<p>Krabble, I am sorry, I thought you were asking distance between two points. For the formula about a point and a line:
if the point is (x1, y1) and the line ax+by+c=0 (you need to rearrange it to make the line in this way if it is y=....)
absolute value of (a<em>x1+b</em>y^1+c) divided by square root of (a^2+b^2). </p>
<p>The logic for this is a little bit complicated. First you find the slope of the given line. Then you take the negative reciprocal of that slope. then you write the equation of the line that has the negative reciprocal slope and passes through the given point. Then you simultaneously solve both equations (the given one and the one you generated). When you solve them, you get a point. The distance between that point and the given point is the result of the question. [The reason to find another line is that you need to find the least distance. To do that, you need to have perpendicular lines.]</p>
<p>So basically I use the equation of my line to find the perpendicular line (with the point given). I find the intersection of the perpendicular line and the original equation line and find the distance between the intersection and the original point?</p>
<p>Thanks!</p>
<p>Krabble88, yeah, that is right. good summary.</p>
<p>Thanks Von Neumann.</p>
<p>How often do polar questions, namely the complex equations with "cis", appear on the test?</p>
<p>Is it possible to graph polar equations in the "r = " form? Or do I have to change it back into rectangular form to graph them (on a graphing calculator)?</p>
<p>Thanks!</p>
<p>What other miscellaneous topics have you noticed on the test?</p>
<p>There was no polar questions in the test, and I don't think they will ask a question on that subject.
It is possible to graph equations in the r= form, I think. Although I have never used it, if you go to "mode" and change sth to pol., you can graph the equations in that way. I left my TI 83+ at school, so I don't remember what you need to change actually.
In the test, there was one limit question that I needed to graph to solve. Also, I solved three questions by using sine rule. There was one question about possible intersections of planes. There was one question on combination, one on possibility and one on statistics (mean, median, mode). I can' remember others, but I remember that I didn't use integral, derivative, matrices, logic and sigma.</p>
<p>Ah, thanks for the info. Would you happen to know how to graph a line parallel to a 3D point? I think that's what the question stated in my last practice test, but I can't remember exactly.</p>
<p>I'm using a ti-89 and tried to change the mode from rectangular to polar, but I don't know what to do next (do I go to "y="?).</p>
<p>What would you say is the most difficult part of the test? I'm a fairly good math student (A+'s throughout high school math and taking Calc BC now), but did horribly on the SAT Math 2 when I took it in June, so I'm worried I'm going to do poorly again (although I didn't get a graphing calculator at the time and forgot a lot of trig identities, etc.). I have a feeling that I may have filled in the wrong bubbles or something, because I probably got less than half right on the actual test.</p>
<p>I am not sure, but I think there are infinite number of parallel lines to a 3D point. There should be some specific details for this question.
To make it into polar form, you go to "mode". (for 3d questions, just draw x-y-z axis, you should be able to see the asnwer, because questions on this subject are very straight-forward in math iic.)
Whatever you do, don't forget your calculator, don't panic and be worried. This is the key to success on math iic. I think you will do very well. The most difficult part was 1 question about trigonometry (the identity was easy (sin^2+cos^2=1) but the algebra of the question was hard) and one question on intersection of planes (The wording was not good, and I think this is the only question I had wrong) . Apart from those 2 questions, the exam was very easy. if you take aadvantage of your calculator, I don't think any reasons for you not getting an 800.</p>
<p>I feel like I'm missing a good 2-3 questions JUST from the 3D-related ones. I don't know how to do these at all and have not done this in school. What should I know?!?</p>
<p>By the way, thank you Von Neumann for all of your help!</p>
<p>P.S.- And each time I come to a 3D question, I feel hopeless and start cursing my math teacher from last year, lol..</p>
<p>b u m p</p>
<p>Can anyone give me an overview of 3D-related questions?</p>
<p>Hey, if you don't want to memorize the formula...Barron's also has that neat little section (ch. 9 I think) with the programs you can put in your calculator. There are ones for the distance btwn a point and a line, the acute angle btwn. two lines, so on and so forth. Even if you end up not being able to plug in the right numbers to execute the program, the equation will be on your calculator for your reference.</p>
<p>Krabble 88, you are welcome. It may be too late, but I am sorry, I am international and was sleeping (Are you taking the test today?) In 3D questions, just think of in terms of real objects if you cannot think in terms of x-y-z planes. Don't worry, in my exam, there were only 2 questions and one of them was very (VERY) easy. If you post some of the questions you miss, I can help you to solve those questions easily. Afterwards, you will be able to solve other questions, I think. Think that a line is a side of your test booklet which extends to infinity :) Then the surface of the booklet, which also extends to infinity in both sides, is your plane. Think of other test booklets which intersect your plane... :)
And I remember that if they ask the distance in 3D, just find the coordinates, use phyto. theorem (but of course in square root, add the squares of the differences of all three parts of the coordinates. (x-y-z))</p>
<p>Hey.
I don't wanna be a pain in the ass, but as a olympiad mathematician, I just had to point something out.
The sine rule is stated in that list as (sin A)/a = (sin B)/b = (sin C)/c, which is of course perfectly correct.
Just as an interesting point though, we can generalise sine rule a little bit.
If we state the rule as a/(sin A) = b/(sin B) = c/(sin C), which is equivalent to the above, but then we can add in a nice extra bit, that a/(sin A) = b/(sin B) = c/(sin C) = 2R.
R here is the circumradius, the radius of the circule that goes through the three vertices of the triangle.
Just something slightly more elegant.
Don't know if you will ever use that in a SAT, but then again, mathematics not just about SAT, is it?</p>
<p>Best wishes
Eric</p>
<p>Yeah, erickangnz is right. (I knew that but never used it except once in a practice exam for Turkish national university entrance exam and once in olympiad)
However, I don't think that it is a part of a sine rule (I mean that 2R equals to all of those, but it isn't a part of the rule)
Nevertheless, it is more elegant as erickangnz pointed out.</p>
<p>Wow thank you very much! Is a 'trace' of a plane just ANY side of the plane? (I can plug in zero for any x, y, or z and it will give me an equation of a trace of the plane?)</p>
<p>I'm leaving in 10 minutes lol.. thanks!!</p>