The points (0, 8/3), (1,3) and (82, 30) are vertices of(a) isoceles triangle(b) an acute angled triangle(c) right angled triangle(d) None of theseExplain pls.

Dear Aditi,

square of one side ,a^2 = { (0-1)^2 + (8/3 -3)^2  } = 10/9 =1.11

b^2 = { (82-1)^2 + (30 -3)^2  } = 7290

 c^2 =  { (82-0)^2 + (30 -8/3)^2  } =7471.11

it shows that , c^2 > a^2 + b^2

so the tiangle would be acute

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Find area of triangle which comes out to be zero. Hence the points are collinear. So option D is correct