Math, asked by Anonymous, 9 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