Course Content
Higher Level (by year)
0/17
Higher Level (by topic)
0/13
Ordinary Level (by year)
0/17
Ordinary Level (by topic)
0/12
Past Papers

## Question 1

The game “Twister” is played using two spinners and a mat with coloured spots.
The two spinners needed for the game are shown below. Both consist of four equal sections.
A player spins both spinners and must then place their selected hand/foot on the selected
coloured spot on the game mat.
The spinners below show the outcome Right Hand on a Yellow Spot.

(a) How many different possible outcomes are there in the game?

$$16$$

Solution

\begin{align}4\times4=16\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) Kate spins each spinner once. What is the probability that she must use:

(i) her left foot?

(ii) a red or yellow coloured spot?

(i) $$\dfrac{1}{4}$$

(ii) $$\dfrac{1}{2}$$

Solution

(i)

\begin{align}\frac{1}{4}\end{align}

(ii)

\begin{align}\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\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!

(c) Jack spins each spinner once. What is the probability that the outcome is:

(i) his right hand and a blue spot?

(ii) his right hand or a blue spot?

(i) $$\dfrac{1}{16}$$

(ii) $$\dfrac{7}{16}$$

Solution

(i)

\begin{align}\frac{1}{4}\times\frac{1}{4}=\frac{1}{16}\end{align}

(ii)

\begin{align}\frac{1}{4}+\frac{1}{4}-\frac{1}{16}=\frac{7}{16}\end{align}

Video Walkthrough

## Question 2

(a) At a stall in a fun fair, the probability of knocking a coconut off its support is $$0.34$$.

(i) What is the probability of not knocking a coconut off its support?

(ii) David is given three attempts to knock the coconut off its support.
What is the probability that he knocks the coconut for the first time, on his third attempt? Give your answer correct to three decimal places.

(i) $$0.66$$

(ii) $$0.148$$

Solution

(i)

\begin{align}P&=1-0.34\\&=0.66\end{align}

(ii)

\begin{align}P&=0.66\times0.66\times0.34\\&\approx0.148\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) The figure $$ABCDE$$ shown in the diagram consists of a large square $$ACDE$$ standing on the diagonal $$[AC]$$ of a smaller square $$ABCF$$.
The smaller square has a side length of $$2\mbox{cm}$$.
Find the area and perimeter of the figure $$ABCDE$$.

$$\mbox{Area}=10\mbox{ cm}^2$$ and $$\mbox{Perimeter}=12.49\mbox{ cm}$$

Solution

\begin{align}|AC|&=\sqrt{|AB|^2+|BC|^2}\\&=\sqrt{2^2+2^2}\\&\sqrt{8}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\mbox{Area}&=(\sqrt{8})^2+\frac{1}{2}(2\times2)\\&=10\mbox{ cm}^2\end{align}

and

\begin{align}\mbox{Perimeter}&=\sqrt{8}+\sqrt{8}+\sqrt{8}+2+2\\&\approx12.49\mbox{ cm}\end{align}

Video Walkthrough

## 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 $$m,c\in\mathbb{Z}$$.

$$y=-4x-2$$

Solution

\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}

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) 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) $$4x+3y-15=0$$

(ii) $$(7,4)$$

Solution

(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}

Video Walkthrough

## 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 $$[AB]$$.

(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 $$c$$.

(i) $$A=(6,2)$$ and $$B=(-2,-4)$$

(ii) $$(2,-1)$$

(iii)

(iv) $$r=5$$ and $$(x-2)^2+(y+1)^2=25$$

Solution

(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}

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) 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.

$$k=4$$

Solution

\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}

Video Walkthrough

## Question 5

(a) The heights of a given population are normally distributed. $$95%$$ of the population fall within the height range $$[147.9\mbox{ cm}\leftrightarrow 178.7\mbox{ cm}]$$.
Using the empirical rule, find the mean and the standard deviation of the distribution of heights in the population.

$$\mbox{Mean}=163.3\mbox{ cm}$$ and $$\mbox{Standard deviation}=7.7\mbox{ cm}$$

Solution

\begin{align}\mbox{Mean}&=\frac{147.9+178.7}{2}\\&=163.3\mbox{ cm}\end{align}

and

\begin{align}\mbox{Standard deviation}&=\frac{178.7-147.9}{4}\\&=7.7\mbox{ cm}\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) The following table shows data on the height and arm span of twenty teenagers.

