Show the triangle with vertices A (6, -7), B (11, -3), and C (2, 2) is a right t

Question

Show the triangle with vertices A (6, -7), B (11, -3), and C (2, 2) is a right triangle and then find its area. There are two ways to show triangle ABC is a right triangle: Use the fact that "slopes of perpendicular lines are negative reciprocals of each other" to show there is a right angle in triangle ABC, or use the converse of the Pythagorean Theorem which states If a and b are the lengths of the sides of a triangle, and c is the length of the hypotenuse, and a+bc ABC is a right triangle." then triangle

Explanation / Answer

distance from A to B =

sqrt [ ( -3 +7)^2 + ( 11-6)^2 ]

= sqrt 41

distance from B to C

sqrt [ ( -2 +3)^2 + ( 2-11)^2 ]

= sqrt 82

distance from C to A

sqrt [ ( -2 + 7 )^2 + ( 2 -6 )^2 ]

= sqrt 41

applying pythagorean theorem

AB^2 + + AC^2 = BC^2

41 + 41 = 82

left side = right side

hence , its a right angled triangle

area = 1/2 * sqrt 41 * sqrt 41

area = 20.5

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