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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 |
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Slope of a Line |
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Length of a Line Segment |
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Midpoint of a Line Segment |
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Equation of a Line |
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Intersection of Two Lines |
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Equation of a Circle |
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Tangent to a Circle |
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Translations |
2022 Paper 2 Question 1
Parts of the lines \(AC\) and \(BC\) are shown in the co-ordinate diagram below (not to scale).
(a)
(i) Find the slope of \(AC\).
(ii) By using slopes, investigate if \(AC\) is perpendicular to \(BC\). Justify your answer.Â
(i)Â \(\dfrac{3}{2}\)
(ii) They are not perpendicular.
(i)
\begin{align}m_{AC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-0}{0-(-2)}\\&=\frac{3}{2}\end{align}
(ii)
\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-0}{0-5}\\&=-\frac{3}{5}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{AC}\times m_{BC}&=\frac{3}{2}\times\left(-\frac{3}{5}\right)\\&=-\frac{9}{10}\\&\neq-1\end{align}
Therefore, they are not perpendicular.
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(b) The triangle \(LMN\) is shown on the co-ordinate diagram below (not to scale).
The point \(M\) has co-ordinates \((9,1)\)
The triangle \(LMN\) is symmetrical about the \(y\)-axis.Â
(i) Find the length \(|LM|\).
(ii) Write down the equation of the horizontal line \(LM\).
(iii) The line \(NM\) has equation \(x+4y-13=0\), where \(x,y\in\mathbb{R}\).
Use this equation to find the co-ordinates of the point \(N\).
(i)Â \(18\)
(ii)Â \(y=1\)
(iii)Â \(\left(0,\dfrac{13}{4}\right)\)
(i)
\begin{align}|LM|&=2\times9\\&=18\end{align}
(ii)
\begin{align}y=1\end{align}
(iii)
\begin{align}0+4y-13=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=\frac{13}{4}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}N=\left(0,\frac{13}{4}\right)\end{align}
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2022 Paper 2 Question 2
(a) The circle \(k\) has equation \((x-4)^2+(y+2)^2=169\).
(i) Write down the centre and radius of the circle \(k\).
(ii) Is the point \((11,10)\) on the circle \(k\), inside the circle \(k\), or outside the circle \(k\)?
Show your working out.
(i)Â Centre: \((4,-2)\) and radius: \(13\)
(ii)Â Outside
(i)Â Centre: \((4,-2)\) and radius: \(13\)
(ii)
\begin{align}(x-4)^2+(y+2)^2&=(11-4)^2+(10+2)^2\\&=49+144\\&=193\\&>169\end{align}
Therefore, this point is outside the circle \(k\).
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(b) The diagram below shows two circles, \(s\) and \(t\). The circle \(s\) has centre \((22,13)\).
The two circles touch at the point \((12,11)\).
(i) Find the co-ordinates of another point on the circle \(s\), other than \((12,11)\).
(ii) The radius of the circle \(t\) is half the radius of \(s\).
Find the co-ordinates of the centre of the circle \(t\).
(i)Â e.g. \((32,15)\)
(ii)Â \((7,10)\)
(i)
\begin{align}(12+10,11+2)\rightarrow(22+10,13+2)\rightarrow(32,15)\end{align}
(ii)
\begin{align}(22-10,13-2)\rightarrow(12-5,11-1)\rightarrow(7,10)\end{align}
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2021 Paper 2 Question 3
(a) A line \(n\) passes through the points \(A(-1,2)\) and \(B(0,-2)\).
Write the equation of \(n\) in the form \(y=mx+c\), where \(m,c\in\mathbb{Z}\).
\(y=-4x-2\)
\begin{align}m&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-2-2}{0-(-1)}\\&=-4\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-2=-4(x-(-1))\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-2=-4x-4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=-4x-2\end{align}
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(b) The diagram below shows the line \(l:3x-4y=5\) and the point \(P(6,-3)\).
(i) Find the equation of the line \(k\) through the point \(P\) that is perpendicular to the line \(l\).
Write your answer in the form \(ax+by+c=0\), where \(a,b,c\in\mathbb{Z}\).
(ii) Find the point of intersection of the lines \(l:3x-4y=5\) and \(h:2x-y=10\).
