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Show that the points $A(3;9), B(-2;-16)$ and $C(0.2;-5)$ lie on the same line.

We can say that three points lie on the same line if the largest segment bounded by two of these points is equal to the sum of the smaller ones. Can you show me why this is sufficient for three points to lie on the same line?

By the distance formula, we can get $AB=\sqrt{650}, BC=\sqrt{125.84}$ and $CA=\sqrt{203.84}$. How to check if $AB=BC+CA$?

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  • $\begingroup$ How about the triangle inequality? $\endgroup$ Commented May 7, 2020 at 10:22
  • $\begingroup$ @R.J.Etienne The problem asker is using the triangle inequality. $\endgroup$ Commented May 7, 2020 at 10:23
  • $\begingroup$ I was referring to : Can you show me why this is sufficient for three points to lie on the same line? $\endgroup$ Commented May 7, 2020 at 10:25

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We can use other ways to check, which is much easier than your method:

$\frac{9-(-16)}{3-(-2)}=\frac{25}5=5$

$\frac{(-16)-(-5)}{-2-0.2}=\frac{-11}{-2.2}=5$

As the two fractions are the same, so the slope of $AB$ is same as the slope of $BC$, and they have a common point $B$, so they lie on the same line.

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    $\begingroup$ It is important that B is in both segments, else the lines could simply be parallel $\endgroup$ Commented May 7, 2020 at 10:23
  • $\begingroup$ @R.J.Etienne Yes, but it is very easy to see that. $\endgroup$ Commented May 7, 2020 at 10:24
  • $\begingroup$ Just to be precise :) $\endgroup$ Commented May 7, 2020 at 10:25
  • $\begingroup$ @R.J.Etienne Thanks $\endgroup$ Commented May 7, 2020 at 10:26
  • $\begingroup$ Thank you for the response! I appreciate it! $\endgroup$ Commented May 7, 2020 at 16:57
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If the points were not colinear, they would form a triangle. Using the triangle inequality, the sum of two smaller sides would always be greater than the biggest side, but in our case they are equal.

Write the distances as fractions:

$AB=\sqrt{650} ,\ BC=\frac{11}{5}\sqrt{26}, \ CA=\frac{14}{5}\sqrt{26}$

Now, $$BC+CA=5\sqrt{26} = \sqrt{25\times 26} =\sqrt{650}=AB$$ We’re done.

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  • $\begingroup$ Thank you for the response! I appreciate it! $\endgroup$ Commented May 7, 2020 at 16:57
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It is sufficient because if three points lie are collinear then there is a middle point, say $M$ and there are two end points, say $E_1$ and $E_2$, then $$E_1E_2=E_1M+E_2M$$ It is same as saying that if your school lies midway between your home and police station, then the distance between your home and police station is the sum of the distances between your home and school and your school and police station.

For your second question, $$BC=\sqrt{125.84}\\ =0.1\times\sqrt{12584}\\ =0.1\times2\times11\times\sqrt{26}\\ =2.2\sqrt{26}$$ Similarly, $$CA=2.8\sqrt{26}$$ and $$AB=5\sqrt{26}$$

Also, there are easier ways to check that three points are collinear, for example note that three points are collinear iff the area of the triangle formed by them is zero, i.e. $$\begin{vmatrix} 3&9&1\\ -2&-16&1\\ 0.2&-5&1 \end{vmatrix}=0$$

Another easy method is that all of them satisfy the linear equation $5x-y=6$, hence they are collinear.

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  • $\begingroup$ Thank you for the response! I appreciate it! $\endgroup$ Commented May 7, 2020 at 16:57
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In mathematics, the triangle inequality states that for any triangle, the sum of the lengths of any two sides must be greater than or equal to the length of the remaining side.

source: https://en.wikipedia.org/wiki/Triangle_inequality

enter image description here

Thus, if $AB+BC=AC$, then the points must be on the same line.

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