L.C. MATHS

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Past Papers

## 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...

Area of a Triangle

Has not appeared

Triangle Concurrencies

Equation of a Line

Features of a Line

Divide Line Segment into Ratio

Perpendicular distance from a Point to a Line

Angle between Two Lines

Equation of a Circle

Tangent to a Circle

Intersection of a Line and a Circle

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 $$4:1$$.

$$\left(\dfrac{4}{5},\dfrac{8}{5}\right)$$

Solution

\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 $$r$$.

$$(0,r-mq)$$

Solution

\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 $$d,e,f\in\mathbb{Z}$$.

$$8+5\sqrt{3}$$

Solution

\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}

Video Walkthrough

## 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 $$k$$.

$$k=-58$$

Solution

\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.

$$2$$

Solution

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 $$(x-13)^2+(y+13)^2=169$$

Solution

\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}

Video Walkthrough

## 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 $$k$$.

$$k=-2$$

Solution

\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 $$t=-\dfrac{43}{11}$$

Solution

\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) $$(23,8)$$

Solution

(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}

(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}

Video Walkthrough

## 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 $$a,b\in\mathbb{N}$$.

$$4\sqrt{2}$$

Solution

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}

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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 $$s$$.

(ii) e.g. $$(16,25)$$

Solution

(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}

Video Walkthrough

## 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.

Solution

\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.

$$33.435^{\circ}$$

Solution

\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}

Video Walkthrough

## 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 $$|BT|$$.

$$|BT|=9$$

Solution

\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 $$(x+2)^2+y^2=25$$

Solution

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}

Video Walkthrough

## 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 $$\dfrac{x}{a}+\dfrac{y}{b}=1$$.

Solution

\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) $$P:\left(\dfrac{18m+25}{3m+4},\dfrac{m}{3m+4}\right)$$

Solution

(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}

Video Walkthrough

## 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 $$k=10$$

Solution

\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 $$s$$.

$$(x-1)^2+(y-1)^2=1$$

Solution

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}

Video Walkthrough

## 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 $$m$$.

Solution

\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.

$$(6,13)$$

Solution

\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.

$$(x+1)^2+(y-2.5)^2=3.25$$

Solution

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}

\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}

Video Walkthrough

## 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 $$P$$.

$$(1,-1)$$

Solution

\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 $$B$$.

$$B:(-2,-4)$$

Solution

\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 $$ABC$$.

Solution

\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.

Video Walkthrough

## 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.

$$x^2+y^2-6.5x-12y=0$$

Solution

\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}

Video Walkthrough
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(b) Find $$|\angle BCA|$$. Give your answer in degrees, correct to $$2$$ decimal places.

$$28.44^{\circ}$$

Solution

\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}

Video Walkthrough

## 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 $$AC$$.

$$3x-2y-9=0$$

Solution

\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}

Video Walkthrough
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(b) Use your answer to part (a) above to find the co-ordinates of the orthocentre of the triangle $$ABC$$.

$$(7,6)$$

Solution

\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}

Video Walkthrough

## 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 $$a,b,c,\in\mathbb{Z}$$.

$$x-7y+43=0$$

Solution

\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}

Video Walkthrough
##### In the meantime, if you feel that you would not be confident if this question (or similar) appeared on your exam, we suggest booking one of our one-to-one grinds!

(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 $$(x+8)^2+(y-5)^2=25$$

Solution

\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}

Video Walkthrough

## 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 $$t$$.

$$t=-5$$

Solution

\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}

Video Walkthrough
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(b) Find, in terms of $$k$$, the distance between the point $$P(10,k)$$ and $$l_1$$.

$$\dfrac{|18-4k|}{5}$$

Solution

\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}

Video Walkthrough
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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) $$3$$

Solution

(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}

Video Walkthrough

## 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: $$6\sqrt{10}$$

Solution

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) $$(x-5)^2+(y-6)^2=40$$

Solution

(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}

Video Walkthrough
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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 $$a,b,c\in\mathbb{Z}$$.

$$x+3y-43=0$$

Solution

\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}

Video Walkthrough