Gender Height (cm) Arm Span (cm)

$$1$$

$$F$$

$$162$$

$$155$$

$$2$$

$$F$$

$$169$$

$$174$$

$$3$$

$$F$$

$$164$$

$$144$$

$$4$$

$$F$$

$$157$$

$$160$$

$$5$$

$$F$$

$$160$$

$$160$$

$$6$$

$$F$$

$$170$$

$$170$$

$$7$$

$$F$$

$$174$$

$$170$$

$$8$$

$$F$$

$$159$$

$$154$$

$$9$$

$$F$$

$$153$$

$$148$$

$$10$$

$$F$$

$$167$$

$$167$$

$$11$$

$$F$$

$$150$$

$$152$$

$$12$$

$$F$$

$$139$$

$$129$$

$$13$$

$$M$$

$$168$$

$$169$$

$$14$$

$$M$$

$$147$$

$$150$$

$$15$$

$$M$$

$$134$$

$$121$$

$$16$$

$$M$$

$$177$$

$$176$$

$$17$$

$$M$$

$$172$$

$$170$$

$$18$$

$$M$$

$$160$$

$$163$$

$$19$$

$$M$$

$$152$$

$$151$$

$$20$$

$$M$$

$$153$$

$$151$$

(i) Complete the scatter plot below by adding in the data for the males (the data in the shaded section of the table).

(ii) Three numbers are shown below.

\begin{align}0.2 && 0.9 && -0.6\end{align}

One of the numbers is the correlation coefficient of the data shown in the table above.
State the one which you think best represents the correlation coefficient between height and arm span for this data set.

(iii) Explain what this correlation coefficient tells us about the relationship between height and arm span for this data set.

(i)

(ii) $$0.9$$

(iii) There is a strong positive linear relationship.

Solution

(i)

(ii) $$0.9$$

(iii) There is a strong positive linear relationship.

Video Walkthrough

## Question 6

(a)

(i) On the diagram below showing the triangle $$ABC$$, construct the perpendicular bisector of the side $$[AC]$$. Show all your construction lines and arcs clearly.

(ii) Construct the circumcircle of the given triangle $$ABC$$.
Show all your construction lines and arcs clearly.

(i)

(ii)

Solution

(i)

(ii)

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) The triangle $$PQR$$ is the image of the triangle $$DEF$$ under an enlargement.
(The diagram is not drawn to scale.)
The scale factor of the enlargement, $$k$$, is $$2.5$$.
$$|EF|=3\mbox{ cm}$$.

(i) Use the scale factor to find $$|QR|$$.

(ii) The area of the triangle $$PQR$$ is $$18.75\mbox{ cm}^2$$.
Find the area of triangle $$DEF$$.

(i) $$7.5\mbox{ cm}$$

(ii) $$3\mbox{ cm}^2$$

Solution

(i)

\begin{align}|QR|&=3\times2.5\\&=7.5\mbox{ cm}\end{align}

(ii)

\begin{align}\frac{18.75}{2.5^2}=3\mbox{ cm}^2\end{align}

Video Walkthrough

## Question 7

(a) The following table shows the age profiles of all of the people in Ireland, aged between $$15$$ and $$65$$, who subscribed to an online streaming platform.

Age (groups) $$15$$ - $$25$$ $$25$$ - $$35$$ $$35$$ - $$45$$ $$45$$ - $$55$$ $$55$$ - $$65$$

Percentage (%) of subscribers

$$24$$

$$27$$

$$25$$

$$16$$

$$8$$

(i) Draw a Histogram to represent the data.

(ii) In a survey of $$1000$$ subscribers, aged between $$15$$ and $$65$$, how many people would you
expect to be between $$25$$ years and $$55$$ years?

(iii) Use the table or the histogram to write down the age group which contains the median age.

