Question

Let $\mathrm{A}(-1,1),\mathrm{B}(3,4)$ and $\mathrm{C}(2,0)$ be given three points. A line $\mathrm{y}=\mathrm{m}\mathrm{x},\mathrm{m}>0$, intersects lines $\mathrm{AC}$ and $\mathrm{BC}$ at point$\mathrm{P}$ and $\mathrm{Q}$ respectively. Let ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$ be the areas of $∆\mathrm{ABC}$ and $∆\mathrm{PQC}$ respectively, such that ${\mathrm{A}}_1=3{\mathrm{A}}_2$, then the value of $\mathrm{m}$ is equal to:

• A. $\frac{4}{15}$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$1$

Explanation for the correct answer:

Finding the slope of the line:

Given three points of a triangle are, $\mathrm{A}(-1,1);\mathrm{B}(3,4)$ and $\mathrm{C}(2,0)$ line $\mathrm{y}=\mathrm{m}\mathrm{x}$cuts line $\mathrm{AC}$and $\mathrm{BC}$ at point $\mathrm{P}$ and $\mathrm{C}$ respectively.
Also, $\text{'}{\mathrm{A}}_1\text{'}$, is the area of triangle $\mathrm{ABC}$ and $\text{'}{\mathrm{A}}_2\text{'}$ is the area of triangle $\mathrm{PQC}$ .

${\mathrm{A}}_1=3{\mathrm{A}}_2$

Step 1: Forming an equation of the type $\mathrm{y}=\mathrm{mx}$

Now $\mathrm{A}$ is equal to,

The figure is shown here

$$\begin{gathered} \Rightarrow A_1=\frac{1}{2}\left|\begin{array}{ccc}-1& 1& 1\\ 3& 4& 1\\ 2& 0& 1\end{array}\right| \\ \Rightarrow A_1=\frac{1}{2}\left|-1(4-0)-1(3-2)+1(0-8)\right| \\ \Rightarrow A_1=\frac{1}{2}\left|-4-1-8\right| \end{gathered}$$

$\Rightarrow {\mathrm{A}}_1=\frac{11-131}{2}$
$\Rightarrow {\mathrm{A}}_1=\frac{13}{2}$units
Now the equation of line $AC$ is,

$\Rightarrow \mathrm{y}-{\mathrm{y}}_1=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\left(\mathrm{x}-{\mathrm{x}}_1\right)$

$\Rightarrow \mathrm{y}-1=\frac{0-1}{2+1}(\mathrm{x}+1)$
$\Rightarrow \mathrm{y}-1=\frac{-1}{3}(\mathrm{x}+1)$

Step 2: Finding the values of $x,y,\mathrm{and}P$

This line intersects with $\mathrm{y}=\mathrm{m}\mathrm{x}$, hence solve these two equation we get the coordinates of point $\mathrm{P}$ . Hence

$$\begin{gathered} \Rightarrow \mathrm{mx}-1=\frac{-1}{3}(\mathrm{x}+1) \\ \Rightarrow 3\mathrm{mx}-3=-\mathrm{x}-1 \\ \Rightarrow 3\mathrm{mx}+\mathrm{x}=2 \\ \Rightarrow \mathrm{x}(3\mathrm{m}+1)=2 \\ \Rightarrow \mathrm{x}=\frac{2}{3\mathrm{m}+1} \end{gathered}$$

Therefore $\mathrm{y}=\frac{2\mathrm{m}}{3\mathrm{m}+1}$

Hence point $\mathrm{P}=\left(\frac{2}{3\mathrm{m}+1},\frac{2\mathrm{m}}{3\mathrm{m}+1}\right)$
Now equation of line $\mathrm{BC}$is,
$\Rightarrow \mathrm{y}-4=\frac{0-4}{2-3}(\mathrm{x}-3)$
,$\Rightarrow \mathrm{y}-4=\frac{-4}{-1}(\mathrm{x}-3)$
$\Rightarrow \mathrm{y}-4=4(\mathrm{x}-3)....\left(1\right)$

Step 3: Finding the values of $x,y,\mathrm{and}Q$

Now, solve this line with $\mathrm{y}=\mathrm{mx}$ we get point Q, hence

$$\begin{gathered} \Rightarrow \mathrm{mx}-4=4(\mathrm{x}-3) \\ \Rightarrow \mathrm{mx}-4=4\mathrm{x}-12 \\ \Rightarrow \mathrm{mx}-4\mathrm{x}=-12+4 \\ \Rightarrow \mathrm{x}(\mathrm{m}-4)=-8 \\ \Rightarrow \mathrm{x}=\frac{-8}{\mathrm{m}-4} \end{gathered}$$

