Question

If $x_{1}, \,x_{2},\, x_{3}$ as well as $y_{1},\, y_{2},\, y_{3}$ are in geometric progression with the same common ratio, then the points $\left(x_{1}, y_{1}\right),\,\left(x_{2}, y_{2}\right),\,\left(x_{3}, y_{3}\right)$ are

• A. vertices of an equilateral triangle
• B. vertices of a right angled triangle
• C. vertices of a right angled isosceles triangle
• D. collinear

Answer 1

Answer: (D)

We have $x_{1},\, x_{2},\, x_{3}$ and $y_{1},\, y_{2},\, y_{3}$ are in $GP$ with the same common ratio.
Let $r$ be the common ratio.
$\therefore x_{1}=x, x_{2}=x r \text { and } x_{3}=x r^{2}$
Similarly, $y_{1}=y$
$y_{2}=y r$ and $y_{3}=y r^{2}$
$\therefore $ Area of $\Delta=\frac{1}{2}\begin{vmatrix}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1\end{vmatrix}$
$=\frac{1}{2}\begin{vmatrix}x & y & 1 \\ x r & y r & 1 \\ x r^{2} & y r^{2} & 1\end{vmatrix}$
$=\frac{1}{2} x y\begin{vmatrix}1 & 1 & 1 \\ r & r & 1 \\ r^{2} & r^{2} & 1\end{vmatrix}=\frac{1}{2} \times 0=0$
$\left[\because C_{1}, C_{2}\right.$ are identical $]$
$\therefore $ The given points are collinear.