Question

If $A=(-3,4),B=(-1,-2),C=(5,6),D=(x,-4)$ are vertices of a quadrilateral such that $\Delta ABD=2\Delta ACD$. Then $x$, is equal to:

• A. $6$
• B. $9$
• C. $69$
• D. $96$

Answer 1

Answer: (C)

$69$

Using $ar\left(\Delta \right)=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
$ar\left(\Delta ABD\right)=\frac{1}{2}|(-3)(-2+4)+(-1)(-4-4)+x(4-(-2\left)\right)|$
$=\frac{1}{2}|2+6x|$
$=|1+3x|⟶\left(i\right)$
$ar\left(\Delta ACD\right)=\frac{1}{2}|(-3)(6+4)+5(-4-4)+x(4-6)|$
$=\frac{1}{2}|-70-2x|$
$=\frac{1}{2}|70+2x|$
$=35+x⟶\left(ii\right)$
$\because |-1|=1$
So,
$ar\left(\Delta ABD\right)=2\Delta \left(ACD\right)$
$1+3x=70+2x$
$69=x$
$\therefore x=69$