Question

Coordinate geometry question #3

Answer 1

I did this question by letting vertices with the knowns (x1,y1), (x2,y2), (x3,y3) and solve it. Is there any faster way to do this question?

Answer 2

looks like your job
http://www.wolframalpha.com/input/?i=%28x1%2Bx2%29%2F2+%3D+3%2C+%28x2%2Bx3%29+%3D+2%2C+%28x1%2Bx3%29+%3D+4

Answer 3

i can't see any other way ... but I'll keep thinking

Answer 4

Alight... Thanks all. Then perhaps that's the only way :(

Answer 5

good old classic brute method always works for me
http://www.wolframalpha.com/input/?i=%28y1%2By2%29%2F2%3D4%2C+%28y2%2By3%29%2F2%3D0%2C+%28y1%2By3%29%2F2%3D2
still there might be some elegant way of doing it.

Answer 6

i was wondering if maybe there was some transformation / resizing that you could do to those points... D:

Answer 7

Draw another triangle with the mid-points? :O

Answer 8

Sorry, question closed for a while. I have an easier question to ask....

Answer 9

Three vertices \(\left(x_1,y_1\right),\left(x_2,y_2\right) \) and \(\left(x_3,y_3\right)\).
\[x_1 + x_2 = 2\cdot3, \quad x_2 + x_3 = 2\cdot2,\quad x_1+x_3=2\cdot 4\]\[y_1 + y_2 = 2\cdot4,\quad y_2 + y_3 = 2\cdot0,\quad y_1+y_3=2\cdot 2\]

Answer 10

\[y_2 = -y_3\]Is the key here.

Answer 11

looks like @Ishaan94 and I think in the similar way.
Genius think alike!!! LOL

Answer 12

Lol hehe

Answer 13

Hmm.. so that's the only way to solve, right?

Answer 14

well, good ol easy method ..! still, there might be other methods. who knows?

Answer 15

I think it would be worth while to determine these vertices points in general terms
=> we would have our own formula

Answer 16

I told my friend to do it in this way as I did. But she was shocked. She couldn't believe that we need to solve the so many equations for a MC question. But she didn't know how to solve it :S

Answer 17

Hmm the equations aren't that hard to solve, yeah they are boring but you can still solve it.

Answer 18

I know they are not ...

Answer 19

Like you should solve the y co-ordinate first, it's much easier and will help you to eliminate some of the options.
\[y_1 -y_3 = 8 \text{ and } y_3 = 4-y_1\]
\[2y_1 = 12 \implies y_1 = 6 \implies (y_2,y_3) =(2,-2) \]

Answer 20

Now you only need the options with 2 as the y co-ordinate.

Answer 21

Hmm.. I know how to solve them, and I told my friend it was easy to solve

Answer 22

The question is that is there any other to do this question...

Answer 23

Solving one of the values of x_1,x_2 and x_3, will give you all the other options easily.

Answer 24

A way of looking at it that I see is to try finding the points that make a parallelogram between two midpoints and the last midpoint + a vertex.

You'd just have to do it three times for each vertex. :O