But it looks like you got the help you needed

Since C is the bisector (halfway between A and B), that means that the distance A-to-C equals the distance B-to-C. We know that the distance from A-to-C is 6 (2 minus -4), so B-to-C must also be 6. Adding these two together, we get a total distance of 12 from A-to-B. -4 plus a distance of 12 gives a result of 8, and thus B is at (8,y)

Next we define y (although you could have started here). Perpendicular requires that A-C be at a 90 degree angle to line C, so since C is horizontal, A-B must be vertical. That means that the y-value of A must equal the y-value of B. Since A has a y-value of 1, B must be (x,1)

Putting the two together, we get the result of B=8,1.

This method is only recommended for people that are horrible at visualization, or for really psychotic problems. The sad thing is I used to do the really psychotic ones, and now I have to look up both bisect and perpendicular

Now, what if they’re not nice right angles? Suprisingly, this isn’t too much harder. You’ll have to have C as either a point (both coordinates), or else an angle and one coordinate (which is harder), since just one coordinate wouln’t yield a single answer anymore.

Assuming that C is a point, we just calculate the following:
[(y-value of C) minus (y-value of A)] * 2 = (y-distance)
Then, the (y-value of A) plus (y-distance) = (y-value of B)
Repeat subsitituting x for y.

To solve the above problem using this method:
1) We first have to compute the y-value of the point where C will intersect A-B. Using that note on perpendiculars again, we know that this must be 1. Therefor, the intersection point (D) will be at 2,1

(2 minus -4) * 2 = a y-distance of 12
(1 minus 1) * 2 = an x-distance of 0

-4 + 12 = a (y-value of B) of 8
1 + 0 = an (x-value of B) of 1

So that gives B = 8,1

That leaves us with the last possibility, which is the angle and only one coordinate for C. This is not as easy, but still not too hard (in fact, the problem above gives us x=2, angle=0, it’s just that angles of 0 or 90 are much easier). What we need to do is figure out the second coordinate through the miracle of Trigonometry. {the italic bits in these braces are sidenotes}

Let’s use an example: Say that C is angle 45, and x=0. We know that A-B is perpendicular, and thus has an angle equal to (angle of C) plus 90. In this case, that gives us the angle of A-B as 135.

Now, the hard part: We can still solve the known-value (in this case x) the same way. In this case, the x-distance is 8. However, we need to figure out how much y changes in respect to x.

As it happens, there’s a geometric function called the Arc-Tangent which gives us the x/y ratio {the Tangent is the y/x ratio, which you’d use instead if you only know y and the angle}.

The formula here would be that y-distance = x-distance * tangent(angle of C)

In this case, the tangent of 45 is 1 {easy!}, so that gives the y-distance = 8 * 1, or just plain 8.

At this point we’re done with the formulas, but here’s some notes on this stuff:

The [Tangent of an angle] is usually referred to as tan(angle). The [ArcTangent of an angle] is referred to as arctan(angle) or atan(angle)

Note that for angle 0, the y-distance will always be 0; and for angle 90 the x-distance will always be 0. Tan(0) and Tan(90) will both crash your calculator
Note also that calculating tangents (or arc-tangents, or sines or cosines… all of the trigometric functions) is something done with a calculator, not by hand.

The windows calculator supports sin, cos and tan. To get arctan, just do tan^-1 (arctan = inverse of tan). Equally, arccos = cos^-1, and arcsin = sin^-1. To access these features, go to View, then select Scientific. {You can also do fun stuff like hexadecimal and binary in scientific mode}

Hopefully that made sense!

Yarha, Bisectual

How good are you with geometry? It’s funny how much of this stuff I don’t remember. All in all, I’m doing better than I thought on this practice test though.