If the area of the triangle is 35 sq units with vertices (2, -6), (5, 4) and (k, 4). Then k is
$
(a){\text{ 12}} \\
(b){\text{ - 2}} \\
(c){\text{ - 12, - 2}} \\
(d){\text{ 12, - 2}} \\
$
Last updated date: 19th Mar 2023
•
Total views: 304.5k
•
Views today: 3.83k
Answer
304.5k+ views
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.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
