Question

The points $(0,\frac{8}{3}),(1,3)$ and $(82,30)$ are the vertices of:

• A. an equilateral triangle
• B. an isosceles triangle
• C. a right angled triangle
• D. none of these

Answer 1

Answer: (A)

an equilateral triangle

According to the problem :

$A{B}^2=(0-1{)}^2+(\frac{8}{3}-3{)}^2$

$=1+\frac{1}{9}=\frac{10}{9}=1.11$

Similarly,

$B{C}^2=(82-1{)}^2+(30-3{)}^2=7290$

and

$A{C}^2=(82-0{)}^2+(30-\frac{8}{3}{)}^2=7471.11$

Therefore,

$A{B}^2+B{C}^2<A{C}^2$

Hence the answer is acute-angled triangle.