(C) Let the points be $A\left( 0, \frac{8}{3} \right)$,$B(1, 3)$,and $C(82, 30)$.Slope of $AB = \frac{3 - \frac{8}{3}}{1 - 0} = \frac{\frac{9-8}{3}}{1

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(C) Let the points be $A\left( 0, \frac{8}{3} \right)$,$B(1, 3)$,and $C(82, 30)$.Slope of $AB = \frac{3 - \frac{8}{3}}{1 - 0} = \frac{\frac{9-8}{3}}{1} = \frac{1}{3}$.Slope of $BC = \frac{30 - 3}{82 - 1} = \frac{27}{81} = \frac{1}{3}$.Since the slope of $AB$ is equal to the slope of $BC$ and they share a common point $B$,the points $A$,$B$,and $C$ are collinear,meaning they lie on a straight line.

Class 11 Mathematics Straight Line Concurrency of three lines

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The points $\left( 0, \frac{8}{3} \right)$,$(1, 3)$,and $(82, 30)$

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A
form an acute angled triangle.
B
form a right angled triangle.
C
lie on a straight line.
D
form an obtuse angled triangle.

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Straight Line

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The points $\left( 0, \frac{8}{3} \right)$,$(1, 3)$,and $(82, 30)$

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