Let A(- 1, 1), B(3, 4) and C(2, 0) be given three points. A line y...

$P ≡ \left({x}_{1}, m{x}_{1}\right)$

$Q ≡ \left({x}_{2}, m{x}_{2}\right)$

{A}_{1} = \dfrac{1}{2}\left|\begin{cases}3 & 4 & 1 \\ 2 & 0 & 1 \\ - 1 & 1 & 1\end{cases}\right| = \dfrac{13}{2}

{A}_{2} = \dfrac{1}{2}\left|\begin{cases}{x}_{1} & m{x}_{1} & 1 \\ {x}_{2} & m{x}_{2} & 1 \\ 2 & 0 & 1\end{cases}\right|

${A}_{2} = \dfrac{1}{2}\left|2\left(m{x}_{1} - m{x}_{2}\right)\right| = m\left|{x}_{1} - {x}_{2}\right|$

${A}_{1} = 3{A}_{2} ⇒ \dfrac{13}{2} = 3m\left|{x}_{1} - {x}_{2}\right|$

$⇒ \left|{x}_{1} - {x}_{2}\right| = \dfrac{16}{6m}$

$AC: x + 3y = 2$

$BC: y = 4x - 8$

$P: x + 3y = 2 \text{ and } y = mx ⇒ {x}_{1} = \dfrac{2}{1 + 3m}$

$Q: y = 4x - 8 \text{ and } y = mx ⇒ {x}_{2} = \dfrac{8}{4 - m}$

\left|{x}_{1} - {x}_{2}\right| = \left|\dfrac{2}{1 + 3m} - \dfrac{8}{4 - m}\right|$</p><p>$= \left|\dfrac{- 26m}{\left(1 + 3m\right)\left(4 - m\right)}\right| = \dfrac{26m}{\left(3m + 1\right)\left|m - 4\right|}$</p><p>$= \dfrac{26m}{\left(3m + 1\right)\left(4 - m\right)}$</p><p>\left|{x}_{1} - {x}_{2}\right| = \dfrac{13}{6m}$

$\dfrac{26m}{\left(3m + 1\right)\left(4 - m\right)} = \dfrac{13}{6m}$

$⇒ 12{m}^{2} = - \left(3m + 1\right)\left(m - 4\right)$

$⇒ 12{m}^{2} = - \left(3{m}^{2} - 11m - 4\right)$

$⇒ 15{m}^{2} - 11m - 4 = 0$

$⇒ 15{m}^{2} - 15m + 4m - 4 = 0$

$⇒ \left(15m + 4\right)\left(m - 1\right) = 0$

$⇒ m = 1$