Are they intersecting at the same point?

Geometry Level 2

There are three lines: $l_1$ , $l_2$ , $l_3$ .

$AB=BC= \alpha$ , $GD=2 \alpha$ , $AD=BE=CF= \beta$ , $AC // GF$ . Do $l_1$ , $l_2$ and $l_3$ intersect at the same point?

Not enouge information No Yes

1 solution

Haoran Wang
May 19, 2018

We can make three lines as three functions: $y=\frac{\beta}{4\alpha}x$ , $y=\frac{\beta}{\alpha}x-2\beta$ , $y=\beta -\frac{\beta}{2\alpha}x$ for $l_1$ , $l_2$ , $l_3$ . Solve $\frac{\beta}{4\alpha}x=\frac{\beta}{\alpha}x-2\beta$ , we can get $x=\frac{4\alpha}{3}$ . Put $x=\frac{4\alpha}{3}$ into $y=\frac{\beta}{4\alpha}x$ , we get $y=\frac{\beta}{3}$ . And put $x=\frac{4\alpha}{3}$ into $y=\beta -\frac{\beta}{2\alpha}x$ , we get $y=\frac{\beta}{3}$ . So answer is yes.

Hey Haoran. Don't get me wrong, bro, I find your explanation for the answer quite good. The only thing I don't understand about it is how you got a denominator of 3 as you were solving y = Bx/4a = Bx/a - 2B

I get the 4a part, but why is 3 the denominator. Because when I was solving the very same thing I got x = 4ay.

Since Bx/a is 4x as much as Bx/4a, if you were to subtract Bx/4a from both sides, you'd be left with with 0 = 3Bx/4a - 2B

To equalize denominator, we would rewrite that as 0 = 3Bx/4a - 8Ba/4a or eventually 8Ba = 3Bx. Wait a second! We would then divide by B and be left with 8a = 3x. And finally, after all of that, I get x = 8a/3

Wow, what an interesting coincidence I came up with after doing all of this. It would end up being x = 4a if it was 8a = 2x as dividing by 2 would then give you x = 4a. I even did all of the steps a second time just to make 100% sure whether 8a ended up being 3x or 2x. But just thinking about all of this really opens my critical think capablities, so thank you if your reading this. If you can reply to what I said, that would be even better, but since this is quite long, it doesn't have to be right away or something like that

Joseph Regassa - 2 years, 2 months ago

Are they intersecting at the same point?

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