(i)Â \(4x+3y-15=0\)
(ii)Â \((7,4)\)
(i)
\begin{align}3x-4y=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=\frac{3}{4}x-5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_l=\frac{3}{4}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_k=-\frac{4}{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-(-3)=-\frac{4}{3}(x-6)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3(y+3)=-4x+24\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3y+9=-4x+24\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4x+3y-15=0\end{align}
(ii)
\begin{align}3x-4y=5\end{align}
\begin{align}2x-y=10\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x-4y=5\end{align}
\begin{align}8x-4y=40\end{align}
\begin{align}\downarrow\end{align}
\begin{align}5x=35\end{align}
\begin{align}\downarrow\end{align}
\begin{align}x=7\end{align}
and
\begin{align}y&=2x-10\\&=2(7)-10\\&=4\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x,y)=(7,4)\end{align}
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2021 Paper 2 Question 4
The co-ordinate diagram below shows two points \(A\) and \(B\).
(a)
(i) Write down the co-ordinates of \(A\) and of \(B\).
(ii) Find the co-ordinates of the midpoint of \([AB]\).Â
(iii) Use a compass to construct the circle \(c\), which has \(AB\) as its diameter on the diagram above.
(iv) Find the length of the radius of the circle \(c\) and hence write down the equation of \(c\).Â
(i)Â \(A=(6,2)\) and \(B=(-2,-4)\)
(ii)Â \((2,-1)\)
(iii)
(iv) \(r=5\) and \((x-2)^2+(y+1)^2=25\)
(i)
\begin{align}A=(6,2)&&B=(-2,-4)\end{align}
(ii)
\begin{align}\mbox{Midpoint}&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{6+(-2)}{2},\frac{2+(-4)}{2}\right)\\&=(2,-1)\end{align}
(iii)
(iv)
\begin{align}r&=\sqrt{(x_2-x_1)^2+(y_1-y_2)^2}\\&=\sqrt{(6-2)^2+(2-(-1))^2}\\&=\sqrt{16+9}\\&=5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x-2)^2+(y+1)^2=25\end{align}
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(b) The point \(P(2,k)\) is in the first quadrant and is on \(c\).
Use algebra to find the value of k and plot the point P on the diagram above.
\(k=4\)
\begin{align}(x-2)^2+(y+1)^2=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(2-2)^2+(k+1)^2=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(k+1)^2=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k+1=\pm 5\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k=4\end{align}
(First quadrant)
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2021 Paper 2 Question 10
A TY maths class has created a game involving a co-ordinate treasure map, as shown below.
The game consists of tasks that involve directions, distances, and locations.
The tasks in the game are based on the map.
(a)
(i) Treasure is hidden at location \(T(-2,-5)\).
Mark \(T\) on the map where this treasure can be found.
(ii) Food is located at a point \(F\) on the map.
The point \(F\) is on a line which contains the point \(A(3,-5)\) and has a slope of \(-1\).
The point \(F\) is also on a line which contains \(B(6,4)\) and has a slope of \(0\).
By drawing appropriate lines on the map above, or otherwise, find the co-ordinates of \(F\).
(i)
(ii)
\begin{align}(-6,4)\end{align}
(i)
(ii)
\begin{align}(-6,4)\end{align}
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(b)
(i) A clue to another treasure is hidden in a locked box at point \(B(6,4)\).
The \(4\)-digit code to open the box is \(d^4\), where \(d\) is the distance from \(B\) to \(C(-3,2)\),
and \(d\in\mathbb{N}\).
Find the 4-digit code (\(d^4\)).
(ii) A meeting point \(P(2,m)\) is below the \(x\)-axis.
\(P\) is a distance of \(\sqrt{41}\) from point \(B(6,4)\).
Find the value of \(m\) and plot \(P\) on the map above.
(i)Â \(6561\)
(ii)Â \(m=-1\)
(i)
\begin{align}d&=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\&=\sqrt{(6-(-3))^2)+(4-2)^2}\\&=\sqrt{85}\\&\approx9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}d^4=6561\end{align}
(ii)
\begin{align}\sqrt{(2-6)^2+(m-4)^2}=\sqrt{41}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}16+(m-4)^2=41\end{align}
\begin{align}\downarrow\end{align}
\begin{align}16+m^2-8m+16=41\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m^2-8m-9=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(m+1)(m-9)=0\end{align}
\begin{align}\downarrow\end{align}
\(m=-1\) or \(m=9\)
\begin{align}\downarrow\end{align}
\begin{align}m=-1\end{align}
(below \(x\)-axis)
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(c)
(i) The line \(k\) has equation \(x-y-3=0\).
Verify, using substitution, that the point \(T(-2,-5)\) is on \(k\).
(ii) Another treasure also needs be somewhere on the line \(k\).
You must pick a spot along \(k\) to contain this treasure.
Use algebra to find another point on \(k\), other than \(T\).
(iii) A spade for digging is hidden on line \(l\) which is parallel to \(k\).
