If 3 points are colinear

You can view as many worked out examples as you want. First you are shown the question. Click Show Answer to view the answer.
Click Show another Example to view another example. The best way to master mathematics is by practice. But practice requires time.
If you don't have the time to practice, you can always view many practice problems completely worked out and become good at it too.

$\text{Determine if the points }A\left(4,-14\right)\text{ , }B\left(-6,26\right)\text{ and }C\left(5,-18\right)\text{ are colinear.}$

**Answer**

$\text{We will find the distance between each pair of points. If one distance is the sum}$

$\text{of the other two distances, the 3 points are colinear otherwise not colinear.}$

$\text{The distance between the points }A\left(4,-14\right)\text{ and }B\left(-6,26\right)$

$AB=\sqrt{(26-(-14))^2+(-6-4)^2}$

$= \sqrt{(40)^2+(-10)^2}$

$= \sqrt{1600+100}$

$= \sqrt{1700}$

$= 41.23$

$\text{The distance between the points }B\left(-6,26\right)\text{ and }C\left(5,-18\right)$

$BC=\sqrt{(-18-26)^2+(5-(-6))^2}$

$= \sqrt{(-44)^2+(11)^2}$

$= \sqrt{1936+121}$

$= \sqrt{2057}$

$= 45.35$

$\text{The distance between the points }C\left(5,-18\right)\text{ and }A\left(4,-14\right)$

$CA=\sqrt{(-14-(-18))^2+(4-5)^2}$

$= \sqrt{(4)^2+(-1)^2}$

$= \sqrt{16+1}$

$= \sqrt{17}$

$= 4.12$

$\text{Since, }BC = AB + CA \text{, the 3 points are colinear.}$