Answer

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Hint – In this question 3 coordinates and area of the triangle is given to us. We have to find the variable in one of the coordinates. Use the direct formula of the area of the triangle using determinant method to get the answer.

Complete step-by-step answer:

Let the vertices of the triangle is

A = (2, -6) $ \equiv \left( {{x_1},{y_1}} \right)$

B= (5, 4) $ \equiv \left( {{x_2},{y_2}} \right)$

C = (k, 4) $ \equiv \left( {{x_3},{y_3}} \right)$

Now it is given that the area of the triangle is 35 sq. units.

Now as we know for the given vertices the area (A) of the triangle is calculated as:

$ \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|$

Now put the values of all the vertices in above formula we have,

$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}

2&{ - 6}&1 \\

5&4&1 \\

k&4&1

\end{array}} \right|$

Now expand the determinant we have,

$ \Rightarrow A = \dfrac{1}{2}\left( {2\left| {\begin{array}{*{20}{c}}

4&4 \\

1&1

\end{array}} \right| - \left( { - 6} \right)\left| {\begin{array}{*{20}{c}}

5&1 \\

k&1

\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}

5&4 \\

k&4

\end{array}} \right|} \right)$

Now again expand the mini determinant we have,

$ \Rightarrow A = \dfrac{1}{2}\left| {2\left( {4 - 4} \right) + 6\left( {5 - k} \right) + \left( {20 - 4k} \right)} \right|$ (Area is dependent on k and area cannot be negative so we have to take the modulus)

Now put the value of area of triangle in above equation we have,

$ \Rightarrow 35 = \dfrac{1}{2}\left| {50 - 10k} \right|$

$ \Rightarrow \left| {50 - 10k} \right| = 70$

$ \Rightarrow 50 - 10k = \pm 70$ [$\because y = \left| x \right| \Rightarrow x = \pm y$]

$ \Rightarrow 50 - 10k = 70{\text{ & }}50 - 10k = - 70$

Now on simplifying we have,

$ \Rightarrow 10k = - 20{\text{ & }}10k = 120{\text{ }}$

$ \Rightarrow k = - 2,12$

So, this is the required value of k for which the area of the triangle is 35 sq. units.

Hence option (d) is correct.

Note – Whenever we face such types of problems the key concept is simply to have the understanding of the basic formula of the determinant form of triangle. The key point is that the value of variables involved can be more than one so take this point into consideration this will help get the right answer.

Complete step-by-step answer:

Let the vertices of the triangle is

A = (2, -6) $ \equiv \left( {{x_1},{y_1}} \right)$

B= (5, 4) $ \equiv \left( {{x_2},{y_2}} \right)$

C = (k, 4) $ \equiv \left( {{x_3},{y_3}} \right)$

Now it is given that the area of the triangle is 35 sq. units.

Now as we know for the given vertices the area (A) of the triangle is calculated as:

$ \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|$

Now put the values of all the vertices in above formula we have,

$ \Rightarrow A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}

2&{ - 6}&1 \\

5&4&1 \\

k&4&1

\end{array}} \right|$

Now expand the determinant we have,

$ \Rightarrow A = \dfrac{1}{2}\left( {2\left| {\begin{array}{*{20}{c}}

4&4 \\

1&1

\end{array}} \right| - \left( { - 6} \right)\left| {\begin{array}{*{20}{c}}

5&1 \\

k&1

\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}

5&4 \\

k&4

\end{array}} \right|} \right)$

Now again expand the mini determinant we have,

$ \Rightarrow A = \dfrac{1}{2}\left| {2\left( {4 - 4} \right) + 6\left( {5 - k} \right) + \left( {20 - 4k} \right)} \right|$ (Area is dependent on k and area cannot be negative so we have to take the modulus)

Now put the value of area of triangle in above equation we have,

$ \Rightarrow 35 = \dfrac{1}{2}\left| {50 - 10k} \right|$

$ \Rightarrow \left| {50 - 10k} \right| = 70$

$ \Rightarrow 50 - 10k = \pm 70$ [$\because y = \left| x \right| \Rightarrow x = \pm y$]

$ \Rightarrow 50 - 10k = 70{\text{ & }}50 - 10k = - 70$

Now on simplifying we have,

$ \Rightarrow 10k = - 20{\text{ & }}10k = 120{\text{ }}$

$ \Rightarrow k = - 2,12$

So, this is the required value of k for which the area of the triangle is 35 sq. units.

Hence option (d) is correct.

Note – Whenever we face such types of problems the key concept is simply to have the understanding of the basic formula of the determinant form of triangle. The key point is that the value of variables involved can be more than one so take this point into consideration this will help get the right answer.

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