If 3 points form a right angle triangle

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$\text{Determine if the points }A\left(-10,3\right)\text{ , }B\left(-18,13\right)\text{ and }C \left(-20,-5\right)\text{ form a right angle triangle.}$

**Answer**

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

$\text{is the sum of the squares of the other two distances, the 3 points will form a right}$

$\text{angle triangle, otherwise not.}$

$\text{The distance between the points }A\left(-10,3\right)\text{ and }B \left(-18,13\right)$

$AB=\sqrt{(13-3)^2+(-18-(-10))^2}$

$= \sqrt{(10)^2+(-8)^2}$

$= \sqrt{100+64}$

$= \sqrt{164}$

$\text{So, }(AB)^2= 164$

$\text{The distance between the points }B \left(-18,13\right)\text{ and }C \left(-20,-5\right)$

$BC=\sqrt{(-5-13)^2+(-20-(-18))^2}$

$= \sqrt{(-18)^2+(-2)^2}$

$= \sqrt{324+4}$

$= \sqrt{328}$

$\text{So, }(BC)^2= 328$

$\text{The distance between the points }C \left(-20,-5\right)\text{ and }A \left(-10,3\right)$

$CA=\sqrt{(3-(-5))^2+(-10-(-20))^2}$

$= \sqrt{(8)^2+(10)^2}$

$= \sqrt{64+100}$

$= \sqrt{164}$

$\text{So, }(CA)^2=164$

$\text{Since, }(BC)^2 = (AB)^2 + (CA)^2 \text{ , the 3 points form a right angle triangle.}$