Question

If $x_1,x_2,x_3$ and $y_1,y_2,y_3$ are both in G.P. with the same common ratio, then the points $\left(x_1,y_1\right)$ , $\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$

• A. Lie on a straight line
• B. Lie on an ellipse
• C. Lie on a circle
• D. Are vertices of a triangle

Answer 1

The correct option is A

Lie on a straight line

Explanation for the correct answer:

Geometric Progression:

Let $r$ be the common ratio of both the geometric progressions

Consider the terms $x_1,x_2,x_3$

As these terms are in geometric progression they can be written as

$x_2=x_1\times r$

$x_3=x_1\times r^2$

Consider the terms $y_1,y_2,y_3$

As these terms are in geometric progression they can be written as

$y_2=y_1\times r$

$y_3=y_1\times r^2$

Let $A=\left(x_1,y_1\right)$, $B=\left(x_2,y_2\right)=(r{x}_1,r{y}_1)$,$C=\left(x_3,y_3\right)=(r^2{x}_1,r^2{y}_1)$

The slope of the line $AB$ can be written as

slope of $AB=\frac{r{y}_1-y_1}{r{x}_1-x_1}$$=\frac{y_1\left(r-1\right)}{x_1\left(r-1\right)}=\frac{y_1}{x_1}$

The slope of the line $BC$ can be written as

slope of $BC=\frac{r^2{y}_1-r{y}_1}{r^2{x}_1-r{x}_1}=\frac{r{y}_1\left(r-1\right)}{r{x}_1\left(r-1\right)}=\frac{y_1}{x_1}$

Hence, the slope of line $AB=$slope of line $BC$

Therefore, points $A,B,C$ are collinear.

Hence, option (A) is the correct answer.