Question

If the vertices of a triangle are $(0,0),(m,0)$ and $(\frac{m}{1+m},\frac{m^2}{1+m})$ where $m$ is a root of the equation $m^3-6m^2+11m-6=0$, then the area of the the triangle can be

• A. $\frac{1}{4}$
• B. $\frac{4}{3}$
• C. $\frac{27}{8}$
• D. $\frac{32}{5}$

Answer 1

Answer: (A), (B), (C)

$\frac{1}{4}$
$\frac{4}{3}$
$\frac{27}{8}$

$m^3-6m^2+11m-6=0$
Implies
$m=1,2,3$
Hence the vertices can be
$(0,0),(1,0),\left(\frac{1}{2},\frac{1}{2}\right)$.
This triangle is isosceles with base $=1$ unit and Height $=0.5$ units.
Hence $Area=\frac{bh}{2}=\frac{1}{4}sq.units$
Or
The vertices can be
$(0,0),(2,0),(\frac{2}{3},\frac{4}{3})$
Using determinant method, we get the area as
$\frac{4}{3}sq.units$.
Or
The vertices could be
$(0,0),(3,0),(\frac{3}{4},\frac{9}{4})$
Again using determinant method, we get the area of the triangle as
$=\frac{27}{8}$sq. units.