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Co-ordinate Geometry — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsCo-ordinate Geometry4 Marks
Question
Find the area of a triangle whose vertices are,
$(\text{at}_1^2, 2\text{at}_1), (\text{at}_2^2, 2\text{at}_2)$ and $(\text{at}_3^2, 2\text{at}_3).$
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Answer

Co-ordinates of $\triangle\text{ABC}$ are $\text{A}(\text{at}_1^2, 2\text{at}_1), \text{B}(\text{at}_2^2, 2\text{at}_2)$ and $\text{C}(\text{at}_3^2, 2\text{at}_3)$
$\therefore$ Area of $\triangle=\frac{1}{2}|\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)|$
$=\frac{1}{2}|\text{at}_1^2(2\text{at}_2-2\text{at}_3)+\text{at}_2^2(2\text{at}_3-2\text{at}_1)+\text{at}_3^2(2\text{at}_1-2\text{at}_2)|$
$=\frac{1}{2}|2\text{a}^2\text{t}_1^2\text{t}_2-2\text{a}^2\text{t}_1^2\text{t}_3+2\text{a}^2\text{t}_2^2\text{t}_3-2\text{a}^2\text{t}_2^2\text{t}_1+2\text{a}^2\text{t}_3^2\text{t}_1-2\text{a}^2\text{t}_3^2\text{t}_2|$
$=\frac{1}{2}\times2\text{a}^2|\text{t}_1^2\text{t}_2-\text{t}_1^2\text{t}_3+\text{t}_2^2\text{t}_3-\text{t}_2^2\text{t}_1+\text{t}_3^2\text{t}_1-\text{t}_3^2\text{t}_2|$
$= \text{a}^2|\text{t}_1^2\text{t}_2 - \text{t}_1^2\text{t}_3 + \text{t}_2^2\text{t}_3 - \text{t}_3^2\text{t}_2 -\text{t}_2^2\text{t}_1 + \text{t}_3^2\text{t}_1|$
$= \text{a}^2|\text{t}_1^2(\text{t}_2 - \text{t}_3) + \text{t}_2\cdot\text{t}_3(\text{t}_2 - \text{t}_3) - \text{t}_1(\text{t}_2^2 -\text{ t}_3^2)|$
$= \text{a}^2|\text{t}_1^2(\text{t}_2 - \text{t}_3) + \text{t}_2\text{t}_3(\text{t}_2 - \text{t}_3) - \text{t}_1(\text{t}_2 +\text{ t}_3)(\text{t}_2-\text{t}_3)|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)[\text{t}_1^2 +\text{t}_2\text{t}_3 - \text{t}_1\text{t}_2 - \text{t}_1\text{t}_3]|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)[\text{t}_1^2 -\text{t}_1\text{t}_2 - \text{t}_1\text{t}_3 +\text{t}_2\text{t}_3]|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)[\text{t}_1(\text{t}_1 - \text{t}_2) - \text{t}_3(\text{t}_1 - \text{t}_2)]|$
$= \text{a}^2|(\text{t}_2 - \text{t}_3)(\text{t}_1 - \text{t}_2)(\text{t}_1 - \text{t}_3)|$
$= \text{a}^2(\text{t}_1 - \text{t}_2)(\text{t}_2 - \text{t}_3)(\text{t}_3 - \text{t}_1)$

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