If 3 points form a right angle triangle

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$\text{Determine if the points }A\left(-6,6\right)\text{ , }B\left(-11,2\right)\text{ and }C \left(-14,16\right)\text{ form a right angle triangle.}$

**Answer**

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

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

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

$\text{The distance between the points }A\left(-6,6\right)\text{ and }B \left(-11,2\right)$

$AB=\sqrt{(2-6)^2+(-11-(-6))^2}$

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

$= \sqrt{16+25}$

$= \sqrt{41}$

$\text{So, }(AB)^2= 41$

$\text{The distance between the points }B \left(-11,2\right)\text{ and }C \left(-14,16\right)$

$BC=\sqrt{(16-2)^2+(-14-(-11))^2}$

$= \sqrt{(14)^2+(-3)^2}$

$= \sqrt{196+9}$

$= \sqrt{205}$

$\text{So, }(BC)^2= 205$

$\text{The distance between the points }C \left(-14,16\right)\text{ and }A \left(-6,6\right)$

$CA=\sqrt{(6-16)^2+(-6-(-14))^2}$

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

$= \sqrt{100+64}$

$= \sqrt{164}$

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

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