Let A(-1,1),B(3,4) and C(2,0) be given three points. A line y=mx,m>0 , intersects lines AC and BC at point P and Q respectively. Let A_1 and A_2 be…

$$P\equiv ({x}_{1},m{x}_{1})$$P≡(x1​,mx1​)

$$Q\equiv ({x}_{2},m{x}_{2})$$Q≡(x2​,mx2​)

$${A}_{1}=\frac{1}{2}|\begin{matrix}3 & 4 & 1 \\ 2 & 0 & 1 \\ -1 & 1 & 1\end{matrix}|=\frac{13}{2}$$A1​=21​∣32−1​401​111​∣=213​

$${A}_{2}=\frac{1}{2}|\begin{matrix}{x}_{1} & m{x}_{1} & 1 \\ {x}_{2} & m{x}_{2} & 1 \\ 2 & 0 & 1\end{matrix}|$$A2​=21​∣x1​x2​2​mx1​mx2​0​111​∣

$${A}_{2}=\frac{1}{2}|2(m{x}_{1}-m{x}_{2})|=m|{x}_{1}-{x}_{2}|$$A2​=21​∣2(mx1​−mx2​)∣=m∣x1​−x2​∣

$${A}_{1}=3{A}_{2}\Rightarrow \frac{13}{2}=3m|{x}_{1}-{x}_{2}|$$A1​=3A2​⇒213​=3m∣x1​−x2​∣

$$\Rightarrow |{x}_{1}-{x}_{2}|=\frac{16}{6m}$$⇒∣x1​−x2​∣=6m16​

$$AC:x+3y=2$$AC:x+3y=2

$$BC:y=4x-8$$BC:y=4x−8

P:x+3y=2&y=mx\Rightarrow {x}_{1}=\frac{2}{1+3m}

Q:y=4x-8&y=mx\Rightarrow {x}_{2}=\frac{8}{4-m}

$$|{x}_{1}-{x}_{2}|=|\frac{2}{1+3m}-\frac{8}{4-m}|$$∣x1​−x2​∣=∣1+3m2​−4−m8​∣

$$=|\frac{-26m}{(1+3m)(4-m)}|=\frac{26m}{(3m+1)|m-4|}$$=∣(1+3m)(4−m)−26m​∣=(3m+1)∣m−4∣26m​

$$=\frac{26m}{(3m+1)(4-m)}$$=(3m+1)(4−m)26m​

$$|{x}_{1}-{x}_{2}|=\frac{13}{6m}$$∣x1​−x2​∣=6m13​

$$\frac{26m}{(3m+1)(4-m)}=\frac{13}{6m}$$(3m+1)(4−m)26m​=6m13​

$$\Rightarrow 12{m}^{2}=-(3m+1)(m-4)$$⇒12m2=−(3m+1)(m−4)

$$\Rightarrow 12{m}^{2}=-(3{m}^{2}-11m-4)$$⇒12m2=−(3m2−11m−4)

$$\Rightarrow 15{m}^{2}-11m-4=0$$⇒15m2−11m−4=0

$$\Rightarrow 15{m}^{2}-15m+4m-4=0$$⇒15m2−15m+4m−4=0

$$\Rightarrow (15m+4)(m-1)=0$$⇒(15m+4)(m−1)=0

$$\Rightarrow m=1$$⇒m=1