Question

Find the value of k if the area of the triangle is 4 sq. units and vertices are ( k , 0) , ( 4 , 0 ) and ( 0 , 2 )

Answer 1

With the given coordinates we know that the formula of area of the triangle is $\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$substituting the given coordinates in the formula and equating it to 4 we get the value of k.

Complete step-by-step answer:
We are given the coordinates of the vertices of the triangle
$\Rightarrow A({x_1},{y_1}) = (k,0) \\\ \Rightarrow B({x_2},{y_2}) = (4,0) \\\ \Rightarrow C({x_3},{y_3}) = (0,2) \\\$
We know that the area of a triangle is given by
$\Rightarrow \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$
And we are given that the area is 4 sq units
$\Rightarrow \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 4sq.units$
Substituting the given points we get
$\Rightarrow \dfrac{1}{2}\left[ {k\left( {0 - 2} \right) + 4\left( {2 - 0} \right) + 0\left( {0 - 0} \right)} \right] = 4sq.units \\\ \Rightarrow \dfrac{1}{2}\left[ { - 2k + 8 + 0} \right] = 4sq.units \\\ \Rightarrow \dfrac{1}{2}\left[ {8 - 2k} \right] = 4sq.units \\\ \Rightarrow 8 - 2k = 8 \\\ \Rightarrow - 2k = 8 - 8 = 0 \\\ \Rightarrow k = \dfrac{0}{2} = 0 \\\$
Hence we get the value of k to be 0.

Additional information
By finding the product of a point's x coordinate times the next point's y coordinate, then subtracting the y coordinate of the first point times the x coordinate of the second coordinate and dividing by two, you will find the area of the polygon.

Note: Since its difficult for many students to remember the formula
The formula is nothing other the value of the determinant formed by the three given coordinates and adding a dummy column with all the entries as 1
The coordinates be $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)$
Then the area of the triangle is given by
\dfrac{1}{2}\left| \begin{gathered} \begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&1 \end{array} \\\ \begin{array}{*{20}{c}} {{x_2}}&{{y_2}}&1 \end{array} \\\ \begin{array}{*{20}{c}} {{x_3}}&{{y_3}}&1 \end{array} \\\ \end{gathered} \right|
Expanding this column wise we get the formula used above