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Question
With out Pythagoras theorem, show that A(4, 4), B(3, 5) and C(-1, -1) are the vertices of a right angled.
Solution
Slope of BC = m1 = `(-1 - 5)/(-1 - 3) = (3)/(2)`
Slope of CA = m2 = `(4 - (-1))/(4 - (-1)` = 1
Also, slope of AB = m3 - `(5 - 4)/(3 - 4)` = -1
Since, m2m3 = 1 x (-1) = -1, So, AB and CA are perpendicular to each.
Thus, ΔABC is a right angled triangle at A.
Hence proved.
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