Question

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

Answer 1

Hint: Problems of this type can be easily solved with the help of the formula of determining the area of the triangle when the three vertices of the triangle are given. We first find the area of the triangle which is formed using the points $A$ , $B$ , $C$ and $D$ , $B$ , $C$ . Upon doing that we substitute the expressions we get from the area of the triangles in the given equation which in turn helps us to get the value of $x$ .

Complete step-by-step answer:
The points we are given are
$A\left( 6,3 \right)$ , $B\left( -3,5 \right)$ , $C\left( 4,-2 \right)$ and $D\left( x,3x \right)$
We know that if a triangle is formed using three points which are known, we can determine the area of that triangle using the formula
 \[Area=\dfrac{1}{2}\left| \begin{matrix}
   {{x}_{1}} & {{y}_{1}} & 1 \\
   {{x}_{2}} & {{y}_{2}} & 1 \\
   {{x}_{3}} & {{y}_{3}} & 1 \\
\end{matrix} \right|\]
Which can also be written as
\[Area=\dfrac{1}{2}\left| x1\left( y2-y3 \right)+x2\left( y3-y1 \right)+x3\left( y1-y2 \right) \right|\]
Hence, if a triangle is formed using the three points $A\left( 6,3 \right)$ , $B\left( -3,5 \right)$ , $C\left( 4,-2 \right)$ the area of that triangle will be
$Area\left( \Delta ABC \right)=\dfrac{1}{2}\left| 6\left( 5-\left( -2 \right) \right)+\left( -3 \right)\left( \left( -2 \right)-3 \right)+4\left( 3-5 \right) \right|$
Further simplifying the above expression, we get
$Area\left( \Delta ABC \right)=\dfrac{1}{2}\left| 6\left( 5+2 \right)-3\left( -2-3 \right)+4\left( 3-5 \right) \right|$
$\Rightarrow Area\left( \Delta ABC \right)=\dfrac{1}{2}\left| 42+15-8 \right|$
$\Rightarrow Area\left( \Delta ABC \right)=\dfrac{1}{2}\cdot 49$
On the other hand, if a triangle is formed using the three points $D\left( x,3x \right)$ , $B\left( -3,5 \right)$ , $C\left( 4,-2 \right)$ the area of that triangle will be
$Area\left( \Delta DBC \right)=\dfrac{1}{2}\left| x\left( 5-\left( -2 \right) \right)+\left( -3 \right)\left( \left( -2 \right)-3x \right)+4\left( 3x-5 \right) \right|$
Further simplifying the above expression, we get
$\Rightarrow Area\left( \Delta DBC \right)=\dfrac{1}{2}\left| x\left( 5+2 \right)-3\left( -2-3x \right)+4\left( 3x-5 \right) \right|$
The above equation can also be written as
\[\Rightarrow Area\left( \Delta DBC \right)=\dfrac{1}{2}\left| 7x+6+9x+12x-20 \right|\]
\[\Rightarrow Area\left( \Delta DBC \right)=\dfrac{1}{2}\left| 28x-14 \right|\]
In this problem we are given that $\dfrac{Area\left( \Delta DBC \right)}{Area\left( \Delta ABC \right)}=\dfrac{1}{2}$
Now, we substitute the expression of $Area\left( \Delta DBC \right)$ and the value of $Area\left( \Delta ABC \right)$ in the above equation as
$\Rightarrow \dfrac{\dfrac{1}{2}\left| 28x-14 \right|}{\dfrac{1}{2}\cdot 49}=\dfrac{1}{2}$
Cancelling the common terms from the numerator and the denominator of the left-hand side of the above equation we get
$\Rightarrow \dfrac{14\left( 2x-1 \right)}{49}=\dfrac{1}{2}$
$\Rightarrow \dfrac{2\left( 2x-1 \right)}{7}=\dfrac{1}{2}$
Executing cross multiplication of the above equation we get
$\Rightarrow 4\left( 2x-1 \right)=7$
$\Rightarrow 8x-4=7$
Further simplifying we get
$\Rightarrow x=\dfrac{11}{8}$

Note: While solving problems of these types we must be very careful about applying the formula for finding the area of triangles to avoid silly mistakes. Also, other calculations must be done properly so that we get the exact value of $x$ .