Geometry

[vampirewitchreine]

Can someone help push me in the right direction for this problem?


Can the three points be the vertices of a triangle? If not, explain why.


24. M(0,0), N(3,2), P(-231, 782)


(The large numbers are throwing me off cause I can't really graph them)

 

[tkhunny]

Subtract consecutive pairs and see where it leads.

N - M = (3,2)

P - N = (-234,780)

M - P = (231,782)

Calculate the three distances and see where it leads.

\(\displaystyle d(M,N) = \sqrt{3^{2} + 2^{2}} = \sqrt{13}\)

d(N,P) = ??

d(P,M) = ???

What say you? Are we getting anywhere?

 

[vampirewitchreine]

Here's what I got for the distances.

\(\displaystyle d(M,N)=\sqrt{3^{2}+2^{2}}=\sqrt{13}~\approx 3.61\)
\(\displaystyle d(N,P)=\sqrt{(3+-231)^{2}+(2+782)^{2}}=\sqrt{-228+784}\) \(\displaystyle =\sqrt{51984+614656}=\sqrt{666640}~\approx 816.48\)
\(\displaystyle d(P,M) = \sqrt{-231^{2} + 782^{2}} = \sqrt{53361+611524} = \sqrt{664885}~\approx 815.4\)


Should I just work the triangle inequality from here? Or should I do something else?

 

[tkhunny]

You must convince yourself that the three points are not colinear.

 

[soroban]

Hello, vampirewitchreine!

Can the three points be the vertices of a triangle?

. . \(\displaystyle M(0,0),\;N(3,2),\;P(-231, 782)\)

Compare the slopes of \(\displaystyle M\!N\) and \(\displaystyle M\!P.\)