The line \(l\) contains the point \(C(-3,2)\).
Find the equation of line \(l\).
(i)Â The answer is already in the question!
(ii)Â e.g. \((0,-3)\)
(iii)Â \(y=x+5\)
(i)
\begin{align}-2-(-5)-3&=-2+5-3\\&=0\end{align}
as required.
(ii)
\begin{align}0-y-3=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(x,y)=(0,-3)\end{align}
(iii)
\begin{align}x-y-3=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=x-3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_k=m_l=1\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-2=1(x-(-3))\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-2=x+3\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y=x+5\end{align}
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2020 Paper 2 Question 2
The points \(A(4,6)\), \(B(-2,2)\) and \(C(10,0)\) are the vertices of the triangle \(ABC\) shown below.
(a) Find \(|AB|\), the length of \([AB]\). Give your answer in the form \(a\sqrt{b}\) units, where \(a,b\in\mathbb{N}\).
\(2\sqrt{13}\)
\begin{align}|AB|&=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\&=\sqrt{(4-(-2)^2+(6-2)^2}\\&=\sqrt{36+16}\\&=\sqrt{52}\\&=\sqrt{4\times13}\\&=\sqrt{4}\times\sqrt{13}\\&=2\sqrt{13}\end{align}
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(b)
(i) Find the coordinates of \(D\), the midpoint of \([AB]\).
(ii) In the triangle \(ABC\), the point \(E(7,3)\) is the midpoint of \([AC]\).
Show that \(DE\) is parallel to \(BC\).
(i)Â \((1,4)\)
(ii)Â The answer is already in the question!
(i)
\begin{align}D&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{4+(-2)}{2},\frac{6+2}{2}\right)\\&=(1,4)\end{align}
(ii)
\begin{align}m_{DE}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{3-4}{7-1}\\&=-\frac{1}{6}\end{align}
\begin{align}m_{BC}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{0-2}{10-(-2)}\\&=-\frac{2}{12}\\&=-\frac{1}{6}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_{DE}=m_{BC}\end{align}
as required.
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(c) Find the area of the triangle \(ABC\).
\(30\mbox{ units}^2\)
\begin{align}(4,6)&&(-2,2)&&(10,0)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(4-10,6)&&(-2-10,2)&&(10-10,0)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(-6,6)&&(-12,2)&&(0,0)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\frac{1}{2}|x_1y_2-x_2y_1|\\&=\frac{1}{2}|(-6)(2)-(6)(-12)|\\&=\frac{1}{2}|60|\\&=30\mbox{ units}^2\end{align}
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2020 Paper 2 Question 4
(a)
(i) The circle \(c\) has equation \((x-1)^2+(y+4)^2=25\).
Find the centre and radius of \(c\).
(ii) The point \((1,k)\) is on \(c\). Find the two possible values of \(k\).
(i)Â Centre: \((1,04)\). Radius: \(5\).
(ii) \(k=-9\) or \(k=1\)
(i)Â Centre: \((1,04)\) and radius: \(\sqrt{25}=5\).
(ii)
\begin{align}(1-1)^2+(k+4)^2=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k^2+8k+16=25\end{align}
\begin{align}\downarrow\end{align}
\begin{align}k^2+8k-9=0\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(k+9)(k-1)=0\end{align}
\begin{align}\downarrow\end{align}
\(k=-9\) or \(k=1\)
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(b) The circle \(s\) has equation \(x^2+y^2=13\).
The point \(A(3,-2)\) is on \(s\).
Find the equation of \(t\), the tangent to the circle at the point \(A\).
Give your answer in the form \(ax+by+c=0\), where \(a,b,c\in\mathbb{Z}\).
\(3x-2y-13=0\)
\begin{align}m_r&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{-2-0}{3-0}\\&=-\frac{2}{3}\end{align}
\begin{align}\downarrow\end{align}
\begin{align}m_t=\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-(-2)=\frac{3}{2}(x-3)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}2y+4=3x-9\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x-2y-13=0\end{align}
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2019 Paper 2 Question 2
The diagram shows the line \(PQ\) and the line \(QR\).
The co-ordinates of the points are \(P(4,2)\), \(Q(8,5)\) and \(R(2,11)\).
(a) Find the slope of \(PQ\).
\(\dfrac{3}{4}\)
\begin{align}m_{PQ}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{5-2}{8-4}\\&=\frac{3}{4}\end{align}
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(b)Â Find the equation of the line \(PQ\).
Give your answer in the form \(ax+by+c=0\), where \(a,b,c\in\mathbb{Z}\).
