Determine whether the points $A$, $B,$ and $C$ are vertices of a right triangle, an isosceles triangle, or both.$A(8,1),B(-3,-1),C(10,5)$

Key Concepts: Right Triangle, Isosceles Triangle, Pythagorean Theorem, Distance Formula Explanation: To determine if $△ABC$ is a right triangle, we need to check if it satisfies the Pythagorean Theorem, which states that if $a$, $b$, and $c$ are the lengths of the sides of a right triangle where $c$ is the hypotenuse, then $a^2+b^2=c^2$. To determine if $△ABC$ is isosceles, we need to check if it has two equal side lengths. Step by Step Solution: Step 1. We first need to find the lengths of the sides of $△ABC$. We can use the distance formula to find the distance between two points. Let $AB$, $BC$, and $AC$ be the sides of $△ABC$. Step 2. Distance between $A$ and $B$: \n$AB=\sqrt{(8-(-3){)}^2+(1-(-1){)}^2}=\sqrt{11^2+2^2}=\sqrt{121+4}=\sqrt{125}$ Step 3. Distance between $B$ and $C$: \n$BC=\sqrt{(10-(-3){)}^2+(5-(-1){)}^2}=\sqrt{13^2+6^2}=\sqrt{169+36}=\sqrt{205}$ Step 4. Distance between $A$ and $C$: \n$AC=\sqrt{(10-8{)}^2+(5-1{)}^2}=\sqrt{2^2+4^2}=\sqrt{20}$ Step 5. Using the Pythagorean Theorem, we see that $A{B}^2+A{C}^2=(\sqrt{125}{)}^2+(\sqrt{20}{)}^2=125+20=145$. While $B{C}^2=(\sqrt{205}{)}^2=205$. Step 6. From step 5, we can see that $A{B}^2+A{C}^2\ne B{C}^2$, so $△ABC$ is not a right triangle. Step 7. We also note that $AB\ne BC$ and $AB\ne AC$ from step 2 and step 4. Thus, $△ABC$ is not an isosceles triangle. Step 8. Therefore, $△ABC$ is neither a right triangle nor an isosceles triangle. Final Answer: $△ABC$ is neither a right triangle nor an isosceles triangle.