Prove that points (3,0) , (6,4) and (-1,3) are vertices of a right angled isosceles triangle.
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rules of an isoceles triangle:
two sides are equal
Given,
A= 3,0
B= 6,4
C= -1,3
distance formula = √ (X2 - X1)^2 + (Y2- Y1)^2
Therefore
AB = √ (6-3)^2 + 4^2
= √ 9+ 16
= √25
= 5
BC = √(-1-6)^2 +(3-4)^2
= √ 49+1
= 5√2
CA = √(3+1)^2 +3^2
= √ 16 +9
= 5
therefore. : AB = CA..
since two sides are equal in length. It is an isosceles triangle
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Answer:
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Step-by-step explanation:
AB and BC are the isosceles sides.
By distance formula, AB=(−1−3)2+(3−0)2=16+9=5
BC=(3−6)2+(0−4)2=9+16=5
Since AB=BC, the ΔABC is isosceles.
For proving ∠ABC is right, we use the relation m1×m2=−1
where m1 is slope of line AB and m2 of BC :
m1=−1−43−0=−43 m2=6−34−0=34
m1×m2=4−3×34=−1
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