(iv) Use the mid-interval values of the groups in the table to estimate the mean age of the subscribers. Give your answer correct to one decimal place.

(i)

(ii) $$680$$

(iii) $$25$$-$$35$$ age group

(iv) $$35.7\mbox{ years old}$$

Solution

(i)

(ii)

\begin{align}1{,}000\times(0.27+0.25+0.16)=680\end{align}

(iii) $$25$$-$$35$$ age group.

(iv)

\begin{align}\mbox{Mean}&=0.24\times20+0.27\times30+0.25\times40+0.16\times50+0.08\times60\\&=35.7\mbox{ years old}\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) Prior to the COVID-19 pandemic, Netflix held the largest share of the online streaming
market with $$65\%$$ of all streaming subscribers in the $$15$$ to $$65$$ age group. Aiden thought that this percentage might have changed during the pandemic and surveyed a sample of subscribers in this age group to get data.
He then carried out a Hypothesis Test to test his theory.

(i) If $$540$$ people responded to his survey, calculate the margin of error of his survey.

(ii) His survey revealed that $$372$$ of the responders said they now have a Netflix account.
Use your answer to Part (b)(i) above to create a $$95\%$$ confidence interval for the percentage of the population that have a Netflix account.

(iii) Use your answer to Part (b)(ii) above to conduct the Hypothesis Test, at the $$5\%$$ level of significance to test Aiden’s claim that the percentage of subscribers to Netflix had changed. State clearly your null hypothesis, your alternative hypothesis and give your conclusion in the context of the question.

(i) $$4.3\%$$

(ii) $$64.6\%<\hat{p}<73.2\%$$

(iii)

Null hypothesis: The percentage of subscribers to Netflix had changed.

Alternative hypothesis: The percentage of subscribers to Netflix had not changed.

Calculations: $$65%$$ is is within the confidence interval above.

Conclusion: The percentage of subscribers to Netflix had not changed.

Solution

(i)

\begin{align}\mbox{Margin of error}&=\frac{1}{\sqrt{n}}\\&=\frac{1}{\sqrt{540}}\\&\approx0.043\\&=4.3\%\end{align}

(ii)

\begin{align}\frac{372}{540}-4.3<\hat{p}<\frac{372}{540}+4.3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}64.6\%<\hat{p}<73.2\%\end{align}

(iii)

Null hypothesis: The percentage of subscribers to Netflix had changed.

Alternative hypothesis: The percentage of subscribers to Netflix had not changed.

Calculations: $$65%$$ is is within the confidence interval above.

Conclusion: The percentage of subscribers to Netflix had not changed.

Video Walkthrough

## Question 8

The diagram shows the plan of a lake.
The line segment $$[PQ]$$ represents the distance from the pier, $$P$$, to the far side of the lake.
At equal intervals of $$10\mbox{ m}$$ along this line, perpendicular measures are made to the sides of the lake as shown.

(a)

(i) Use the Trapezoidal Rule to estimate the surface area of the lake.

(ii) If the lake is on average $$8\mbox{ m}$$ deep, estimate the volume of water in the lake.

(i) $$2{,}050\mbox{ m}^2$$

(ii) $$16{,}400\mbox{ m}^3$$

Solution

(i)

\begin{align}A&=\frac{10}{2}\left[0+0+2(110)\right]+\frac{10}{2}\left[10+0+2(90)\right]\\\&=2{,}050\mbox{ m}^2\end{align}

(ii)

\begin{align}V&=(2{,}050)(8)\\&=16{,}400\mbox{ m}^3\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)

(i) Calculate the volume , in $$\mbox{cm}^3$$, of a hemisphere of radius $$21\mbox {cm}$$.
Give your answer terms of $$\pi$$.

(ii) A buoy on the lake is in the shape of a hemisphere of radius $$21\mbox{ cm}$$ surmounted by a cone.
The volume of the cone is equal to the volume of the hemisphere.
Find $$h$$, the total height of the buoy.

