Let A(-1, 1),B(3, 4),and C(2, 0) be three giv | vedclass

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(B) Let $P \equiv (x_1, mx_1)$ and $Q \equiv (x_2, mx_2)$.The area of $\Delta ABC$ is given by:$A_1 = \frac{1}{2} |3(0 - 1) + 2(1 - 4) + (-1)(4 - 0)|

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(B) Let $P \equiv (x_1, mx_1)$ and $Q \equiv (x_2, mx_2)$.The area of $\Delta ABC$ is given by:$A_1 = \frac{1}{2} |3(0 - 1) + 2(1 - 4) + (-1)(4 - 0)| = \frac{1}{2} |-3 - 6 - 4| = \frac{13}{2}$.The area of $\Delta PQC$ with vertices $(x_1, mx_1)$,$(x_2, mx_2)$,and $(2, 0)$ is:$A_2 = \frac{1}{2} |x_1(mx_2 - 0) + x_2(0 - mx_1) + 2(mx_1 - mx_2)| = \frac{1}{2} |mx_1x_2 - mx_1x_2 + 2m(x_1 - x_2)| = m|x_1 - x_2|$.Given $A_1 = 3A_2$,we have $\frac{13}{2} = 3m|x_1 - x_2|$,so $|x_1 - x_2| = \frac{13}{6m}$.The equation of line $AC$ is $y - 0 = \frac{1 - 0}{-1 - 2}(x - 2)$ $\Rightarrow y = -\frac{1}{3}(x - 2)$ $\Rightarrow x + 3y = 2$.Intersection $P$ with $y = mx$: $x + 3(mx) = 2 \Rightarrow x_1 = \frac{2}{1 + 3m}$.The equation of line $BC$ is $y - 0 = \frac{4 - 0}{3 - 2}(x - 2)$ $\Rightarrow y = 4(x - 2)$ $\Rightarrow y = 4x - 8$.Intersection $Q$ with $y = mx$: $mx = 4x - 8$ $\Rightarrow x(4 - m) = 8$ $\Rightarrow x_2 = \frac{8}{4 - m}$.$|x_1 - x_2| = |\frac{2}{1 + 3m} - \frac{8}{4 - m}| = |\frac{8 - 2m - 8 - 24m}{(1 + 3m)(4 - m)}| = \frac{26m}{(3m + 1)(4 - m)}$.Equating to $\frac{13}{6m}$:$\frac{26m}{(3m + 1)(4 - m)} = \frac{13}{6m} \Rightarrow 12m^2 = (3m + 1)(4 - m) = -3m^2 + 11m + 4$.$15m^2 - 11m - 4 = 0 \Rightarrow (15m + 4)(m - 1) = 0$.Since $m > 0$,we get $m = 1$.

Let $A(-1, 1)$,$B(3, 4)$,and $C(2, 0)$ be three given points. $A$ line $y = mx$,$m > 0$,intersects lines $AC$ and $BC$ at points $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $\Delta ABC$ and $\Delta PQC$ respectively,such that $A_1 = 3A_2$. Then the value of $m$ is equal to:

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