Question

Suppose that the three points \(A\), \(B\) and \(C\) in the plane are such that their \(x\)-coordinates as well as \(y\)-coordinates are in G.P. with the same common ratio. Then the points \(A\), \(B\) and \(C\)

• A. constitute a right angled triangle
• B. form an isosceles triangle
• C. lie on a straight line
• D. form an equilateral triangle

Hint:

If the slopes between consecutive pairs of points are equal, then all the points are collinear and lie on the same straight line.

Answer 1

The Correct Option is C

Step 1: Set up coordinates using the GP condition.
Let $A=(x_1,y_1)$, $B=(x_1r,y_1r)$, $C=(x_1r^2,y_1r^2)$ with common ratio $r\ne 1$.
Step 2: Compute slope of AB.
\[m_{AB}=\frac{y_1r-y_1}{x_1r-x_1}=\frac{y_1(r-1)}{x_1(r-1)}=\frac{y_1}{x_1}.\]
Step 3: Compute slope of BC.
\[m_{BC}=\frac{y_1r^2-y_1r}{x_1r^2-x_1r}=\frac{y_1r(r-1)}{x_1r(r-1)}=\frac{y_1}{x_1}.\]
Step 4: Compare slopes.
$m_{AB}=m_{BC}$ and the lines share point $B$, so $A$, $B$, $C$ are collinear.
Step 5: Verify with area formula.
Area of $\triangle ABC = \frac{x_1y_1}{2}|r-r^2+r^3-r+r^2-r^3|=0$. Zero area confirms collinearity.
Step 6: State the conclusion.
\[ \boxed{\text{lie on a straight line}} \]