Question

Let two points be $A\left(1,-1\right)$ and $B\left(0,2\right)$. If a point $P\left(x\text{'},y\text{'}\right)$ be such that the area of $∆PAB=5\text{sq. units}$ and it lies on the line, $3x+y-4\lambda =0$, then the value of lambda is

• A. $4$
• B. $1$
• C. $-3$
• D. $3$

Answer 1

The correct option is D

$3$

Explanation for the correct option:

Step 1: Calculating the area of triangle

Given two points as $A\left(1,-1\right)$ and $B\left(0,2\right)$.

Also given that, the area of $∆PAB=5\text{sq. units}$, we know that the area of $∆PAB$ is,

$\text{area of}∆PAB=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_2-y_1\right)+x_3\left(y_1-y_2\right)\right|$

Substituting $\left(x_1,y_1\right)$ as $\left(x\text{'},y\text{'}\right)$, $\left(x_2,y_2\right)$ as $\left(0,2\right)$, $\left(x_3,y_3\right)$ as $\left(1,-1\right)$ and $∆PAB$ as $5$ in area of $∆PAB$. [given]

$\begin{array}{rcl}5& =& \frac{1}{2}\left|x\text{'}\left(2-\left(-1\right)\right)+0\left(2-y\text{'}\right)+1\left(y\text{'}-2\right)\right|\\ & \Rightarrow & 10=\left|x\text{'}\left(2+1\right)+0+y\text{'}-2\right|\\ & \Rightarrow & 10=\left|x\text{'}\left(3\right)+0+y\text{'}-2\right|\\ & \Rightarrow & 10=\left|3x\text{'}+y\text{'}-2\right|\end{array}$

By taking off the modulus, we get

$$\begin{gathered} 3x\text{'}+y\text{'}-2=10 \\ 3x\text{'}+y\text{'}-12=0 \end{gathered}$$

And the negative as,

$\begin{array}{rcl}3x\text{'}+y\text{'}-2& =& -10\\ 3x\text{'}+y\text{'}+8& =& 0\end{array}$

Step 2: Finding the value of $\lambda$

Now by comparing the given line of equations $3x+y-4\lambda =0$ and $3x\text{'}+y\text{'}-12=0$ we get,

$\begin{array}{rcl}-4\lambda & =& -12\\ \lambda & =& 3\end{array}$

Also by comparing the given line of equations $3x+y-4\lambda =0$ and $3x\text{'}+y\text{'}+8=0$ we get,

$\begin{array}{rcl}-4\lambda & =& 8\\ \lambda & =& -2\end{array}$

The obtained values of $\lambda$ are $3\text{and}-2$, where $3$ is the positive value.

Therefore, the value of $\lambda$ is $3$.

Hence, option (D) is the correct answer.