Question

Four points $A(6,3),B(-3,5),C(4,-2)$ and $D(x,3x)$ are given in such a way that
$\frac{area△DBC}{area△ABC}=\frac{1}{2}$, find $x$.

Answer 1

We have,

Area of triangle $=\frac{1}{2}\mid y_1(x_2-x_3)+y_2(x_3-x_1)+y_3(x_1-x_2)\mid$

Consider $\Delta DBC$

Here, $x_1=x,y_1=3x,x_2=-3,y_2=5,x_3=4,y_3=-2$

Area of $△DBC$ $=\frac{1}{2}\mid (5x+6+12x)-(-9x+20-2x)\mid$

$=\frac{1}{2}\mid (28x-14)\mid =\mid 14x-7\mid =7\mid 2x-1\mid$

Now,

In $\Delta ABC$

$x_1=6,y_1=3,x_2=-3,y_2=5,x_3=4,y_3=-2$

Area of $△ABC$

$=\frac{1}{2}\mid (30+6+12)-(-9+20-12)\mid$

$=\frac{1}{2}\mid (48+1)\mid =\frac{49}{2}$

We kow that,

$\Rightarrow$ $\frac{\text{Area of}△DBC}{\text{Area of}△ABC}=\frac{1}{2}$

$\Rightarrow \frac{7\mid 2x-1\mid }{\frac{49}{2}}=\frac{1}{2}$

$\Rightarrow \mid 2x-1\mid =\frac{7}{4}$

$\Rightarrow 2x-1=\pm \frac{7}{4}$

$\Rightarrow$ $2x=\frac{11}{4}$ or $2x=-\frac{3}{4}$

$\Rightarrow$ $x=\frac{11}{8}$ or $x=-\frac{3}{8}$