Prove that the points (7,10), (-2,5) and (3,-4) are the vertices of an isosceles right triangle.
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Find the distance between ab, bc, ca by distance formula also in a isosceles triangle all sides are equal and collinear ab+bc=ac
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As it is isosceles right triangle So perpendicular and base should be equal
let's fight distance between any 2 points given
(7,10) and -2,5)
√10-5)^2 + ( 7+2)^2 = √25 + 81 = √106
distance between (-2,5)( 3,-4)
√(3+2)^2 + ( -4-5)^2 = √25 +81 = √106
distance between (7,10) and (3,-4)
√(7-3)^2 + ( 10 +4)^2 = √16 + 14^2 = √196 +16
= √212
As in right angle p^2 + B^2 = H^2
106+ 106 = 212
212 = 212
As perpendicular and base are equal to √106
So its isosceles right triangle
let's fight distance between any 2 points given
(7,10) and -2,5)
√10-5)^2 + ( 7+2)^2 = √25 + 81 = √106
distance between (-2,5)( 3,-4)
√(3+2)^2 + ( -4-5)^2 = √25 +81 = √106
distance between (7,10) and (3,-4)
√(7-3)^2 + ( 10 +4)^2 = √16 + 14^2 = √196 +16
= √212
As in right angle p^2 + B^2 = H^2
106+ 106 = 212
212 = 212
As perpendicular and base are equal to √106
So its isosceles right triangle
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