So is anybody good at math? In order to take calculus, I need to take a math placement test. I’m trying to brush up using a review packet, and I came across a question that puzzles me a bit. I was able to draw out the answer in my head, but I cannot figure out the actual math formula to answer the question.
Point A (-4, 1) is in the standard (x,y) coordinate plane. What must be the coordinates of point B so that the line x=2 is the perpendicular bisector of line AB?
a. (-6, 1)
b. (-4, -1)
c. (-4, 3)
d. (-2, 1)
e. (8, 1)
I will post the answer to the question in a comment, in case you’d rather not know. Is there a formula that makes this easier than drawing the graph out in my head and plotting the points?
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The answer is E (8, 1)
Comment by jaxia — June 14, 2005 @ 11:55 am
Not sure of the exact math term but as the line x=2 is purely vertical you dont have to fek with Y at all. Lets call the the x=2 line C instead or this gets way to confusing.
Bx = 2C - Ax
Although that only works if A is to the left of C… I think…
*more confused*.
Comment by tarss — June 14, 2005 @ 12:10 pm
I’m not convinced there is a formula that covers this A is either side of the line. Frankly I suspect they were looking for you to work it out in your head.
Its been toooo long since I tried thinking of this stuff
Comment by tarss — June 14, 2005 @ 12:20 pm
Although that only works if A is to the left of C… I think…
That is why the answer is the 8 one. Because I need an answer to the right of -4 AND 2 with a y of 1. BUT, isn’t (5, 1) a good answer, too (but just not an option?)
/still confused
Comment by jaxia — June 14, 2005 @ 12:23 pm
(8,1) is the only answer, because the x=2 line must both bisect and be perpendicular to the line AB.
Yarha, Details, Details..
Comment by yarha — June 14, 2005 @ 12:32 pm
Oh! Bisect = exactly half?!
It’s only been 9 years since I’ve had a math class…
Comment by jaxia — June 14, 2005 @ 12:52 pm
Time to put my math degree to good use…
The line x=2 is a vertical line that crosses the x-axis at (2,0). Since the perpendicular bisector of a line meets that line at a 90-degree angle, the line with endpoint (-4,1) must be a horizontal line; specifically, it is a segment of the line y=1.
We went to find out how far the point (-4, 1) is from the line x=2, because the answer to this problem will be a point which is the same distance from the line x=2, only on the other side of the line. Since AB is horizontal, and the bisector is vertical, all you need to do is count how many units A is away from x=2 along the x-axis. The point at which x=2 bisects AB must be (2,1), since the bisector is the line x=2, and AB is on the line y=1. Well, (-4,1) is 6 units away from (2,1) [2-(-4)=6], so B must be 6 further units away from (2,1).
Thu, point B is at (2+6,1), or (8,1).
Did that make sense?
Comment by thecanadian_97 — June 14, 2005 @ 12:58 pm
And, since I missed the part about wanting a formula, no, it’s much easier to just draw it out.
Comment by thecanadian_97 — June 14, 2005 @ 1:02 pm
Yeah, I forgot that bisect meant exactly half. I was thinking “intersect.”
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.
Comment by jaxia — June 14, 2005 @ 1:06 pm
Yeppers, ‘bisect’ passes through exactly half of the thing bisected, hence ‘bi’ and ’sect’. This, of course, sets me up for the tag-line…
Yarha, Bisectual
Comment by yarha — June 14, 2005 @ 1:24 pm
Okies I could be miss understanding your answer here but the formula I gave made Bx as 8 not 5
Bx = 2C - Ax is
Bx = ( 2 times C ) - -4
thus Bx = (4) - -4
Bx = 4 +4
Comment by tarss — June 14, 2005 @ 1:31 pm
Graphing problems tend to be something you’d work out visually, but there is a formula. Given that I couldn’t even understand what the question was asking until I got the answer and then Yarha’s explanation of the answer, however, I can’t be the best person to ask
We’ll start by computing the x value
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!
Comment by banditangel — June 15, 2005 @ 1:24 am
Ahhh, yet another stellar example of why I am not a mathematician.
But it looks like you got the help you needed
Comment by grizzilla — June 15, 2005 @ 3:17 am
OW.
Comment by maskedfencer — June 15, 2005 @ 9:09 am
Yeah, I basically took one look at the trig section of the practice test and decided I’d need to retake it. Once we got to sin, cos and tan, I quit paying attention in class
Thanks for your help!
Comment by jaxia — June 15, 2005 @ 12:51 pm
Yeah, I’d rather stick to the basics myself. This is a prime example of why I’m getting a BA instead of a BS
Comment by jaxia — June 15, 2005 @ 12:52 pm
Trigonometry was my biggest headache in High School - I actually had to relearn most of what I just explained to me, since it never quite stuck with me back then. It’s one of the less “intuitive” areas of math, at least in my opinion. Once you get a good grip on the basic formulas, though, it’s a fairly useful branch since it’s the stuff you use on all the graphs that aren’t nice, neat right angles
Comment by banditangel — June 15, 2005 @ 11:44 pm