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Straight Lines and Pair of Straight Lines question

2003 · Shift 0 · Q111

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Straight Lines and Pair of Straight Lines
2003 · Shift 0 · Q111

Straight Lines and Pair of Straight Lines question

2003 · Shift 0 · Q111

JEE MainMathematicsStraight Lines and Pair of Straight LinesMCQ+4 / −1

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
are vertices of a triangle
B
lie on a straight line
C
lie on an ellipse
D
lie on a circle

View written solutionFree

Correct answer: B

Let the common ratio of both G.P.s be $r$ .

Since $x_1,x_2,x_3$ are in G.P., we can write
$\qquad x_3=x_1r^2.$$ Similarly, since $y_1,y_2,y_3$ are also in G.P. with the same common ratio, $$y_2=y_1r, \qquad y_3=y_1r^2.$$$
So the three given points are $P_1=(x_1,y_1),\quad P_2=(x_1r,y_1r),\quad P_3=(x_1r^2,y_1r^2).$
Observe that each point is a scalar multiple of $(x_1,y_1)$ : $P_2=r(x_1,y_1),\qquad P_3=r^2(x_1,y_1).$

Hence all three points lie on the same line through the origin and $(x_1,y_1)$ .
We can also verify by slope: $\text{slope of }P_1P_2=\frac{y_1r-y_1}{x_1r-x_1}=\frac{y_1(r-1)}{x_1(r-1)}=\frac{y_1}{x_1},$ and $\text{slope of }P_2P_3=\frac{y_1r^2-y_1r}{x_1r^2-x_1r}=\frac{y_1r(r-1)}{x_1r(r-1)}=\frac{y_1}{x_1}.$

Since the slopes are equal, the three points are collinear.
Therefore, the points lie on a straight line.

Option check

A: False, because three collinear points do not form a triangle.
B: True.
C: Not the intended necessary result.
D: Not the intended necessary result.

Thus the correct answer is B.

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