Put $\mathrm{x}=\frac{-8}{\mathrm{m}-4}$ in equation $\left(1\right)$

$\Rightarrow \mathrm{y}-4=4\left(\frac{-8}{\mathrm{m}-4}-3\right)$

$\Rightarrow \mathrm{y}-4=\frac{-32}{\mathrm{m}-4}-12$
$$\begin{gathered} \Rightarrow \mathrm{y}=\frac{-32}{\mathrm{m}-4}-12+4 \\ \Rightarrow \mathrm{y}=\frac{-32}{\mathrm{m}-4}·8 \\ \Rightarrow \mathrm{y}=\frac{-32-8(\mathrm{m}-4)}{\mathrm{m}-4} \\ \Rightarrow \mathrm{y}=\frac{-32-8\mathrm{m}+32}{\mathrm{m}-4} \\ \Rightarrow \mathrm{y}=\frac{-8\mathrm{m}}{\mathrm{m}-4} \end{gathered}$$
Therefore point $\mathrm{Q}=\left(\frac{-8}{\mathrm{m}-4},\frac{-8\mathrm{m}}{\mathrm{m}-4}\right)$

Step 4: Finding the value of $m$
Area of $∆PQC$ is
$$\begin{gathered} \Rightarrow {\mathrm{A}}_2=\frac{1}{2}\left|\begin{array}{ccc}2& 0& 1\\ \frac{2}{3\mathrm{m}+1}& \frac{2\mathrm{m}}{3\mathrm{m}+1}& 1\\ \frac{-8}{\mathrm{m}-4}& \frac{-8\mathrm{m}}{\mathrm{m}-4}& 1\end{array}\right| \\ \Rightarrow {\mathrm{A}}_2=\frac{1}{2}\left|2\left(\frac{2\mathrm{m}}{3\mathrm{m}+1}+\frac{8\mathrm{m}}{\mathrm{m}-4}\right)-0+1\left(\frac{-16\mathrm{m}}{(3\mathrm{m}+1)(\mathrm{m}-4)}\frac{16\mathrm{m}}{(\mathrm{m}-4)(3\mathrm{m}+1)}\right)\right| \end{gathered}$$

$\Rightarrow {\mathrm{A}}_2=\frac{1}{2}\left|\frac{4\mathrm{m}}{3\mathrm{m}+1}+\frac{16\mathrm{m}}{\mathrm{m}-4}\right|$
$\Rightarrow {\mathrm{A}}_2=\frac{1}{2}\left|\frac{4{\mathrm{m}}^2-16\mathrm{m}+48{\mathrm{m}}^2+16\mathrm{m}}{(3\mathrm{m}+1)(\mathrm{m}-4)}\right|$
$\Rightarrow {\mathrm{A}}_2=\frac{1}{2}\left|\frac{52{\mathrm{m}}^2}{(3\mathrm{m}+1)(\mathrm{m}-4)}\right|$

$$\begin{gathered} \Rightarrow {\mathrm{A}}_2=\left|\frac{26{\mathrm{m}}^{2}1}{3{\mathrm{m}}^2-12\mathrm{m}+\mathrm{m}-4}\right| \\ \Rightarrow {\mathrm{A}}_2=\left|\frac{26{\mathrm{m}}^2}{3{\mathrm{m}}^2-11\mathrm{m}-4}\right| \\ \Rightarrow \frac{{\mathrm{A}}_1}{3}=\left|\frac{26{\mathrm{m}}^2}{3{\mathrm{m}}^2-11\mathrm{m}-4}\right|\left[\because {\mathrm{A}}_1=3{\mathrm{A}}_2\right] \\ \Rightarrow \frac{13}{6}=\left|\frac{26{\mathrm{m}}^2}{3{\mathrm{m}}^2-11\mathrm{m}-4}\right| \\ \Rightarrow \frac{26{\mathrm{m}}^2}{3{\mathrm{m}}^2-11\mathrm{m}-4}=\pm \frac{13}{6} \end{gathered}$$

$\Rightarrow 12{\mathrm{m}}^2=\pm \left(3{\mathrm{m}}^2-112\mathrm{m}-4\right)$

Taking positive sign we get

$$\begin{gathered} \Rightarrow 12{\mathrm{m}}^2=3{\mathrm{m}}^2-11\mathrm{m}-4 \\ \Rightarrow 9{\mathrm{m}}^2=-11\mathrm{m}-4 \\ \Rightarrow 9{\mathrm{m}}^2+11\mathrm{m}+4=0 \end{gathered}$$

solving this equation we get value of $\mathrm{m}$ as imaginary value, Hence it is rejected.
taking negative sign we get,
$\Rightarrow 12{\mathrm{m}}^2=-3{\mathrm{m}}^2+11\mathrm{m}+4$
$$\begin{gathered} \Rightarrow 15{\mathrm{m}}^2-11\mathrm{m}-4=0 \\ \Rightarrow 15{\mathrm{m}}^2-15\mathrm{m}+4\mathrm{m}-4=0 \\ \Rightarrow 15\mathrm{m}(\mathrm{m}-1)+4(\mathrm{m}-1)=0 \\ \Rightarrow (15\mathrm{m}+4)(\mathrm{m}-1)=0 \\ \Rightarrow \mathrm{m}=\frac{-4}{15},1 \end{gathered}$$

But is given in the question that $\mathrm{m}>0$, hence $\mathrm{m}=1$

Hence, option (B) is the correct answer.