CUET · MATHS · PYQ PAPER 2023
Area of the triangle ABC with vertices \(\left(a, a^2\right),\left(b, b^2\right)\) and \(\left(c, c^2\right)\) is equal to :
- A \(\frac{1}{2}|a b c(a-b)(b-c)(c-a)|\)
- B \(\frac{1}{2}\left|\frac{1}{a b c}(a-b)(b-c)(c-a)\right|\)
- C \(\frac{1}{2}|(a-b)(b-c)(c-a)|\)
- D \(\frac{1}{2}[(a-b)(b-c)(c-a)]^2\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}|(a-b)(b-c)(c-a)|\)
Step-by-step Solution
Detailed explanation
Area \( = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \) Area \( = \frac{1}{2} \left| \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} \right| \) Area \( = \frac{1}{2} |(a-b)(b-c)(c-a)| \)
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