Question

If four points $A(6,3),B(-3,5),C(4,-2)$ and $D(x,3x)$ are given in such a way that $\frac{Area\left(\Delta DBC\right)}{Area\left(\Delta ABC\right)}=\frac{1}{2}$ then the value of $x$ is

• A. $\frac{3}{8}or\frac{-14}{8}$
• B. $2$ or $-3$
• C. $\frac{11}{8}or\frac{-3}{8}$
• D. None of these

Answer 1

The correct option is C $\frac{11}{8}or\frac{-3}{8}$

Given,
$A(6,3),B(-3,5),C(4,-2)$ and $D(x,3x)$
Area of DBC $=\left|\frac{\left(x\right)(5+2)+(-3)(-2-3x)+4(3x-5)}{2}\right|=\left|\frac{7x+6+9x+12x-20}{2}\right|=\left|\frac{28x-14}{2}\right|$
Area of ABC $=\left|\frac{\left(6\right)(5+2)+(-3)(-2-3)+4(3-5)}{2}\right|=\left|\frac{42+15-8}{2}\right|=\frac{49}{2}$

Given: $\frac{AreaofDBC}{AreaofABC}=\frac{1}{2}$

$\frac{\left|\frac{28x-14}{2}\right|}{\frac{49}{2}}=\frac{1}{2}$
$|28x14|=\frac{49}{2}$
$28x14=\pm \frac{49}{2}$

Taking positive sign
$28x14=\frac{49}{2}$
$56x-28=49$
$56x=77$

$x=\frac{77}{56}=\frac{11}{8}$

Taking negative sign

$28x14=-\frac{49}{2}$
$56x-28=-49$
$56x=-21$
$x=\frac{21}{56}=\frac{3}{8}$

So, $x=\frac{11}{8}orx=\frac{3}{8}$