Specialist Mathematics · Past QCAA questions All simple familiar questions
2024 · Paper 1 · Simple familiar

Question 11

[4 marks] Technology-free
Topic guidePremiumUnit 3 Topic 3.2 · Equations of lines
Game planPremium3.2 Writing the equation of a line
The question

The vector equation of a straight line is given by $\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2\\0\end{pmatrix}+k\begin{pmatrix}-1\\2\end{pmatrix}$, where $k$ is a scalar.

a)

Express the equation of the line as a pair of parametric equations.

[1 mark] b)

Use your result from Question 11a) to express the equation of the line as a Cartesian equation.

[1 mark] c)

Determine the coordinates of the point that the line passes through when $k=5$.

[1 mark] d)

Determine the value of $k$ when the line intersects the $y$-axis.

[1 mark]
Watch the situation first

Start at one point, then take steps of the direction vector

Watch the point move along the line as $k$ grows. Every value of $k$ gives one point on the line.

When k = 0, the line is at (2, 0) Each step of k moves 1 left and 2 up At k = 5 the point is (−3, 10) The line meets the y-axis where x = 0 −3−2−113246810xy (−1, 2) (2, 0) (−3, 10) (0, 4) x = 2 − k y = 2k k = 5: x = −3, y = 10 x = 0: 2 − k = 0, so k = 2

The point $(2,0)$ is where $k=0$, and the vector $(-1,2)$ is one step of $k$.

Now for the mathematics

Choose a step along the line and find the point

Move the slider to change $k$. The readouts use the parametric equations, and the last line checks that the point fits the Cartesian equation.

−4−2224681012xy

This is part c). With $k=5$ the point is $(-3,10)$.

This is part d). The point is on the $y$-axis, because $x=0$. That happens when $k=2$.

With $k=0$ you are at the starting point $(2,0)$, the position vector in the equation.

Every value of $k$ gives a point on the line, and every point fits $y=4-2x$.

QCAA marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. Each part is worth one mark.

Step 1 · part a)1 mark

Read the vector equation one row at a time. The top row gives $x$ and the bottom row gives $y$.

$$x=2+k(-1)=2-k\qquad y=0+k(2)=2k$$

Marker: correctly express the equation of the line as a pair of parametric equations

Step 2 · part b)1 mark

A Cartesian equation has no $k$. Make $k$ the subject of one equation and substitute it into the other.

$$k=2-x\quad\Rightarrow\quad y=2(2-x)$$

The expanded form $y=4-2x$ is also accepted.

Marker: expresses equation of line as a Cartesian equation

Step 3 · part c)1 mark

Substitute $k=5$ into the vector equation.

$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}2\\0\end{pmatrix}+5\begin{pmatrix}-1\\2\end{pmatrix}=\begin{pmatrix}-3\\10\end{pmatrix}$$

$(-3,\ 10)$

Marker: determines coordinates of point

Step 4 · part d)1 mark

Every point on the $y$-axis has $x=0$. Set the $x$ equation to 0 and solve for $k$.

$$0=2-k$$

$k=2$

The question asks for $k$, not the point. The point itself is $(0,4)$.

Marker: determines value of $k$

Putting it all together

Each phrase of the question gave you something

“a pair of parametric equations”
One equation for $x$ and one for $y$, both in terms of $k$.
“a Cartesian equation”
An equation in $x$ and $y$ only, so you eliminate $k$.
“when $k=5$”
You substitute into the vector equation and give the coordinates.
“intersects the $y$-axis”
On the $y$-axis, $x=0$.
What makes this simple familiar

Each part is one routine step for lines, and part b) tells you to use part a). The numbers are small enough to do in your head.

If I had to guess where you might lose marks, it would be part d). The question asks for the value of $k$, so writing only the point $(0,4)$ doesn’t answer it. In part c), give a point, not just the value of $x$.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2024 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2024, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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