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Distance Formula

Find the area...

Question

Find the area of the triangle formed by the straight lines whose equations are
$y = m_{1} x + \frac{a}{m_{1}}$ , $y = m_{2} x + \frac{a}{m_{2}}$ , and $y = m_{3} x + \frac{a}{m_{3}}$ .

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Solution

$y = m_{1} x + \frac{a}{m_{1}} . . . . . . . . (i) y = m_{2} x + \frac{a}{m_{2}} . . . . . . . . (i i) y = m_{3} x + \frac{a}{m_{3}} . . . . . . . . (i i i)$

solving $(i)$ and $(i i)$ by $y$ from $(i i)$ in $(i)$

$m_{1} x + \frac{a}{m_{1}} = m_{2} x + \frac{a}{m_{2}} (m_{2} - m_{1}) x = \frac{a}{m_{1}} - \frac{a}{m_{2}} (m_{2} - m_{1}) x = a (\frac{1}{m_{1}} - \frac{1}{m_{2}}) = a \frac{m_{2} - m_{1}}{m_{1} m_{2}} \Rightarrow x = \frac{a}{m_{1} m_{2}} y = m_{1} \times \frac{1}{m_{1} m_{2}} + \frac{a}{m_{1}} \Rightarrow y = a (\frac{1}{m_{2}} + \frac{1}{m_{1}})$

So the point of intersection is $A (\frac{a}{m_{1} m_{2}}, a (\frac{1}{m_{1}} + \frac{1}{m_{2}}))$

solving $(i i)$ and $(i i i)$

$\Rightarrow B (\frac{a}{m_{2} m_{3}}, a (\frac{1}{m_{2}} + \frac{1}{m_{3}}))$

solving $(i i i)$ and $(i)$

$\Rightarrow C (\frac{a}{m_{3} m_{1}}, a (\frac{1}{m_{3}} + \frac{1}{m_{1}}))$

So the vertices of triangle are $A (\frac{a}{m_{1} m_{2}}, a (\frac{1}{m_{1}} + \frac{1}{m_{2}})), B (\frac{a}{m_{2} m_{3}}, a (\frac{1}{m_{2}} + \frac{1}{m_{3}})), C (\frac{a}{m_{3} m_{1}}, a (\frac{1}{m_{3}} + \frac{1}{m_{1}}))$

Area of $△ = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{a}{m_{1} m_{2}} & a (\frac{1}{m_{1}} + \frac{1}{m_{2}}) & 1 \frac{a}{m_{2} m_{3}} & a (\frac{1}{m_{2}} + \frac{1}{m_{3}}) & 1 \frac{a}{m_{3} m_{1}} & a (\frac{1}{m_{3}} + \frac{1}{m_{1}}) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

Simplifying using properties of determinant

$△ = \frac{1}{2} a^{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{m_{1} m_{2}} & \frac{1}{m_{1}} + \frac{1}{m_{2}} & 1 \frac{1}{m_{2} m_{3}} & \frac{1}{m_{2}} + \frac{1}{m_{3}} & 1 \frac{1}{m_{3} m_{1}} & \frac{1}{m_{3}} + \frac{1}{m_{1}} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ △ = \frac{1}{2} a^{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{1}{m_{1} m_{2}} - \frac{1}{m_{2} m_{3}} & \frac{1}{m_{1}} + \frac{1}{m_{2}} - \frac{1}{m_{2}} - \frac{1}{m_{3}} & 1 \frac{1}{m_{2} m_{3}} - \frac{1}{m_{3} m_{1}} & \frac{1}{m_{2}} + \frac{1}{m_{3}} - \frac{1}{m_{3}} - \frac{1}{m_{1}} & 1 \frac{1}{m_{3} m_{1}} & \frac{1}{m_{3}} + \frac{1}{m_{1}} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

$△ = \frac{1}{2} a^{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{m_{3} - m_{1}}{m_{1} m_{2} m_{3}} & \frac{m_{3} - m_{1}}{m_{3} m_{1}} & 0 \frac{m_{1} - m_{2}}{m_{1} m_{2} m_{3}} & \frac{m_{1} - m_{2}}{m_{2} m_{1}} & 0 \frac{1}{m_{3} m_{1}} & \frac{1}{m_{3}} + \frac{1}{m_{1}} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$

Expanding along $C_{3}$

$△ = \frac{1}{2} a^{2} {0 + 0 + 1 (\frac{m_{3} - m_{1}}{m_{1} m_{2} m_{3}} \times \frac{m_{1} - m_{2}}{m_{2} m_{1}} - \frac{m_{3} - m_{1}}{m_{3} m_{1}} \times \frac{m_{1} - m_{2}}{m_{1} m_{2} m_{3}})} △ = \frac{1}{2} a^{2} (m_{3} - m_{1}) (m_{1} - m_{2}) (\frac{1}{m_{1}^{2} m_{2}^{2} m_{3}} - \frac{1}{m_{1}^{2} m_{3}^{2} m_{2}}) △ = \frac{1}{2} a^{2} (m_{3} - m_{1}) (m_{1} - m_{2}) (\frac{m_{3} - m_{2}}{m_{1}^{2} m_{2}^{2} m_{3}^{2}}) △ = a^{2} (m_{3} - m_{1}) (m_{1} - m_{2}) (m_{3} - m_{2}) \div 2 m_{1}^{2} m_{2}^{2} m_{3}^{2}$

Hence proved.

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