Question

Solve the following system of linear equations graphically: $x - y + 2 = 0,{\text{ }}4x - y - 4 = 0$. Find the vertices of the triangle formed by the above two lines and the x-axis. Also find the area of the triangle.

Answer 1

Hint:Plot graph for the equations: $x - y + 2 = 0,{\text{ }}4x - y - 4 = 0$
The solution for these systems of linear equations will be the point of intersection of these lines, since that point lies on both lines and that means it satisfies both of the linear equations.
Then, find the vertices of the triangle formed by these equations and the x-axis just by reading the graph.
And area of a triangle is given by formula: $\dfrac{1}{2} \times base \times height$

Complete step-by-step solution:
Given pair of equation of lines are:
$
  x - y + 2 = 0 - - - - - - - - - - - - [1] \\
  4x - y - 4 = 0 - - - - - - - - - - - [2] \\
$
For plotting graph of these linear equations:
Graph of $x - y + 2 = 0$:
$
  x - y + 2 = 0 \\
   \Rightarrow y = x + 2{\text{ or }}x = y - 2 \\
$
 When x =0, we get:
$y = 0 + 2 = 2$
And when y =0, we get:
$x = 0 - 2 = - 2$$$
Thus, we have the following table for the equation $x - y + 2 = 0$

x	0	-2
y	2	0

Plot points $A(0,2){\text{ and }}B( - 2,0)$for the graph of equation $x - y + 2 = 0$.
Graph of $4x - y - 4 = 0:$
$
  4x - y - 4 = 0 \\
   \Rightarrow y = 4x - 4{\text{ or }}x = \dfrac{{y + 4}}{4} \\
$
When x = 0, we get:
$y = 4\left( 0 \right) - 4 = 0 - 4 = - 4$
When y = 0, we get:
$x = \dfrac{{0 + 4}}{4} = \dfrac{4}{4} = 1$
Thus, we have the following table for the equation $4x - y - 4 = 0:$

x	0	1
y	-4	0

Plot points $C\left( {0, - 4} \right){\text{ and }}D\left( {1,0} \right)$for the graph of equation $4x - y - 4 = 0$.

For the intersection point solving the two linear equations $x - y + 2 = 0,{\text{ }}4x - y - 4 = 0$ by elimination:
$
  x - y + 2 = 0 - - - - - - - - - - - - [1] \\
  4x - y - 4 = 0 - - - - - - - - - - - [2] \\
$
Multiplying eq(1) by 4 and then subtracting it from eq(2):
$
   \Rightarrow (4x - y - 4) - 4(x - y + 2) = 0 \\
   \Rightarrow 4x - y - 4 - 4x + 4y - 8 = 0 \\
   \Rightarrow 3y - 12 = 0 \\
   \Rightarrow y = 4 \\
$
Putting y=4 in eq(1) we get:
$
  x - 4 + 2 = 0 \\
   \Rightarrow x = 2 \\
$
Vertices of the triangle formed by the two lines $x - y + 2 = 0,{\text{ }}4x - y - 4 = 0$ and x-axis are:
$B( - 2,0),D(1,0){\text{ and }}E(2,4)$
Now, area of triangle is given by:
A = $\dfrac{1}{2} \times base \times height$
Here, base is 3 units and height is 4 units putting these values in the formula for area:
$A = \dfrac{1}{2} \times 3 \times 4 = \dfrac{1}{2} \times 12 = 6{\text{ sq}}{\text{.units}}$
Hence, the area of the triangle is\[6{\text{ }}sq.units\].

Note:Alternative method: area of a triangle can also be calculated by the formula:
$Area = \dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$
when the coordinates of vertices of the triangle are taken in a clockwise order.
Here, in above question coordinates of the vertices of the triangle are:
$
  \left( {{x_1},{y_1}} \right) = \left( { - 2,0} \right) \\
  \left( {{x_2},{y_2}} \right) = \left( {1,0} \right) \\
  \left( {{x_3},{y_3}} \right) = \left( {2,4} \right) \\
$
Calculating area:
\[\begin{gathered}
  Area = \dfrac{1}{2}\left| { - 2\left( {0 - 4} \right) + 1\left( {4 - 0} \right) + 2\left( {0 - 0} \right)} \right| \\
  Area = \dfrac{1}{2}\left| {8 + 4} \right| = 6{\text{ sq}}{\text{.units}} \\
\end{gathered} \]