If the vertices of a triangle be $(a m_{1}^{2}, 2 a m_{1}), (a m_{2}^{2}, 2 a m_{2}) a n d (a m_{3}^{2}, 2 a m_{3})$ , then the area of the triangle is

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

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

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

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Solution

The correct option is C
$a^{2} (m_{2} - m_{3}) (m_{3} - m_{1}) (m_{1} - m_{2})$

Area = $\frac{1}{2}$ $∣ ∣ ∣ ∣ ∣ \begin{matrix} a m_{1}^{2} & 2 a m_{1} & 1 a m_{2}^{2} & 2 a m_{2} & 1 a m_{3}^{2} & 2 a m_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ = $\frac{1}{2}$ $a^{2}$ × 2 $∣ ∣ ∣ ∣ ∣ \begin{matrix} m_{1}^{2} & m_{1} & 1 m_{2}^{2} & m_{2} & 1 m_{3}^{2} & m_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

= $a^{2}$ $∣ ∣ ∣ ∣ ∣ \begin{matrix} m_{1}^{2} - m_{2}^{2} & m_{1} - m_{2} & 0 m_{2}^{2} - m_{3}^{2} & m_{2} - m_{3} & 0 m_{3}^{2} & m_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ , by $R_{1}$ $\to$ $R_{1} - R_{2}$

$R_{2}$ $\to$ $R_{2} - R_{3}$

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

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

Trick: Let a = 2, $m_{1}$ = 0, $m_{2}$ = 1, $m_{3}$ = 2, then the coordinates are (0, 0), (2, 4), (8, 8).

$∴$ $△$ = $\frac{1}{2}$ $∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 2 & 8 & 1 4 & 8 & 1 \end{matrix} ∣ ∣ ∣ ∣$ = $\frac{1}{2}$ (16 - 32) = 8 sq. units.

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Similar questions

If the vertices of a triangle be $(a m_{1}^{2}, 2 a m_{1}), (a m_{2}^{2}, 2 a m_{2}) a n d (a m_{3}^{2}, 2 a m_{3})$ , then the area of the triangle is

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(a m_{1} m_{2}, a (m_{1} + m_{2})) (a m_{2} m_{3}, a (m_{2} + m_{3}))

(a m_{3} m_{1}, a (m_{3} + m_{1}))

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(a m_{1}^{2}, 2 a m_{1})

and

(a m_{2}^{2}, 2 a m_{2})

Q. If the coordinates of the vertices of a triangle are (0,0) , (0,2) and(3,1) , then area of the triangle is

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