \(A\) and \(C\) are \((0,6)\) and \((4,2)\) respectively.
\(G\left(\dfrac{2}{3},\dfrac{4}{3}\right)\) is the centroid of the triangle \(ABC\).
\(AG\) intersects \(BC\) at the point \(P\).
\(|AG|:|GP|=2:1\).
(a) Find the co-ordinates of \(P\).
\((1,-1)\)
\begin{align}\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)=\left(\frac{2}{3},\frac{4}{3}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left(\frac{(1)(0)+(2)x_2}{1+2},\frac{(1)(6)+(2)y_2}{1+2}\right)=\left(\frac{2}{3},\frac{4}{3}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left(\frac{2x_2}{3},\frac{6+2y_2}{3}\right)=\left(\frac{2}{3},\frac{4}{3}\right)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2x_2=2\end{align}
\begin{align}6+2y_2=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x_2=1\end{align}
\begin{align}y_2=-1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}P:(1,-1)\end{align}
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(b) Find the co-ordinates of \(B\).
\(B:(-2,-4)\)
\begin{align}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=(1,-1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\left(\frac{4+x_2}{2},\frac{2+y_2}{2}\right)=(1,-1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{4+x_2}{2}=1\end{align}
\begin{align}\frac{2+y_2}{2}=-1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x_2&=(1)(2)-4\\&=-2\end{align}
\begin{align}y_2&=(-1)(2)-2\\&=-4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}B:(-2,-4)\end{align}
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(c) Prove that \(C\) is the orthocentre of the triangle \(ABC\).
The answer is already in the question!
\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{6-2}{0-4}\\&=-1\end{align}
and
\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-4-2}{-2-4}\\&=1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{AC}\times m_{BC}&=(-1)(1)\\&=-1\end{align}
Therefore, as the two lines are perpendicular to each other, \(C\) must be the orthocentre of the triangle.
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2017 Paper 2 Question 4
\(A(0,0)\), \(B(6.5,0)\) and \(C(10,7)\) are three points on a circle.
(a) Find the equation of the circle.
\(x^2+y^2-6.5x-12y=0\)
\begin{align}x^2+y^2+2gx+2fy+c=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}0^2+0^2+2g(0)+2f(0)+c=0\end{align}
and
\begin{align}6.5^2+0^2+2g(6.5)+2f(0)+c=0\end{align}
and
\begin{align}10^2+7^2+2g(10)+2f(7)+c=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}c=0\end{align}
\begin{align}42.25+13g+c=0\end{align}
\begin{align}149+20g+14f+c=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}42.25+13g=0\end{align}
\begin{align}149+20g+14f=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g=-3.25\end{align}
\begin{align}149+20g+14f=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}149+20(-3.25)+14f=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f&=\frac{20(3.25)-149}{14}\\&=-6\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x^2+y^2-6.5x-12y=0\end{align}
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(b) Find \(|\angle BCA|\). Give your answer in degrees, correct to \(2\) decimal places.
\(28.44^{\circ}\)
\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-7}{6.5-10}\\&=2\end{align}
and
\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-7}{0-10}\\&=\frac{7}{10}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\tan|\angle BCA|&=\pm\frac{m_{BC}-m_{AC}}{1+m_{BC}m_{AC}}\\&=\pm\frac{2-\frac{7}{10}}{1+(2)(\left(\frac{7}{10}\right)}\\&=\frac{\frac{13}{10}}{\frac{24}{10}}\\&=\frac{13}{24}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|\angle BCA|&=\tan^{-1}\left(\frac{13}{24}\right)\\&\approx28.44^{\circ}\end{align}
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2016 Paper 2 Question 1
The points \(A(6,-2)\), \(B(5,3)\) and \(C(-3,4)\) are shown on the diagram.
(a) Find the equation of the line through \(b\) which is perpendicular to \(AC\).
\(3x-2y-9=0\)
\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{4-(-2)}{-3-6}\\&=-\frac{2}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{\perp}=\frac{3}{2}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-3=\frac{3}{2}(x-5)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2y-6=3x-15\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x-2y-9=0\end{align}
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(b) Use your answer to part (a) above to find the co-ordinates of the orthocentre of the triangle \(ABC\).
\((7,6)\)
\begin{align}m_{AB}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-(-2)}{5-6}\\&=-5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{\perp}=\frac{1}{5}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-4=\frac{1}{5}(x-(-3))\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5y-20=x+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x-5y+23=0\end{align}
We therefore have the following two equations with two unknowns:
\begin{align}3x-2y-9=0\end{align}
\begin{align}x-5y+23=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x-2y-9=0\end{align}
\begin{align}3x-15y+69=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}13y-78=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y&=\frac{78}{13}\\&=6\end{align}
and
\begin{align}x-5y+23=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x&=5y-23\\&=5(6)-23\\&=7\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x,y)=(7,6)\end{align}
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2016 Paper 2 Question 2
A point \(X\) has co-ordinates \((-1,6)\) and the slope of the line \(XC\) is \(\dfrac{1}{7}\).
(a) Find the equation of \(XC\). Give your answer in the form \(ax+by+c=0\), where \(a,b,c,\in\mathbb{Z}\).
\(x-7y+43=0\)
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-6=\frac{1}{7}(x-(-1))\end{align}
\begin{align}\downarrow\end{align}
\begin{align}7y-42=x+1\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x-7y+43=0\end{align}
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(b) \(C\) is the centre of a circle \(S\), of radius \(5\mbox{ cm}\). The line \(l:3x+4y-21=0\) is a tangent to \(s\) and passes through \(X\), as shown. Find the equation of one such circle \(s\).
