Question

Find the area of the triangle formed by the lines $ y - x = 0 $ , $ x + y = 0 $ and $ x - k = 0 $ .

Answer 1

Hint: In this question, we need to find the area the triangle formed the lines $ y - x = 0 $ , $ x + y = 0 $ and $ x - k = 0 $ . By using these line equations we can find the vertices of the triangle and we know that if the vertices of the triangle are given then we can use the area of the triangle formula, by which we can determine the area of the triangle.

Complete step-by-step answer:
The equations of the given lines are,
Let $ y - x = 0 $ , $ x + y = 0 $ and $ x - k = 0 $ be equation 1, 2, 3 respectively.
Let us solve the equations 1 and 2,
 $
  y - x = 0 \\
   \Rightarrow x + y = 0 \;
  $
By adding these two equations, we get,
 $
  2y = 0 \\
   \Rightarrow y = 0 \;
  $
Substituting $ y = 0 $ in equation 2,
 $
  x + 0 = 0 \\
   \Rightarrow x = 0 \;
  $
Therefore, from $ x = 0 $ , $ y = 0 $ , we get a point $ \left( {0,0} \right) $ of the triangle.
Now, let us solve equations 2 and 3,
 $
  x + y = 0 \\
   \Rightarrow x - k = 0 \;
  $
By subtracting these two equations, we get,
 $
  y + k = 0 \\
   \Rightarrow y = - k \;
  $
Substituting $ y = - k $ in equation 2,
 $
  x - k = 0 \\
   \Rightarrow x = k \;
  $
Therefore, from $ x = k $ and $ y = - k $ , we get a point $ \left( {k, - k} \right) $ of the triangle.
Now, let us solve the equations 3 and 1,
 $
  x - k = 0 \\
   \Rightarrow y - x = 0 \;
  $
By adding these two equations, we get,
 $
  y - k = 0 \\
   \Rightarrow y = k \;
  $
Substituting $ y = k $ in equation 1,
 $
  k - x = 0 \\
   \Rightarrow x = k \;
  $
Therefore, from $ x = k $ and $ y = k $ , we get a point $ \left( {k,k} \right) $ of the triangle.
Now, we have the three points $ \left( {0,0} \right) $ , $ \left( {k, - k} \right) $ and $ \left( {k,k} \right) $ of the triangle
We know that, if three points of the triangle are known then, we can use the formula, $ Area = \dfrac{1}{2}\left( {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right) $
Therefore,
  $ Area = \dfrac{1}{2}\left( {0\left( { - k - k} \right) + k\left( {k - 0} \right) + k\left( {0 + k} \right)} \right) $
$
  = \dfrac{1}{2}\left( {0 + {k^2} + {k^2}} \right) \\
  = \dfrac{{2{k^2}}}{2} \\
  = {k^2} \;
  $
Hence, the area of the triangle is $ {k^2} $ .
So, the correct answer is “ $ {k^2} $ sq.units”.

Note: It is important to note here that, the area of the triangle formula, $ Area = \dfrac{1}{2}\left( {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right) $ , which can be only used when the three of the triangle are given. Further, we can also use Heron's formula, $ Area = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ where, $ s = \dfrac{{a + b + c}}{2} $ to find the area of the triangle, by determining the sides of the triangle by using the distance formula.