Question

If the vertices of a triangle be $(am_{1}^{2},2am_{1}),(am_{2}^{2},2am_{2})$ and $(am_{3}^{2},2am_{3}),$ then the area of the triangle is.

• A. $a(m_{2} - m_{3})(m_{3} - m_{1})(m_{1} - m_{2})$
• B. $(m_{2} - m_{3})(m_{3} - m_{1})(m_{1} - m_{2})$
• C. $a^{2}(m_{2} - m_{3})(m_{3} - m_{1})(m_{1} - m_{2})$
• D. None of these

Answer 1

Answer: (C)

$a^{2}(m_{2} - m_{3})(m_{3} - m_{1})(m_{1} - m_{2})$

Answer 2

Area

$= \frac { 1 } { 2 } \left| \begin{array} { l l l } a m _ { 1 } ^ { 2 } & 2 a m _ { 1 } & 1 \\ a m _ { 2 } ^ { 2 } & 2 a m _ { 2 } & 1 \\ a m _ { 3 } ^ { 2 } & 2 a m _ { 3 } & 1 \end{array} \right| = \frac { 1 } { 2 } a ^ { 2 } \times 2 \left| \begin{array} { l l l } m _ { 1 } ^ { 2 } & m _ { 1 } & 1 \\ m _ { 2 } ^ { 2 } & m _ { 2 } & 1 \\ m _ { 3 } ^ { 2 } & m _ { 3 } & 1 \end{array} \right|$

$= a ^ { 2 } \left| \begin{array} { c c c } m _ { 1 } ^ { 2 } - m _ { 2 } ^ { 2 } & m _ { 1 } - m _ { 2 } & 0 \\ m _ { 2 } ^ { 2 } - m _ { 3 } ^ { 2 } & m _ { 2 } - m _ { 3 } & 0 \\ m _ { 3 } ^ { 2 } & m _ { 3 } & 1 \end{array} \right|$ , by $R _ { 1 } \rightarrow R _ { 1 } - R _ { 2 }$ $R _ { 2 } \rightarrow R _ { 2 } - R _ { 3 }$

$= a ^ { 2 } \left( m _ { 2 } ^ { 2 } - m _ { 3 } ^ { 2 } \right) \left( m _ { 1 } - m _ { 2 } \right) - \left( m _ { 2 } - m _ { 3 } \right) \left( m _ { 1 } ^ { 2 } - m _ { 2 } ^ { 2 } \right)$ $= a ^ { 2 } \left( m _ { 1 } - m _ { 2 } \right) \left( m _ { 2 } - m _ { 3 } \right) \left( m _ { 3 } - m _ { 1 } \right)$.

Trick : Let $a = 2 , m _ { 1 } = 0 , m _ { 2 } = 1 , m _ { 3 } = 2$ then the

coordinates are (0, 0), (2, 4), (8, 8).

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