Question

If x1, x2, x3 as well as y1, y2, y3 are in GP with the same common ratio,

Answer 1

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Answer 2

All three point are co-linear.

Step-by-step explanation:

1. Let coordinate of point (A) $=(x_{1},y_{1})$

        coordinate of point (B) $=(x_{2},y_{2})$

        coordinate of point (C) $=(x_{3},y_{3})$

2. Given that $x_{1},x_{2},x_{3}$ and $y_{1},y_{2},y_{3}$ are in GP with same common ratio.

$\frac{x_{2}}{x_{1}}=\frac{x_{3}}{x_{2}}=\frac{y_{2}}{y_{1}}=\frac{y_{3}}{y_{2}}=k$       ...1)

 

3.  So with help of equation 1)

 $x_{2}=kx_{1}$       ...2)

 $x_{3}=kx_{2}=k\times kx_{1}=k^{2}x_{1}$      ...3)

 $y_{2}=ky_{1}$       ...4)

 $y_{3}=ky_{2}=k\times ky_{1}=k^{2}y_{1}$      ...5)

4. Now slope($m_{1}$) of AB

   $m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{ky_{1}-y_{1}}{kx_{1}-x_{1}}=\frac{(k-1)y_{1}}{(k-1)x_{1}}=\frac{y_{1}}{x_{1}}$      ...6)

5. Now slope($m_{2}$) of AC

   $m_{1}=\frac{y_{3}-y_{1}}{x_{3}-x_{1}}=\frac{k^{2}y_{1}-y_{1}}{k^{2}x_{1}-x_{1}}=\frac{(k^{2}-1)y_{1}}{(k^{2}-1)x_{1}}=\frac{y_{1}}{x_{1}}$   ...7)

6. Now we see that slope of AB and AC are equal. So ABC are co-linear point, means A,B and C are point are on straight line.