CBSE BoardEnglish MediumSTD 10MathsCo-ordinate Geometry5 Marks
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.
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Answer
Let $(a, a^2), (b, b^2), (c, c^2)$ he 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.
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