derive a relationship between a and b where P (12,8), Q (a,b) and R (6,0) are the vertices of right angled triangle with angle R= right angle
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Samaritan
Taking A(12, 8), B(-2, 6) and C(6, 0)
AB² = (12--2)² + (8-6)²
= 196+4
=200
AC² = (12-6)² + (8-0)²
= 36+64
=100
BC² = (-2-6)² + (6-0)²
=64+36
=100
By Pythogoras theorem
Since AB² = AC² + BC², the points (12, 8), (-2, 6) and (6, 0) are vertices of a right angled triangle.
AB is the hypotenuse.
Mid-point of AB = [(12+-2)/2 , (8+6)/2]
= (5, 7)
Let the mid-point be M (5, 7)
AM = √(12-5)² + (8-7)²
= √49+1
= √50
= 5√2
MB = √(5- -2)² + (7-6)²
= √49+1
= 5√2
AM = MB = 5√2
This proves that the midpoint of the hypotenuse is equidistant from the angular points.
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