Question

If $A=(-3,4),B=(-1,-2),C=(5,6),D(x,-4)$are vertices of a quadrilateral such that$ar(∆ABD)=2ar(∆ACD)$ then $x$ is equal to

• A. $6$
• B. $9$
• C. $69$
• D. $96$

Answer 1

The correct option is C

$69$

The explanation for the correct option:

Step 1. Writing the given data:

Given,

$A=(-3,4),B=(-1,-2),C=(5,6),D(x,-4)$

and, $∆ABD=2∆ACD$

Step 2. Find the value of $x$:

Area $\begin{array}{rcl}(∆ABD)& =& \frac{1}{2}\left|\right(-3\left)\right(-2+4)+(-1\left)\right(-4-4)+x(4-(-2)|\end{array}$ $\left[\because Area(∆)=\left(\frac{1}{2}\right)\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\right]$

$\begin{array}{rcl}& =& \frac{1}{2}|2+6x|\\ & =& |1+3x|\end{array}$

Similarly,

Area $\begin{array}{rcl}(∆ACD)& =& \frac{1}{2}\left|\right(-3\left)\right(6+4)+5(-4-4)+x(4-6\left)\right|\end{array}$

$\begin{array}{rcl}& =& \frac{1}{2}|70+2x|\end{array}$

$\therefore ∆ABD=2∆ACD$

$\Rightarrow$$1+3x=2\left(\frac{1}{2}\right)(70+2x)$

$\Rightarrow$$1+3x=70+2x$

$\mathbf{\therefore }\mathit{x}\mathbf{=}\mathbf{}\mathbf{69}$

Hence, the correct option is (C)