Answer
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Hint: In this question use the direct formula for area of triangle in determinant form when three coordinates are given that is $A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
{{x_1}}&{{y_1}}&1 \\
{{x_2}}&{{y_2}}&1 \\
{{x_3}}&{{y_3}}&1
\end{array}} \right|$. This will help to get the area.
Complete step-by-step answer:
The vertices of the triangle are (a, b + c), (a, b – c) and (-a, c).
Let A = $(x_1, y_1)$ = (a, b + c).
B = $(x_2, y_2)$ = (a, b - c).
C= $(x_3, y_3)$ = (-a, c).
Now as we know that the area (A) of the triangle when all the three vertices are given is
$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
{{x_1}}&{{y_1}}&1 \\
{{x_2}}&{{y_2}}&1 \\
{{x_3}}&{{y_3}}&1
\end{array}} \right|$ Sq. units.
Now substitute the values of the vertices in above formula we have,
$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
a&{b + c}&1 \\
a&{b - c}&1 \\
{ - a}&c&1
\end{array}} \right|$
Now apply the determinant property
I.e. ${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$ we have,
$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
a&{b + c}&1 \\
{a - a}&{b - c - b - c}&{1 - 1} \\
{ - a - a}&{c - b - c}&{1 - 1}
\end{array}} \right|$
Now simplify we have,
\[ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
a&{b + c}&1 \\
0&{ - 2c}&0 \\
{ - 2a}&{ - b}&0
\end{array}} \right|\]
Now expand the determinant we have,
\[ \Rightarrow A = \dfrac{1}{2}\left[ {a \times 0 + \left( {b + c} \right) \times 0 + 1\left|
{\begin{array}{*{20}{c}}
0&{ - 2c} \\
{ - 2a}&{ - b}
\end{array}} \right|} \right]\]
\[ \Rightarrow A = \dfrac{1}{2}\left[ {0 + 0 + 1\left( {0 - 4ac} \right)} \right]\]
\[ \Rightarrow A = \dfrac{1}{2}\left[ { - 4ac} \right] = - 2ac\]
As we know area cannot be negative so we take the absolute value of the area
$ \Rightarrow \left| A \right| = \left| { - 2ac} \right|$
$ \Rightarrow A = 2ac$
So this is the required area of the triangle.
Hence option (A) is correct.
Note: It is always advised to remember the direct formula for the area of the triangle in this form. Area is not a vector quantity but it can be represented as one because we often represent area as a vector whose length is actual scalar area and whose direction is perpendicular to the plane. The basic determinant properties like ${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$ are applied to make maximum possible zeros inside the determinant as it helps in determinant simplification.
{{x_1}}&{{y_1}}&1 \\
{{x_2}}&{{y_2}}&1 \\
{{x_3}}&{{y_3}}&1
\end{array}} \right|$. This will help to get the area.
Complete step-by-step answer:
The vertices of the triangle are (a, b + c), (a, b – c) and (-a, c).
Let A = $(x_1, y_1)$ = (a, b + c).
B = $(x_2, y_2)$ = (a, b - c).
C= $(x_3, y_3)$ = (-a, c).
Now as we know that the area (A) of the triangle when all the three vertices are given is
$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
{{x_1}}&{{y_1}}&1 \\
{{x_2}}&{{y_2}}&1 \\
{{x_3}}&{{y_3}}&1
\end{array}} \right|$ Sq. units.
Now substitute the values of the vertices in above formula we have,
$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
a&{b + c}&1 \\
a&{b - c}&1 \\
{ - a}&c&1
\end{array}} \right|$
Now apply the determinant property
I.e. ${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$ we have,
$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
a&{b + c}&1 \\
{a - a}&{b - c - b - c}&{1 - 1} \\
{ - a - a}&{c - b - c}&{1 - 1}
\end{array}} \right|$
Now simplify we have,
\[ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}
a&{b + c}&1 \\
0&{ - 2c}&0 \\
{ - 2a}&{ - b}&0
\end{array}} \right|\]
Now expand the determinant we have,
\[ \Rightarrow A = \dfrac{1}{2}\left[ {a \times 0 + \left( {b + c} \right) \times 0 + 1\left|
{\begin{array}{*{20}{c}}
0&{ - 2c} \\
{ - 2a}&{ - b}
\end{array}} \right|} \right]\]
\[ \Rightarrow A = \dfrac{1}{2}\left[ {0 + 0 + 1\left( {0 - 4ac} \right)} \right]\]
\[ \Rightarrow A = \dfrac{1}{2}\left[ { - 4ac} \right] = - 2ac\]
As we know area cannot be negative so we take the absolute value of the area
$ \Rightarrow \left| A \right| = \left| { - 2ac} \right|$
$ \Rightarrow A = 2ac$
So this is the required area of the triangle.
Hence option (A) is correct.
Note: It is always advised to remember the direct formula for the area of the triangle in this form. Area is not a vector quantity but it can be represented as one because we often represent area as a vector whose length is actual scalar area and whose direction is perpendicular to the plane. The basic determinant properties like ${R_2} \to {R_2} - {R_1},{R_3} \to {R_3} - {R_1}$ are applied to make maximum possible zeros inside the determinant as it helps in determinant simplification.
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