Math, asked by Anonymous, 10 months ago

show that the points A(4,0) B(10,0)C(4,8) are the vertices of right angled triangle and also show that the midpoint of the hypotenuse is equidistant from the angular point

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Answered by srikanthn711

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Taking A(4, 0), B(10, 0) and C(4, 8)

AB² = (4-10)² + (0-0)²

= 36+0

=36

AC² = (4 -4)² + (0-8)²

= 0+64

=64

BC² = (10-4)² + (0-8)²

=64+36

=100

By Pythogoras theorem

Since AB² = AC² + BC², the points A(4, 0), B(10, 0) and C(4, 8) are vertices of a right angled triangle.

AB is the hypotenuse.

Mid-point of AB = [(10+4)/2 , (0+0)/2]

= (7, 0)

Let the mid-point be M (7, 0)

AM = √(4-5)² + (0)²

= √-1+0

= √-1

MB = √(7-10)² + (0-0)²

= √1+0

AM = MB = √1

This proves that the midpoint of the hypotenuse is equidistant from the angular points.

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Answered by Anonymous

Step-by-step explanation:

Step-by-step explanation:

x^2 - 4x - 5 = 0

x^2 -5x + x -5 = 0

x( x -5 ) + 1( x - 5) = 0

( x + 1)( x -5 ) = 0

x = - 1 and x = 5

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show that the points A(4,0) B(10,0)C(4,8) are the vertices of right angled triangle and also show that the midpoint of the hypotenuse is equidistant from the angular point​

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show that the points A(4,0) B(10,0)C(4,8) are the vertices of right angled triangle and also show that the midpoint of the hypotenuse is equidistant from the angular point