Question

The points \[\left( {0,\dfrac{8}{3}} \right),\left( {1,3} \right)\] and \[\left( {82,30} \right)\] are the vertices of,
\[\left( {\text{A}} \right)\]An obtuse angled triangle
\[\left( {\text{B}} \right)\] An acute angled triangle
\[\left( {\text{C}} \right)\] A right-angled triangle
\[\left( {\text{D}} \right)\]None of these

Answer 1

Hint:- Find slope of each line and check for collinearity.

As, three vertices of the triangle are given,
$ \Rightarrow $Let, ${\text{A}}\left( {0,\dfrac{8}{3}} \right),{\text{ }}B\left( {1,3} \right){\text{ }}$and $C\left( {82,30} \right)$be the vertices of a triangle.
And the triangle will be $\Delta {\text{ABC}}$
Let, ${m_1}$be the slope of side AB.
 $ \Rightarrow $So, ${m_1} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right){\text{ }}$where $({x_1},{y_1})$ and $({x_2},{y_2})$ are the points A and B.
$ \Rightarrow $So, ${m_1} = \left( {\dfrac{{3 - \dfrac{8}{3}}}{{1 - 0}}} \right) = \dfrac{1}{3}$
Let, ${m_2}$ be the slope of side BC.
$ \Rightarrow $So, ${m_2} = \left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)$ where $({x_1},{y_1})$ and $({x_2},{y_2})$ are the points B and C.
$ \Rightarrow $So, ${m_2} = \left( {\dfrac{{30 - 3}}{{82 - 1}}} \right) = \dfrac{{27}}{{81}} = \dfrac{1}{3}$
$ \Rightarrow $ As we have proved above that, ${m_1} = {m_2} = \dfrac{1}{3}$
And we know that if slopes of two lines are same and passes through same point (here B)
Then the lines are collinear.
So, therefore A, B and C are not the vertices of any triangle.
Because A, B and C lie on the same line. Hence, they are collinear.
Hence, the correct option will be D.

Note:- In such type of questions the easiest and efficient way to find the type of triangle
is by finding the slope of each line. So, first we had to find the slope of each line then we can
also find the length of each side by using distance formula and then we will easily get which type
triangle is given.