Question

the points (0,8/3), (1,3) and (82,30) are of which triangle

Answer 1

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

Answer 2

Answer:

(0,8/3), (1,3) and (82,30) are co-linear

Step-by-step explanation:

Let A( 0,8/3), B(1,3) and C( 82,30)

AREA OF ΔABC

=1/2[ x₁(y₂ - y₃ + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

=1/2[0*(3 - 30) + 1*(30 - 8/3) + 82*(8/3 - 3)]

=1/2[0+ 30-8/3 +656/3-82*3)

=1/2[ 30-8/3 + 656/3 -246]

=1/(2*3)( 90-8+656-738)

=1/6(746-746)

=1/6*0

=0

Hence Area of ΔABC=0

As the area of triangle formed by these points is zero

hence these three points

(0,8/3), (1,3) and (82,30) are co-linear