Prove that the points (2, −2), (−2, 1) and (5, 2) are the vertices of a right angled triangle.
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A ( 2, -2 )
B (-2 , 1 )
C ( 5, 2 )
distance between AB by using distance formula
AB = 5
similarly
BC = 5 underoot 2
CA = 5
using pythagores theorem
BC^2 = AB^2 + AC^2
hence proved
also see the attachment
i hope its help ☆☆ :)
B (-2 , 1 )
C ( 5, 2 )
distance between AB by using distance formula
AB = 5
similarly
BC = 5 underoot 2
CA = 5
using pythagores theorem
BC^2 = AB^2 + AC^2
hence proved
also see the attachment
i hope its help ☆☆ :)
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Answered by
1
Answer:
Step-by-step explanation:
A(2,-2) B(-2,1) C(5,2)
AB^2= (X1-X2)^2+(Y1-Y2)^2
= (2-(-2))^2+(-2-1)^2
= 4^2+(-3)^2
=16+9
=25
BC^2=(X1-X2)^2+(Y1-Y2)^2
=(-2-5)^2+(1-2)^2
=49+1
=50
AC^2=(X1-X2)^2+(Y1-Y2)^2
=(2-5)^2+(-2-2)^2
=9+16
=25
Now we can tell that
AB^2+AC^2=BC^2
By converse Pythagorean theorem
Triangle ABC is a right angled triangle with angle A 90
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