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