fercho81 wrote:Hey guys, need a little help with the below, let me know what you think:
In the xy-plane, does the line in question y=3x+2 contain the point (r,s)?
(1) (3r+2-s)(4r+9-s)=0
(2) (4r-6-s)(3r+2-s)=0
Sorry if it was posted before, I couldn't find it. The answer is C. Thank you,
Uuuugly quadratics in the statements: a clear sign that we do NOT need to do complicated algebra to solve.
So, we just need to understand the basics of products of 0 and coordinate geometry.
When two terms multiply to 0, at least one of them must be 0. Let's look at the statements in those terms.
(1) either (3r+2-s)= 0 or (4r+9-s)= 0
Since r is our x coordinate and s is our y coordinate, we can rewrite in "y=mx+b" form:
either:
s = 3r + 2; or
s = 4r + 9
In the first case, if s = 3r + 2, does s = 3r + 2? Definitely YES.
In the second case, if s = 4r + 9, could s = 3r + 2? Sure, we really have no clue, it depends on the values of s and r.
Therefore, (1) is insufficient.
(2) Jumping ahead to our final calculations, we get either:
s = 3r + 2; or
s = 4r - 6
We have the same first case as for (1), so we know we can get a "YES".
In the second case, if s = 4r - 6, do we know if s = 3r + 2? Nope, no clue. Insufficient.
Eliminate A, B and D.
Now we need to combine the statements. We now know that:
either:
s = 3r + 2; or
s = 4r + 9
AND
s = 3r + 2; or
s = 4r - 6
Now, could both second equations be true? That is, could it be true that:
s = 4r + 9
AND
s = 4r - 6?
Well, if we subtract the second equation from the first, we get:
0 = 15
which is patently absurd. Therefore, it's impossible for both of those equations to be true.
Since both of them cannot be true, we know for sure that, to make the original statements correct, it must be true that:
s = 3r + 2,
giving us a definite "YES" answer to the original question: choose (C).
Wasn't that fun?
