Question

Find the area of the triangle whose vertices are (a, b+c), (a, b-c) and (-a, c).
$(a){\text{ 2ac}} \\\ (b){\text{ 2bc}} \\\ (c){\text{ b(a + c)}} \\\ (a){\text{ c(a - b)}} \\\$

Answer 1

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,

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.