Question

The vertices of a triangle are (8,-10) (7,-3) (0,-4) prove angle abc is a right angle

Answer 1

Step-by-step explanation:

If A(8,-10), B(7,-3), C(0-4)

AB= √(7−8)^2+(−3−−10)^2

    =√50

AC=√(0−8)^2+(−4−−10)^2

     =√100

BC=√(0−7)^2+(−4−−3)^2

    =√50

If its a right triangle

Pythagoras Theorm

AC^2 = BC^2 + AC^2

100 = 50 + 50

100 = 100

Hence its a right triangle

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Answer 2

Step-by-step explanation:

Let three points be A(8,-10),B(7,-3),C(0,-4).

AB=√(x2-x1)^2+(y2-y1)^2

=√(7-8)^2+(-3+10)^2

=√50

BC=√(0-7)^2+(-4+3)^2

√50

AC =√(0-8)^2+(-4+10)^2

=√100

If it is right angled triangle then using hypotenuse

AC^2=AB^2+BC^2