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# The converse of the Pythagorean theorem is also a true statement: If the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, then the triangle is a right triangle. Use the distance formula and the Pythagorean theorem to determine whether the set of points could be vertices of a right triangle. (-3,1), (2,-1), and (6,9) # The converse of the Pythagorean theorem is also a true statement: If the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, then the triangle is a right triangle. Use the distance formula and the Pythagorean theorem to determine whether the set of points could be vertices of a right triangle. (-3,1), (2,-1), and (6,9)

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Right triangles and trigonometry asked 2021-02-02
The converse of the Pythagorean theorem is also a true statement: If the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, then the triangle is a right triangle. Use the distance formula and the Pythagorean theorem to determine whether the set of points could be vertices of a right triangle.
(-3,1), (2,-1), and (6,9)

## Answers (1) 2021-02-03
We know that the square of distance between two points $$\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}{\quad\text{and}\quad}{\left({x}_{{2}},{y}_{{2}}\right)}{i}{s}{\left({x}_{{2}}-{x}_{{1}}\right)}^{{2}}+{\left(_{y}{2}-{y}_{{1}}\right)}^{{2}}$$ and the converse of pythagorus theorem , we have if sum of square of any two side is equal to square of third side of triangle, then it is right angled triangle.
Using distance formula and converse of pythagorus theorem , we find that ABC is right angled triangle right angled at A
A(-3,1), B(2,-1), C(6,9)
$$\displaystyle{c}^{{2}}={\left|{A}{B}\right|}^{{2}}={\left({2}+{3}\right)}^{{2}}+{\left(-{1}-{1}\right)}^{{2}}={25}+{4}={29}$$
$$\displaystyle{a}^{{2}}={\left|{B}{C}\right|}^{{2}}={\left({6}-{2}\right)}^{{2}}+{\left({9}+{1}\right)}^{{2}}={16}+{100}={116}$$
$$\displaystyle{b}^{{2}}={\left|{C}{A}\right|}^{{2}}={\left({6}+{3}\right)}^{{2}}+{\left({9}-{1}\right)}^{{2}}={81}+{64}={145}$$
$$\displaystyle{a}^{{2}}\ne{b}^{{2}}+{c}^{{2}}$$
145=116+29

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