Show that the following sets of points are the
vertices of right isosceles
triangle
(i)(0, -4), (3/2,-1) and (3,0).
Answers
Answered by
1
Answer:
Distance between 2 sides must be equal.
(I) √[(3/2)²+(3)²]=√[9/4+9]=√(45/4) units
(II)√[(3)²+(4)²]=√[9+16]=5 units
(III) √[(3/2–3)²+(1)²]=√[13/4] units
As, none of the pair of sides are equal, sp the given vertices are not of isosceles ∆
Answered by
1
Answer:
AB= √(x^2-x^1)^2 +√(y^2-y^1)^2
√(3/2-0)^2 +√(-1-(-4))^2
√9/4+9
√18
BC=√(3-3/2)^2+(0--1)^2
√3/2^2+1^2
√9/4+1
=9
AC=√(0-3)^2+(-4-0)^2
√(-3)^2+(-4)^2
√9+16
√25
=5
Step-by-step explanation:
since none of the them are equal .
it is not an isoceles triangle
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