Question
If $a \neq b \neq 0$, prove that the points $\left(a, a^2\right),\left(b, b^2\right)$, $(0,0)$ will not be collinear.
✓
Answer
We know that three points are collinear if the area of the triangle formed by them is zero.
The area of triangle can be calculated using the formula
$A=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
Here, $x_1=a, y_1=a^2, x_2=b, y_2=b^2, x_3=0, y_3=0$
$A=\frac{1}{2}\left|a\left(b^2-0\right)+b\left(0-a^2\right)+0\left(a^2-b^2\right)\right|$
$A=\frac{1}{2}\left|a b^2-a^2 b\right|$
According to the question $a \neq b \neq 0$
$\therefore A=\frac{1}{2}\left|a b^2-a^2 b\right| \neq 0$
Hence, the points $\left(a, a^2\right),\left(b, b^2\right),(0,0)$ are non collinear.
Hence proved.
The area of triangle can be calculated using the formula
$A=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
Here, $x_1=a, y_1=a^2, x_2=b, y_2=b^2, x_3=0, y_3=0$
$A=\frac{1}{2}\left|a\left(b^2-0\right)+b\left(0-a^2\right)+0\left(a^2-b^2\right)\right|$
$A=\frac{1}{2}\left|a b^2-a^2 b\right|$
According to the question $a \neq b \neq 0$
$\therefore A=\frac{1}{2}\left|a b^2-a^2 b\right| \neq 0$
Hence, the points $\left(a, a^2\right),\left(b, b^2\right),(0,0)$ are non collinear.
Hence proved.
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