Let the area of the triangle with vertices A(1,), B(,0) and C(0,) be 4 sq.

Question

Let the area of the triangle with vertices $A(1,\alpha),\ B(\alpha,0)$ and $C(0,\alpha)$ be 4 sq. units. If the points $(\alpha,-\alpha),\ (-\alpha,\alpha)$ and $(\alpha^2,\beta)$ are collinear, then $\beta$ is equal to:

Accepted Answer

$$\because$$ A(1, $$\alpha$$), B($$\alpha$$, 0) and C(0, $$\alpha$$) are the vertices of $$\Delta$$ABC and area of $$\Delta$$ABC = 4

$$\therefore$$ $$\left| {{1 \over 2}\left| {\begin{pmatrix} 1 & \alpha  & 1  \
   \alpha  & 0 & 1  \
   0 & \alpha  & 1  \

\end{pmatrix} } \right|} \right| = 4$$

$$ \Rightarrow \left| {1(1 - \alpha ) - \alpha (\alpha ) + {\alpha ^2}} \right| = 8$$

$$ \Rightarrow \alpha  = \, \pm \,8$$

Now, $$(\alpha ,\, - \alpha ),\,( - \alpha ,\alpha )$$ and $$({\alpha ^2},\beta )$$ are collinear

$$\therefore$$ $$\left| {\begin{pmatrix} 8 & { - 8} & 1  \
   { - 8} & 8 & 1  \
   {64} & \beta  & 1  \

\end{pmatrix} } \right| = 0 = \left| {\begin{pmatrix} { - 8} & 8 & 1  \
   8 & { - 8} & 1  \
   {64} & \beta  & 1  \

\end{pmatrix} } \right|$$

$$ \Rightarrow 8(8 - \beta ) + 8( - 8 - 64) + 1( - 8\beta  - 8 \times 64) = 0$$

$$ \Rightarrow 8 - \beta  - 72 - \beta  - 64 = 0$$

$$ \Rightarrow \beta  =  - 64$$

Solutions · 1