Let $△_{1}$ denotes the area of the triangle formed by the vertices $(a m_{1}^{2}, 2 a m_{1}) (a m_{2}^{2}, 2 a m_{2}), (a m_{3}^{2}, 2 a m_{3})$ & $△_{2}$ denotes the area of the triangle formed by the vertices $(a m_{1} m_{2}, a (m_{1} + m_{2})) (a m_{2} m_{3}, a (m_{2} + m_{3}))$ & $(a m_{3} m_{1}, a (m_{3} + m_{1}))$ . Find $\frac{△_{1}}{△_{2}}$ .

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Solution

Given vertices are $(a m_{1}^{2}, 2 a m_{1}) (a m_{2}^{2}, 2 a m_{2})$ and $(a m_{3}^{2}, 2 a m_{3})$
$= ∣ ∣ ∣ \frac{a m_{1}^{2} (2 a m_{2} - 2 a m_{3}) + a m_{2}^{2} (2 a m_{3} - 2 a m_{1}) + a m_{3}^{2} (2 a m_{1} - 2 a m_{2})}{2} ∣ ∣ ∣$
$△_{1} = ∣ ∣ a^{2} (m_{1}^{2} (m_{2} - m_{3}) + m_{2}^{2} (m_{3} - m_{1}) + m_{3}^{2} (m_{1} - m_{2})) ∣ ∣$
Given vertices of second triangle are $(a m_{1} m_{2}, a (m_{1} + m_{2})) (a m_{2} m_{3}, a (m_{2} + m_{3})), (a m_{3} m_{1}, a (m_{3} + m_{1}))$
$= ∣ ∣ ∣ \frac{a m_{1} m_{2} (a m_{2} - a m_{1}) + a m_{2} m_{3} (a m_{3} - a m_{2}) + a m_{3} m_{1} (a m_{1} - a m_{3})}{2} ∣ ∣ ∣$
$△_{2} = \frac{a^{2}}{2} ∣ ∣ m_{1} m_{2}^{2} - m_{1}^{2} m_{2} + m_{2} m_{3}^{2} - m_{2}^{2} m_{3} + m_{3} m_{1}^{2} - m_{3}^{2} m_{1} ∣ ∣ = \frac{△_{1}}{2}$
$∴ \frac{△_{1}}{△_{2}} = 2$

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