(i) $$6{,}174\pi\mbox{ cm}^3$$

(ii) $$63\mbox{ cm}$$

Solution

(i)

\begin{align}V&=\frac{2}{3}\pi r^3\\&=\frac{2}{3}\pi(21^3)\\&=6{,}174\pi\mbox{ cm}^3\end{align}

(ii)

\begin{align}\frac{1}{3}\pi r^2H=6{,}174\pi\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{1}{3} (21^2)H=6{,}174\end{align}

\begin{align}\downarrow\end{align}

\begin{align}H&=\frac{(3)(6{,}174)}{(21^2)}\\&=42\mbox{ cm}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}h&=H+r\\&=42+21\\&=63\mbox{ cm}\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!

(c) The buoy is situated at $$B$$, $$43a\mbox{ m}$$ from the pier $$P$$ and $$26\mbox{ m}$$ from the point $$Q$$, as shown in the
diagram above.
Find $$|\angle QBP|$$, the angle at the buoy shown on the diagram.

$$118.74^{\circ}$$

Solution

\begin{align}60^2=43^2+26^2-2(43)(26)\cos|\angle QBP|\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|\angle QBP|&=\cos^{-1}\left(\frac{43^2+26^2-60^2}{2(43)(26)}\right)\\&\approx118.74^{\circ}\end{align}

Video Walkthrough

## Question 9

The line segment $$[SE]$$, shown below, represents an airport runway.
The point $$S$$ and the point $$E$$ represent the start and end points of the runway respectively.
The diagram is drawn to a scale of $$1\mbox{ unit}=250\mbox{ metres}$$.

(a)

(i) Find the length of the runway. Give your answer in $$\mbox{km}$$.

(ii) An aircraft starts at the point S and travels $$1250\mbox{ m}$$ to a point $$L$$ where it lifts off.
From the point $$L$$, the aircraft makes an angle of $$14^{\circ}$$ with the ground, $$[LE]$$.
The aeroplane travels along this path until it is directly above $$E$$.
Plot and label the lift-off point $$L$$, and draw the aircraft’s flight path for this part of its journey, on the diagram above.

(iii) Find the total distance the aeroplane has travelled when it is directly above $$E$$.

(i) $$2.5\mbox{ km}$$

(ii)

(iii) $$2{,}538\mbox{ m}$$

Solution

(i)

\begin{align}l&=250\times 10\\&=2{,}500\mbox{ m}\\&=2.5\mbox{ km}\end{align}

(ii)

(iii)

\begin{align}d&=1{,}250+\frac{1{,}250}{\cos 14^{\circ}}\\&\approx2{,}538\mbox{ m}\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) The aircraft flies from airport $$A$$ to airport $$B$$, and then on to airport $$C$$, at the same altitude.
The pilot records the flight summary on the given diagram.

(i) Find the distance from airport $$B$$ to airport $$C$$. Give your answer correct to the nearest $$\mbox{km}$$.

(ii) When the plane was directly over airport $$C$$, the pilot was instructed to “circle” until a runway was available. She therefore flew $$10\mbox{ km}$$ away from $$C$$ before turning and flying along a circular arc of $$70^{\circ}$$ and then returning to the airport.
This path was all flown at the same altitude.
Find the total distance travelled.
Give your answer, in $$\mbox{km}$$, correct to two decimal places.

(i) $$324\mbox{ km}$$

(ii) $$32.22\mbox{ km}$$

Solution

(i)

\begin{align}\frac{|BC|}{\sin47^{\circ}}=\frac{260}{\sin36^{\circ}}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}|BC|&=\frac{260(\sin47^{\circ})}{\sin36^{\circ}}\\&\approx324\mbox{ km}\end{align}

(ii)

\begin{align}d&=10+10+2\pi(10)\left(\frac{70^{\circ}}{360^{\circ}}\right)\\&\approx32.22\mbox{ km}\end{align}

Video Walkthrough

## 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}

Solution

(i)

(ii)

\begin{align}(-6,4)\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)

(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) $$6561$$

(ii) $$m=-1$$

Solution

(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)

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!

(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 $$l$$.

(ii) e.g. $$(0,-3)$$

(iii) $$y=x+5$$

Solution

(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}

Video Walkthrough