MCQ
If the points $(k, 2k), (3k, 3k)$ and $(3, 1)$ are collinear, then $k$ :
- A$\frac{1}{3}$
- ✓$\frac{-1}{3}$
- C$\frac{2}{3}$
- D$\frac{-2}{3}$
✓
Answer
Correct option: B.
$\frac{-1}{3}$
We have three collinear points $A(k, 2k), B(3k, 3k)$ and $C(3, 1).$
In general if $A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ are collinear then, area of the triangle is $0$.
$\frac{1}{2}\Big[\text{x}_1(\text{y}_2 - \text{y}_3) + \text{x}_2(\text{y}_3 - \text{y}_1) + \text{x}_3(\text{y}_1 - \text{y}_2)\Big] = 0$
So $, k(3k - 1) + 3k(1 - 2k) + 3(2k - 3k) = 0$
So $, -3k^2- k = 0$
Take out the common terms,
$- k(3k + 1) = 0$
Therefore,
$\text{k}=-\frac{1}{3}$
In general if $A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ are collinear then, area of the triangle is $0$.
$\frac{1}{2}\Big[\text{x}_1(\text{y}_2 - \text{y}_3) + \text{x}_2(\text{y}_3 - \text{y}_1) + \text{x}_3(\text{y}_1 - \text{y}_2)\Big] = 0$
So $, k(3k - 1) + 3k(1 - 2k) + 3(2k - 3k) = 0$
So $, -3k^2- k = 0$
Take out the common terms,
$- k(3k + 1) = 0$
Therefore,
$\text{k}=-\frac{1}{3}$
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