Question

Four points A(6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are given in such a way that $\dfrac{Area\left( \Delta DBC \right)}{Area\left( \Delta ABC \right)}=\dfrac{1}{2}$ find x
(The question has multiple correct options)
(a) $\dfrac{11}{8}$
(b) $\dfrac{-3}{8}$
(c) $\dfrac{9}{8}$
(d) None of these

Answer 1

Hint:Calculate area of both the triangles involved in the problem. Area of a triangle with the coordinates $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is given by the relation
$area=\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$
Don’t ignore the modulus sign given with the formula of area to get multiple values of $x$ . Any function f(x) with mod sign i.e. $\left| f\left( x \right) \right|$ will open as f(x) and -f(x) both. Use this property to get the answer.

Complete step-by-step answer:
We know area of a triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ can be given as
Area of triangle = $=\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|.......\left( i \right)$
Now, coming to the question, we are given four points on the coordinate plane, that are $A(6, 3)$, $B(-3, 5)$, $C(4, -2)$ and $D\left( x,3x \right)$ .And hence, we need to find value of $x$ if it is given that
$\dfrac{Area\left( \Delta DBC \right)}{Area\left( \Delta ABC \right)}=\dfrac{1}{2}........\left( ii \right)$
So, let us calculate area of triangle ABC with the help of coordinates A, B, C given in the problem and using equation (i) we get
$area\text{ }of\text{ }\Delta ABC\text{=}\dfrac{1}{2}\left| \begin{matrix} 6 & 3 & 1 \\\ -3 & 5 & 1 \\\ 4 & -2 & 1 \\\ \end{matrix} \right|$
Now on expanding the determinant along row 1, we get
$\begin{aligned} & area\left( \Delta ABC \right)=\dfrac{1}{2}\left| \left( 6\left( 5\times 1-\left( -2\times 1 \right) \right)-3\left( \left( -3\times 1 \right)-4\times 1 \right)+1\left( \left( -3 \right)\times \left( -2 \right)-4\times 5 \right) \right) \right| \\\ & =\dfrac{1}{2}\left| \left( 6\left( 5+2 \right)-3\left( -3-4 \right)+1\left( 6-20 \right) \right) \right| \\\ & =\dfrac{1}{2}\left| 42+21-14 \right| \\\ & =\dfrac{1}{2}\left| \left( 63-14 \right) \right| \\\ & =\dfrac{49}{2} \end{aligned}$
$area\left( \Delta ABC \right)=\dfrac{49}{2}........\left( iii \right)$
Now similarly we can calculate the area of $\Delta DBC$ with the help of given coordinates of D, B and C and applying the formula given in equation (i). So, we get
$\begin{aligned} & area\left( \Delta DBC \right)=\dfrac{1}{2}\left| \begin{matrix} x & 3x & 1 \\\ -3 & 5 & 1 \\\ 4 & -2 & 1 \\\ \end{matrix} \right| \\\ & =\dfrac{1}{2}\left| \left( x\left( 5\times 1-\left( -2\times 1 \right) \right)-3x\left( \left( -3\times 1 \right)-4\times 1 \right)+1\left( \left( -3\times -2 \right)-4\times 5 \right) \right) \right| \\\ & =\dfrac{1}{2}\left| \left( x\left( 5+2 \right)-3x\left( -3-4 \right)+1\left( 6-20 \right) \right) \right| \\\ & =\dfrac{1}{2}\left| 7x+21x-14 \right| \\\ & =\dfrac{1}{2}\left| 28x-14 \right| \end{aligned}$
$area\left( \Delta DBC \right)=\left| 14x-7 \right|.......\left( iv \right)$
Now, we know the ratio of area of $\Delta DBC$ and area of $\Delta ABC$ is given as $\dfrac{1}{2}$ from the equation (ii). So, we get
$\dfrac{\left| 14x-7 \right|}{\left( \dfrac{49}{2} \right)}=\dfrac{1}{2}$
On cross-multiplying the above equation, we get
$2\left| 14x-7 \right|=\dfrac{49}{2}.........\left( v \right)$
We know the modulus function can be opened both ways, if any negative term is lying in the modulus function, it will become positive by putting one more negative sign in front of it and positive number will open as positive number. Hence, mod of $14x-7$ can be opened by both ways, depending on the sign of $14x-7$ .If it is positive, then no effect at $14x-7$, but if it will be negative, then we need to put a negative sign in order to make it positive. So, there will be two cases for solving equation (v).

Case 1: If $14x-7\ge 0$ i.e. modulus value is positive, we get
$\begin{aligned} & 2\left( 14x-7 \right)=\dfrac{49}{2} \\\ & 28x-14=\dfrac{49}{2} \\\ & 28x=14+\dfrac{49}{2}=\dfrac{28+49}{2}=\dfrac{77}{2} \\\ & x=\dfrac{77}{2\times 28}=\dfrac{11}{28} \\\ \end{aligned}$
So, we get $x=\dfrac{11}{28}$

Case 2: If $14x-7<0$ , then i.e. modulus value is negative. So, we get
$\begin{aligned} & -2\left( 14x-7 \right)=\dfrac{49}{2} \\\ & -28x+14=\dfrac{49}{2} \\\ & -28x=\dfrac{49}{2}-14=\dfrac{49-28}{2} \\\ & -28x=\dfrac{21}{2} \\\ & x=\dfrac{-21}{2\times 28} \\\ & x=\dfrac{-3}{8} \\\ \end{aligned}$
So, values of $x$ are $\dfrac{11}{8}$ and $\dfrac{-3}{8}$ .
So option (a) and option (b) is correct answer

Note: Another formula to get area of triangle would be given as
$\dfrac{1}{2}\left[ \left| \begin{matrix} {{x}_{1}} & {{y}_{1}} \\\ {{x}_{2}} & {{y}_{2}} \\\ \end{matrix} \right|+\left| \begin{matrix} {{x}_{2}} & {{y}_{2}} \\\ {{x}_{3}} & {{y}_{3}} \\\ \end{matrix} \right|+\left| \begin{matrix} {{x}_{3}} & {{y}_{3}} \\\ {{x}_{1}} & {{y}_{1}} \\\ \end{matrix} \right| \right]$
Where $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right),\left( {{x}_{3}},{{y}_{3}} \right)$ are the coordinates of the triangle.
One may try to use heron’s formula i.e. $\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$ , where (a,b,c) are sides and s is $\dfrac{a+b+c}{2}$ and $\Delta$ is representing the area of triangle. But it will be a longer approach and might be complex for one, so always use the identity as per the requirement and given conditions.
Take care of mod sign with the term $\left| 14x-7 \right|$ , one may miss the modulus sign and hence will get only a single value of ‘x’. So take care and solve it.