Answers

2015-03-14T07:18:26+05:30

This Is a Certified Answer

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Distance between AB=
 \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
 \sqrt{(-3-5)^2+(2+1)^2}
 \sqrt{(-8)^2+(3)^2}
 \sqrt{64+9}
 \sqrt{73} units

Distance between BC= 
 \sqrt{(1+3)^2+(6-2)^2}
 \sqrt{4^2+4^2}
 \sqrt{16+16}
 \sqrt{32}
 \sqrt{16X2}
4 \sqrt{2} units

Distance between CA=

 \sqrt{(5-1)^2+(-1-6)^2}
 \sqrt{4^2+(-7)^2}
 \sqrt{16+49}
 \sqrt{65} units

In an isosceles triangle, any two sides are equal. The distance of any two sides is not equal here. So, the given points do not form an isosceles triangle.
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Comment has been deleted
good
nice answer
2015-03-14T11:45:09+05:30

This Is a Certified Answer

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
A(5,-1)
b(-3,2)
c(1,6)
distance²=(x2-x)²+(y²-y)²
ab²=(5--3)²+(-1-2)²
ab²=64+9=73
ab=√73
bc²=(-3-1)²+(2-6)²
=16+16
bc=√32
ac²=(5-1)²+(-1-6)²=16+49
ac=√65

there aren't any same sides...so it is not a isosceles triangle
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