Question

Find the value of k so that the area of triangle ABC with A \[\left( {k + 1,{\text{ }}1} \right),{\text{ }}B{\text{ }}\left( {4,{\text{ }} - 3} \right),{\text{ }}and{\text{ }}C{\text{ }}\left( {7,{\text{ }} - k} \right)\], is \[6\] square units.

Answer 1

Hint: In this question to find the value of we will use this formula to calculate area of triangle\[ = \]
 $ \dfrac{1}{2}|{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})| $ and get the value of k.

Complete step-by-step answer:
 We are given that area of triangle \[ABC\]
\[ = {\text{ }}6{\text{ }}sq.{\text{ }}units.\]

And co-ordinate are as follows
 $
  {x_1} = k + 1 \\
  {x_2} = 4 \\
  {x_3} = 7 \\
  {y_1} = 1 \\
  {y_2} = - 3 \\
  {y_3} = - k \\
 $
where, A \[\left( {k + 1,{\text{ }}1} \right),{\text{ }}B{\text{ }}\left( {4,{\text{ }} - 3} \right),{\text{ }}and{\text{ }}C{\text{ }}\left( {7,{\text{ }} - k} \right)\]
$ \Rightarrow \text{Area of triangle ABC} = \dfrac{1}{2}|{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right) $
$ \Rightarrow 6 = \dfrac{1}{2}\left| {\left( {k + 1} \right)\left( { - 3 + k} \right) + 4\left( { - k - 1} \right) + 7\left( {1 + 3} \right)} \right| $
 $ 12 = \left| {{k^2} - 2k - 3 - 4k + 24} \right| $
 $ 12 = \left| {{k^2} - 6k + 21} \right| $
 $ case:\left( 1 \right) $
 $ 12 = {k^2} - 6k + 21 $
 $ {k^2} - 6k + 9 = 0 $
 $ k = 3 $
 $ case:\left( 2 \right) $
 $ - 12 = {k^2} - 6k + 21 $
 $ {k^2} - 6k + 33 = 0 $
 $ D = {b^2} - 4ac $
 $ = \left( {36 - 4 \times 1 \times 33} \right) < 0 $
So, it does not have real roots
Hence value of k \[ = {\text{ }}3.\]

Note: Students should keep in mind that area is given \[6{\text{ }}sq.\]units, but we must use \[ + 6\] and \[ - 6\] both the values to calculate k. Mostly, students miss this. Also, formulas for the area of the triangle should be learned.