Question
If $\text{a}\neq\text{b}\neq\text{c},$ prove that the points $(a, a^2), (b, b^2), (c, c^2)$ can never be collinear.
✓
Answer
Let $A(a, a^2), B(b, b^2), C(c, c^2)$ the given points.
Three points are collinear if area enclosed by three points is zero.
Area of $\triangle\text{ABC}=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
$=\frac{1}{2}|\text{a}(\text{b}^2-\text{c}^2)+\text{b}(\text{c}^2-\text{a}^2)+\text{c}(\text{a}^2-\text{b}^2)|$
$=\frac{1}{2}|\text{ab}^2-\text{ac}^2+\text{bc}^2-\text{a}^2\text{b}+\text{a}^2\text{c}-\text{b}^2\text{c}|$
$=\frac{1}{2}|(\text{a}^2\text{c}-\text{a}^2\text{b})+(\text{ab}^2-\text{ac}^2)+(\text{bc}^2-\text{b}^2\text{c})|$
$=\frac{1}{2}|(-\text{a}^2)(\text{b}-\text{c})+\text{a}(\text{b}^2-\text{c}^2)-\text{bc}(\text{b}-\text{c})|$
$=\frac{1}{2}|(\text{b}-\text{c})(-\text{a}^2+\text{a}(\text{b}+\text{c})-\text{bc})|$
$=\frac{1}{2}|(\text{b}-\text{c})(-\text{a}^2+\text{ab}+\text{ac}-\text{bc})|$
$=\frac{1}{2}|(\text{b}-\text{c})[(-\text{a})(\text{a}-\text{b})+\text{c}(\text{a}-\text{b})]|$
$=\frac{1}{2}|(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}-\text{b})|$
It is given that $\text{a}\neq\text{b}\neq\text{c}$ Hence area of triangle made by three points is never zero.
Hence given points are never collinear.
Three points are collinear if area enclosed by three points is zero.
Area of $\triangle\text{ABC}=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
$=\frac{1}{2}|\text{a}(\text{b}^2-\text{c}^2)+\text{b}(\text{c}^2-\text{a}^2)+\text{c}(\text{a}^2-\text{b}^2)|$
$=\frac{1}{2}|\text{ab}^2-\text{ac}^2+\text{bc}^2-\text{a}^2\text{b}+\text{a}^2\text{c}-\text{b}^2\text{c}|$
$=\frac{1}{2}|(\text{a}^2\text{c}-\text{a}^2\text{b})+(\text{ab}^2-\text{ac}^2)+(\text{bc}^2-\text{b}^2\text{c})|$
$=\frac{1}{2}|(-\text{a}^2)(\text{b}-\text{c})+\text{a}(\text{b}^2-\text{c}^2)-\text{bc}(\text{b}-\text{c})|$
$=\frac{1}{2}|(\text{b}-\text{c})(-\text{a}^2+\text{a}(\text{b}+\text{c})-\text{bc})|$
$=\frac{1}{2}|(\text{b}-\text{c})(-\text{a}^2+\text{ab}+\text{ac}-\text{bc})|$
$=\frac{1}{2}|(\text{b}-\text{c})[(-\text{a})(\text{a}-\text{b})+\text{c}(\text{a}-\text{b})]|$
$=\frac{1}{2}|(\text{b}-\text{c})(\text{c}-\text{a})(\text{a}-\text{b})|$
It is given that $\text{a}\neq\text{b}\neq\text{c}$ Hence area of triangle made by three points is never zero.
Hence given points are never collinear.
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