\(3x-4y-4=0\)
\begin{align}y-y_1=m(x-x_1)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}y-2=\frac{3}{4}(x-4)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}4y-8=3x-12\end{align}
\begin{align}\downarrow\end{align}
\begin{align}3x-4y-4=0\end{align}
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(c) Write down the slope of any line perpendicular to \(PQ\).
\(-\dfrac{3}{4}\)
\begin{align}m_{\perp}=-\frac{3}{4}\end{align}
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(d) Find the area of the triangle \(PQR\).
\(21\mbox{ units}^2\)
\begin{align}(4,2)&&(8,5)&&(2,11)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(4-4,2-2)&&(8-4,5-2)&&(2-4,11-2)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(0,0)&&(4,3)&&(-2,9)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\frac{1}{2}|x_1y_2-x_2y_1|\\&=\frac{1}{2}|(4)(9)-(-2)(3)|\\&=21\mbox{ units}^2\end{align}
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2019 Paper 2 Question 4
The circle \(c\) is enclosed in the square \(PQRS\) and touches all four sides, as shown in the diagram.
The co-ordinates of three of the vertices are \(P(2,3)\), \(R(4,17)\), and \(S(-4,11)\).
(a) Find the co-ordinates of Q.
\((10,9)\)
\begin{align}Q&=(4+6,17-8)\\&=(10,9))\end{align}
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(b) Find the co-ordinates of the centre of \(c\).
\((3,10)\)
\begin{align}\mbox{Centre}&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{-4+10}{2},\frac{11+9}{2}\right)\\&=(3,10)\end{align}
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(c) Find the length of the radius of \(c\).
\(5\mbox{ units}\)
\begin{align}r&=\frac{1}{2}|RS|\\&=\frac{1}{2}\sqrt{(-4-4)^2+(11-17)^2}\\&=5\mbox{ units}\end{align}
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(d) Find the equation of circle \(c\).
\((x-3)^2+(y-10)^2=25\)
\begin{align}(x-3)^2+(y-10)^2=25\end{align}
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2018 Paper 2 Question 2
The points \(P(7,10)\), \(Q(1,2)\) and \(R(11,4)\) are the vertices of the triangle shown.
The point \(U(4,6)\) is the midpoint of \([PQ]\) and the point \(V\) is the midpoint of \([PR]\).
(a) Find the co-ordinates of \(V\).
\((9,7)\)
\begin{align}V&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{7+11}{2},\frac{10+4}{2}\right)\\&=(9,7)\end{align}
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(b) Show, by using slopes, that \(UV\) is parallel to \(QR\).
The answer is already in the question!
\begin{align}m_{UV}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{7-6}{9-4}\\&=\frac{1}{5}\end{align}
and
\begin{align}m_{QR}&=\frac{y_2-y_1}{x_2-x_1}\\&=\frac{4-2}{11-1}\\&=\frac{1}{5}\end{align}
as required.
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(c) Find the area of the triangle \(PQR\).
\(34\mbox{ units}^2\)
\begin{align}(7,10)&&(1,2)&&(11,4)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(7-7,10-10)&&(1-7,2-10)&&(11-7,4-10)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}(0,0)&&(-6,-8)&&(4,-6)\end{align}
\begin{align}\downarrow\end{align}
\begin{align}A&=\frac{1}{2}|x_1y_2-x_2y_1|\\&=\frac{1}{2}|(-6)(-6)-(4)(-8)|\\&=34\mbox{ units}^2\end{align}
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(d) The point \(S\) is the image of the point \(Q\) under the translation \(\vec{UV}\).
Find the coordinates of \(S\).
\((6,3)\)
\begin{align}S&=(1+5,2+1)\\&=(6,3)\end{align}
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2018 Paper 2 Question 4
The points \(A(1,8)\) and \(B(9,0)\) are the end-points of a diameter of the circle \(w\), as shown in the diagram.
(a) Find the co-ordinates of the centre of \(w\).
\((5,4)\)
\begin{align}\mbox{Centre}&=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\&=\left(\frac{1+9}{2},\frac{8+0}{2}\right)\\&=(5,4)\end{align}
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(b) Find the length of the radius of \(w\). Give your answer in the form \(p\sqrt{q}\), where \(p,q\in\mathbb{N}\).
\(4\sqrt{2}\mbox{ units}\)
\begin{align}r&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\&=\sqrt{(5-1)^2+(4-8)^2}\\&=\sqrt{32}\\&=\sqrt{16\times2}\\&=\sqrt{16}\times\sqrt{2}\\&=4\sqrt{2}\mbox{ units}\end{align}