Question

The points of $\left( {0,\dfrac{8}{3}} \right), \left( {1,3} \right){\text{ and }}\left( {82,30} \right)$ are the vertices of:
$(a)$ Obtuse angled triangle
$(b)$ Right angled triangle
$(c)$ Isosceles triangle
$(d)$ None of these

Answer 1

Hint – In this question consider the given points as A($x_1$, $y_1$), B($x_2$, $y_2$) and C($x_3$, $y_3$). Find the slopes of the line AB, BC using the concept that $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$, where ($x_1$, $y_1$), ($x_2$, $y_2$) are the coordinates of the points. Use these slopes to find the correct option.

Complete step-by-step answer:
Given points $\left( {0,\dfrac{8}{3}} \right),\left( {1,3} \right){\text{ and }}\left( {82,30} \right)$

Let A = ($x_1$, $y_1$) = $\left( {0,\dfrac{8}{3}} \right)$
B = ($x_2$, $y_2$) = $\left( {1,3} \right)$
C = ($x_3$, $y_3$) = $\left( {82,30} \right)$
As we know slope (m) of two points ($x_1$, $y_1$) and ($x_2$, $y_2$) is calculated as
$ \Rightarrow m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$
So the slope ($m_1$) of line AB is
$ \Rightarrow {m_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{3 - \dfrac{8}{3}}}{{1 - 0}} = \dfrac{{\dfrac{{9 - 8}}{3}}}{1} = \dfrac{1}{3}$
Now the slope ($m_2$) of line BC is
$ \Rightarrow {m_2} = \dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}} = \dfrac{{30 - 3}}{{82 - 1}} = \dfrac{{27}}{{81}} = \dfrac{1}{3}$
So as we see that the slope of line AB is equal to the slope of line BC.
So this is the condition of collinear points.
So the given points $\left( {0,\dfrac{8}{3}} \right),\left( {1,3} \right){\text{ and }}\left( {82,30} \right)$ are collinear.
Hence option (D) none of these is correct.

Note – An obtuse triangle is a triangle with one obtuse angle that is angle is greater than ${90^\circ}$ and two acute angles, whereas in acute triangle all the three angles are acute angles (that is angles are less than ${90^\circ}$). Isosceles triangle is one in which two sides are equal out of the three sides.