Question

The area of the triangle with vertices $A\left( 3,7 \right),B\left( -5,2 \right),$ and $C\left( 2,5 \right)$ is denoted by $\Delta$ . If ${{\Delta }_{A}},{{\Delta }_{B}},{{\Delta }_{C}}$ denote the area of the triangles with vertices $OBC,AOC\;$ and $ABO\;$ respectively, $O$ being the origin, then
A. ${{\Delta }_{A}}+{{\Delta }_{B}}=\Delta +{{\Delta }_{C}}$
B. ${{\Delta }_{A}}+{{\Delta }_{B}}={{\Delta }_{C}}-\Delta$
C. ${{\Delta }_{A}}+{{\Delta }_{B}}=2{{\Delta }_{C}}$
D. ${{\Delta }_{A}}+{{\Delta }_{B}}+{{\Delta }_{C}}=2\Delta$

Answer 1

We need to find the area of all the triangles given in the question. Firstly, we need to find the values of $\Delta ,{{\Delta }_{A}},{{\Delta }_{B}},{{\Delta }_{C}}$ using the area of the triangle formula. Then, we need to find the relationship between the areas to get the desired result.

Complete step by step solution:
We are given the vertices of the triangle and need to find the area of the triangles. We will be solving the given question using the area of the triangle formula.
The area of the triangle with vertices $\left( {{x}_{1}},{{y}_{1}} \right)$ , $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ is given by $\alpha$ .
$\Rightarrow \alpha =\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & {{y}_{3}} & 1 \\\ \end{matrix} \right|$The value of the determinant is given by
$\Rightarrow \alpha =\dfrac{1}{2}\left| {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right|$
The area of the triangle with vertices $\left( 3,7 \right),\left( -5,2 \right),$ and $\left( 2,5 \right)$ is given by $\Delta$ .
Find the value of $\Delta$ using the above formula, we get,
$\Rightarrow \Delta =\dfrac{1}{2}\left| \begin{matrix} 3 & 7 & 1 \\\ -5 & 2 & 1 \\\ 2 & 5 & 1 \\\ \end{matrix} \right|$The value of the determinant is given by
$\Rightarrow \Delta =\dfrac{1}{2}\left| 3\left( 2-5 \right)-7\left( \left( -5 \right)-2 \right)+1\left( \left( -25 \right)-4 \right) \right|$
Simplifying the above equation, we get,
$\Rightarrow \Delta =\dfrac{1}{2}\left| 3\left( -3 \right)-7\left( -7 \right)+1\left( -29 \right) \right|$
Evaluating the equation further,
$\Rightarrow \Delta =\dfrac{1}{2}\left| -9+49-29 \right|$
$\therefore \Delta =\dfrac{11}{2}$
The area of the triangle with vertices $\left( 0,0 \right),\left( -5,2 \right),$ and $\left( 2,5 \right)$ is given by ${{\Delta }_{A}}$
Find the value of ${{\Delta }_{A}}$ using the above formula, we get,
$\Rightarrow {{\Delta }_{A}}=\dfrac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\\ -5 & 2 & 1 \\\ 2 & 5 & 1 \\\ \end{matrix} \right|$The value of the determinant is given by
$\Rightarrow {{\Delta }_{A}}=\dfrac{1}{2}\left| 0\left( 2-5 \right)-0\left( \left( -5 \right)-2 \right)+1\left( \left( -25 \right)-4 \right) \right|$
Simplifying the above equation, we get,
$\Rightarrow {{\Delta }_{A}}=\dfrac{1}{2}\left| \left( \left( -25 \right)-4 \right) \right|$
$\Rightarrow {{\Delta }_{A}}=\dfrac{1}{2}\left| -29 \right|$
$\therefore {{\Delta }_{A}}=\dfrac{29}{2}$
The area of the triangle with vertices $\left( 3,7 \right),\left( 0,0 \right),$ and $\left( 2,5 \right)$ is given by ${{\Delta }_{B}}$
Find the value of ${{\Delta }_{B}}$ using the above formula, we get,
$\Rightarrow {{\Delta }_{B}}=\dfrac{1}{2}\left| \begin{matrix} 3 & 7 & 1 \\\ 0 & 0 & 1 \\\ 2 & 5 & 1 \\\ \end{matrix} \right|$The value of the determinant is given by
$\Rightarrow {{\Delta }_{B}}=\dfrac{1}{2}\left| 3\left( 0-5 \right)-7\left( 0-2 \right)+1\left( 0-0 \right) \right|$
Simplifying the above equation, we get,
$\Rightarrow {{\Delta }_{B}}=\dfrac{1}{2}\left| 3\left( -5 \right)-7\left( -2 \right)+0 \right|$
$\Rightarrow {{\Delta }_{B}}=\dfrac{1}{2}\left| -15+14 \right|$
$\Rightarrow {{\Delta }_{B}}=\dfrac{1}{2}\left| -1 \right|$
$\therefore {{\Delta }_{B}}=\dfrac{1}{2}$
The area of the triangle with vertices $\left( 3,7 \right),\left( -5,2 \right),$ and $\left( 0,0 \right)$ is given by ${{\Delta }_{C}}$
$\Rightarrow {{\Delta }_{C}}=\dfrac{1}{2}\left| \begin{matrix} 3 & 7 & 1 \\\ -5 & 2 & 1 \\\ 0 & 0 & 1 \\\ \end{matrix} \right|$The value of the determinant is given by
$\Rightarrow {{\Delta }_{C}}=\dfrac{1}{2}\left| 3\left( 2-0 \right)-7\left( -5-0 \right)+1\left( 0-0 \right) \right|$
Simplifying the above equation, we get,
$\Rightarrow {{\Delta }_{C}}=\dfrac{1}{2}\left| 6+35+1 \right|$
$\therefore {{\Delta }_{C}}=\dfrac{41}{2}$
From the above,
$\Rightarrow {{\Delta }_{A}}+{{\Delta }_{B}}+\Delta =\dfrac{29}{2}+\dfrac{1}{2}+\dfrac{11}{2}$
$\Rightarrow {{\Delta }_{A}}+{{\Delta }_{B}}+\Delta =\dfrac{29+1+11}{2}$
$\Rightarrow {{\Delta }_{A}}+{{\Delta }_{B}}+\Delta =\dfrac{41}{2}$
We know that ${{\Delta }_{C}}=\dfrac{41}{2}$ . Substituting the same, we get,
$\Rightarrow {{\Delta }_{A}}+{{\Delta }_{B}}+\Delta ={{\Delta }_{C}}$
Moving the term $\Delta$ to the other side of the equation, we get,
$\Rightarrow {{\Delta }_{A}}+{{\Delta }_{B}}={{\Delta }_{C}}-\Delta$

So, the correct answer is “Option B”.

Note: We need to keep in mind that the correct formula has to be applied and the values substituted should be correct while solving these questions. The result of the given question can be cross-checked using the equation
${{\Delta }_{A}}+{{\Delta }_{B}}={{\Delta }_{C}}-\Delta$
LHS:
$\Rightarrow {{\Delta }_{A}}+{{\Delta }_{B}}$
$\Rightarrow \dfrac{29}{2}+\dfrac{1}{2}$
$\Rightarrow \dfrac{30}{2}$
$\Rightarrow 15$
RHS:
$\Rightarrow {{\Delta }_{C}}-\Delta$
$\Rightarrow \dfrac{41}{2}-\dfrac{11}{2}$
$\Rightarrow \dfrac{30}{2}$
$\Rightarrow 15$
LHS $=\;$ RHS
The result attained is correct.