If 3 points form a right angle triangle

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$\text{Determine if the points }A\left(-9,0\right)\text{ , }B\left(-24,-9\right)\text{ and }C \left(-15,10\right)\text{ form a right angle triangle.}$

**Answer**

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

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

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

$\text{The distance between the points }A\left(-9,0\right)\text{ and }B \left(-24,-9\right)$

$AB=\sqrt{(-9-0)^2+(-24-(-9))^2}$

$= \sqrt{(-9)^2+(-15)^2}$

$= \sqrt{81+225}$

$= \sqrt{306}$

$\text{So, }(AB)^2= 306$

$\text{The distance between the points }B \left(-24,-9\right)\text{ and }C \left(-15,10\right)$

$BC=\sqrt{(10-(-9))^2+(-15-(-24))^2}$

$= \sqrt{(19)^2+(9)^2}$

$= \sqrt{361+81}$

$= \sqrt{442}$

$\text{So, }(BC)^2= 442$

$\text{The distance between the points }C \left(-15,10\right)\text{ and }A \left(-9,0\right)$

$CA=\sqrt{(0-10)^2+(-9-(-15))^2}$

$= \sqrt{(-10)^2+(6)^2}$

$= \sqrt{100+36}$

$= \sqrt{136}$

$\text{So, }(CA)^2=136$

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