\((x-6)^2+(y-7)^2=25\) or \((x+8)^2+(y-5)^2=25\)
\begin{align}x-7y+43=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-g-7(-f)+43=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-g+7f+43=0\end{align}
and
\begin{align}\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{|(3)(-g)+4(-f)-21|}{\sqrt{3^2+4^2}}=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{|-3g-4f-21|}{5}=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}|-3g-4f-21|=25\end{align}
Choosing the positive value:
\begin{align}-3g-4f-21=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3g+4f+46=0\end{align}
We therefore have the following two equations for two unknowns:
\begin{align}-g+7f+43=0\end{align}
\begin{align}3g+4f+46=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-3g+21f+129=0\end{align}
\begin{align}3g+4f+46=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}25f+175=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}f&=-\frac{175}{25}\\&=-7\end{align}
and
\begin{align}-g+7f+43=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}g&=7f+43\\&=7(-7)+43\\&=-6\end{align}
\begin{align}\downarrow\end{align}
Centre: \((6,7)\)
\begin{align}\downarrow\end{align}
\begin{align}(x-6)^2+(y-7)^2=25\end{align}
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2015 Paper 2 Question 3
(a) The co-ordinates of two points are \(A(4,-1)\) and \(B(7,t)\).
The line \(l_1:3x-4y-12=0\) is perpendicular to \(AB\). Find the value of \(t\).
\(t=-5\)
\begin{align}3x-4y-12=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=\frac{3}{4}x-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{l_1}=\frac{3}{4}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{AB}=-\frac{4}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{t-(-1)}{7-4}=-\frac{4}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{t+1}{3}=-\frac{4}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t+1=-4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}t=-5\end{align}
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(b) Find, in terms of \(k\), the distance between the point \(P(10,k)\) and \(l_1\).
\(\dfrac{|18-4k|}{5}\)
\begin{align}d&=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\\&=\frac{|3(10)-4k-12|}{\sqrt{3^2+4^2}}\\&=\frac{|18-4k|}{5}\end{align}
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(c) \(P(10,k)\) is on a bisector of the angles between the lines \(l_1\) and \(l_2:5x+12y-20=0\).
(i) Find he possible values of \(k\).
(ii) If \(k>0\), find the distance from \(P\) to \(l_1\).
(i) \(k=\dfrac{3}{4}\) or \(k=-48\)
(ii) \(3\)
(i)
\begin{align}\frac{|18-4k|}{5}=\frac{|5(10)+12k-20|}{\sqrt{5^2+12^2}}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}\frac{|18-4k|}{5}=\frac{|30+12k|}{13}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}13|18-4k|=5|30+12k|\end{align}
\begin{align}\downarrow\end{align}
\begin{align}13(18-4k)=5(30+12k)\end{align}
or
\begin{align}13(18-4k)=-5(30+12k)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}-122k=-84\end{align}
or
\begin{align}8k=-384\end{align}
\begin{align}\downarrow\end{align}
\(k=\dfrac{3}{4}\) or \(k=-48\)
(ii)
\begin{align}d&=\frac{|18-4k|}{5}\\&=\frac{|18-4\left(\frac{3}{4}\right)|}{5}\\&=3\end{align}
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2015 Paper 2 Question 4
Two circles \(s\) and \(c\) touch internally at \(B\), as shown.
(a) The equation of the circle \(s\) is
\begin{align}(x-1)^2+(y+6)^2=360\end{align}
Write down:
- the co-ordinates of the centre of \(s\)
- the radius of \(s\) in the form \(a\sqrt{10}\), where \(a\in\mathbb{N}\).
Centre: \((1,-6)\)
Radius: \(6\sqrt{10}\)
Centre: \((1,-6)\)
\begin{align}r&=\sqrt{360}\\&=\sqrt{36\times10}\\&=\sqrt{36}\times\sqrt{10}\\&=6\sqrt{10}\end{align}
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(b)
(i) The point \(K\) is the centre of circle \(c\).
The radius of \(c\) is one-third the radius of \(s\).
The co-ordinates of B are \((7,12)\).
Find the co-ordinates of \(K\).
(ii) Find the equation of \(c\).
(i) \((5,6)\)
(ii) \((x-5)^2+(y-6)^2=40\)
(i)
\begin{align}K&=\left(\frac{bx_1+ax_2}{b+a},\frac{by_1+ay_2}{b+a}\right)\\&=\left(\frac{(2)(7)+?(1)(1)}{2+1},\frac{(2)(12)+(1)(-6)}{2+1}\right)\\&=(5,6)\end{align}
(ii)
\begin{align}(x-5)^2+(y-6)^2=\left(\frac{6\sqrt{10}}{3}\right)^2\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-5)^2+(y-6)^2=40\end{align}
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(c) Find the equation of the common tangent at \(B\).
Give your answer in the form \(ax+by+c=0\), where \(a,b,c\in\mathbb{Z}\).
\(x+3y-43=0\)
\begin{align}m_{AB}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{12-(-6)}{7-1}\\&=3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{\perp}=-\frac{1}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-12=-\frac{1}{3}(x-7)\end{align}
\begin{align}3y-36=-x+7\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x+3y-43=0\end{align}
