Question
If $\text{a}\neq\text{b}\neq0,$ prove that the points $(a, a^2), (b, b^2), (0, 0)$ will not be collinear.
✓
Answer
Let $A(a, a^2), B(b, b^2)$ and $C(0, 0)$ be the coordinates of the given points.
We know that the area of triangle having vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is $\Big|\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]\Big|$ square units.
So, Area of $\triangle\text{ABC}=\Big|\frac{1}{2}[\text{a}(\text{b}^2-0)+\text{b}(0-\text{a}^2)+0(\text{a}^2-\text{b}^2)]\Big|$
$=\Big|\frac{1}{2}(\text{ab}^2-\text{a}^2\text{b})\Big|$
$=\frac{1}{2}|\text{ab}(\text{b}-\text{a})|$
$\neq0\ \ (\because\ \text{a}\neq\text{b}\neq0)$
Since the area of the triangle formed by the points $(a, a^2), (b, b^2)$ and $(0, 0)$ is not zero, so the given points are not collinear.
We know that the area of triangle having vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is $\Big|\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]\Big|$ square units.
So, Area of $\triangle\text{ABC}=\Big|\frac{1}{2}[\text{a}(\text{b}^2-0)+\text{b}(0-\text{a}^2)+0(\text{a}^2-\text{b}^2)]\Big|$
$=\Big|\frac{1}{2}(\text{ab}^2-\text{a}^2\text{b})\Big|$
$=\frac{1}{2}|\text{ab}(\text{b}-\text{a})|$
$\neq0\ \ (\because\ \text{a}\neq\text{b}\neq0)$
Since the area of the triangle formed by the points $(a, a^2), (b, b^2)$ and $(0, 0)$ is not zero, so the given points are not collinear.
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