1. Show that the points (12, 8), (-2, 6) and (6, 0) are the vertices o

1.	1. Show that the points (12, 8), (-2, 6) and (6, 0) are the vertices of a right angled triangle.Also, show that the mid-point of the hypotenuse is equidistant from the angular points
Answer» 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 angledtriangle. 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 thatthe midpoint of the hypotenuse is equidistant from the angular points. thanks dear

1. Show that the points (12, 8), (-2, 6) and (6, 0) are the vertices of a right angled triangle.Also, show that the mid-point of the hypotenuse is equidistant from the angular points

Answer»

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 angledtriangle.

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 thatthe midpoint of the hypotenuse is equidistant from the angular points.

thanks dear