Hello once again everybody, time once again for another guide to get you all set for Trials and HSC! This one is going to cover plane geometry including the basic formulae, and questions on locus. I'll also do a run down of parametric equations for Extension students.
Remember to register for an account and ask any questions you have below! Also, be sure to take advantage of the notes for 2/3 Unit , which go into fantastic detail and will give you that bit of extra assistance and revision, if you need it 
Okay, so plane geometry. Depending on which level of mathematics you did in Year 9-10, you almost definitely would have seen the formulas for distance, midpoint, and gradient:
^2+(y_2-y_1)^2} )
 )
These actually prove more important than you may think! They pop up a lot. In 2 unit we were introduced to a few more, which pop up here:
The first part requires the use of the two point formula (at least that is the easiest way):
The second part requires use of the perpendicular distance formula:
And the third combines this result with the distance formula to find the area of the triangle.
}^2+{\left(6-0\right)}^2}\\\\=\frac{1}{2}\times\sqrt{5}\times3\sqrt{5}\\\\=\frac{15}{2}{u}^{2} )
Questions like this involving shapes, particularly triangles, parallelograms, or kites, are common. There is a lot of room to make a mistake, but the questions are designed so that
your answer should, in 99% of cases, be rational . In the question above, the root 5 cancels with itself in the final part. Look for this in your exam, if you get an irrational answer, check it, because it could mean you have made an error.
This type of question has appeared in almost every HSC and CSSA Trial exam in the past five years, so make sure you are ready for it.
Another question common in the multiple choice is a question asking for the angle made with the x axis. This is a simple formula linking angle with gradient, sub and go:
Extension students have one more formula, a division in given ratios formula:
Spoiler
The point dividing a given interval in the ratio m:n is given by:
)
This is likely to be in your multiple choice.
Remember, if the division is EXTERNAL, simply make either m or n negative, and it works in exactly the same manner. Again, sub and go, if you know the formula this won't give you any trouble.
Of course, you'll also have to use the number plane for graphing. The calculus guides section(s) covers the process I recommend for such questions. But some other, more generic pointers for graphing:
- Make sure you label your axes, or this will cost you a mark (a lot of the time). The same is true for lines, label the equations!
- Use a ruler to make sure it is neat, or otherwise, take your time! If your lines are messy and all over the place the marker won't be happy
- Think about your scale before you draw, otherwise your graph may not be clear.
- Draw asymptotes clearly with a dotted line and a sharp pencil, and LABEL THEM!
For 2 unit, the last thing you really have to worry about on the number plane is locus.
A locus is the path taken by a point which obeys a given condition . You should be familiar with at least these two common loci:
- The locus of a point which remains a set distance from another point is a circle.
- The locus of a point which remains equidistant from a line and a point is a parabola.
You study that second one in length. The point is called the
focus , and the line is called the
directrix . A parabola can be defined using these geometric entities using the formula:
.
The variable 'a' is the focal length. The focus is located 'a' units above the vertex, so in the simplest case, has coordinates (0,a). The directrix is a horizontal line located 'a' units below the vertex, and has the equation y=-a. Of course, these coordinates and equations change slightly as the position of the parabola changes, according to the more general formula:
}^{2}=4a\left(y-b\right))
where (a,b) is the vertex of the parabola.
Chords are intervals from one arm of the parabola to the other. A focal chord is a chord which passes through the focus.
2 Unit students are mostly given students concerning interpretation of these formulae.
Example: A parabola has focus (5,0) and a directrix x=1. What is the equation of the parabola. You should practice looking at questions like these and interpreting the information. The first thing you should notice, the directrix is in terms of x (i.e., it is vertical, not horizontal). This means the parabola is sideways! This is okay, we now just look at the formulas the 'opposite' way:
}^{2}=4a\left(x-b\right))
We are given the focus and the directrix, not the vertex. To find it, remember this:
The vertex of a parabola is halfway between its directrix and its focus. Halfway between x coordinate 5 and x coordinate 1, is three, so the vertex has coordinates (3, 0). The focal length is half of this difference, 2.
So the equation is:
)
Interestingly, my 2 Unit HSC paper had zero questions on locus. It is a smaller part of the course, but given that it didn't appear in 2014, expect it in 2015!
Extension students go into a lot more depth here, and it is linked to parametric representation of curves.
Spoiler
Parametric representation of parabolas in the 3U course use the following parametric equations:
Questions in extension exams usually give you two coordinates on the parabola with some other details, and ask you to prove various geometric facts or reach certain conclusions. There are many formulas linking these points you can remember, for gradient, midpoint etc, but I find it is much easier to just derive them in the exam. For example, for gradient:
)
The only thing I suggest you remember is
if PQ is a focal chord! This is a result you are allowed to quote and may save you some work in the exam.
There are literally unlimited things they could ask, but lets walk through one which also utilises interval division:
For the first part, we use the interval division formula to get the coordinates of Q. Look at the answers to guide which point should be considered first, and which second. I'll skip the algebra:
\\\\=\left(\frac{2at}{1+{t}^{2}},\frac{2a{t}^{2}}{1+{t}^{2}}\right))
Expressing the gradient of OQ is a snap:
This next bit threw me for a loop for longer than I'd like when I sat this paper last year. But it highlights an important tip,
never forget the basics . The slope OQ is just the y coordinate of q over the x coordinate of q. Therefore,
. This is sort of strange, but think about it, the gradient is t, and the rise on run is just whatever the y coordinate of Q is, divided by the x coordinate. Hence the relationship above. Therefore, and I almost didn't even think to do this, we substitute this expression in the place of t in the x or y coordinate for Q found in part (i). Probably better to see it, and the x coordinate is easier:
}^{2}={a}^{2} )
This is a WEIRD proof, but be prepared for things similar to this (that process utilised completing the square, in case it is unclear). I absolutely advise leaving these questions till last unless you get them right off the bat, they are noodle scratchers. Look where you are heading, and the information you have. How can you bring it all together? Just relax and the answer will come.
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