Draw the graphs of the pair of linear equations $x-y+2=0$ and $4x-y-4=0$. Calculate the area of the triangle formed by the lines so drawn and the x-axis.

Given:

The given equations are:

$x-y+2=0$ and $4x-y-4=0$.

To do:

We have to find the area of the triangle formed by the lines so drawn and the x-axis.

Solution:

To represent the above equations graphically we need at least two solutions for each of the equations.

For equation $x-y+2=0$,

$y=x+2$

If $x=-2$ then $y=-2+2=0$

If $x=2$ then $y=2+2=4$

$x$	$-2$	$2$
$y$	$0$	$4$

For equation $4x-y-4=0$,

$y=4x-4$

If $x=1$ then $y=4(1)-4=4-4=0$

If $x=2$ then $y=4(2)-4=8-4=4$

$x$	$1$	$2$
$y$	$0$	$4$

The equationof x-axis is $y=0$.

The above situation can be plotted graphically as below:

The lines AB, CD and AC represent the equations $x-y+2=0$, $4x-y-4=0$ and x-axis respectively.

As we can see, the points of intersection of the lines AB, CD and AC taken in pairs are the vertices of the given triangle.

Hence, the vertices of the given triangle are $(-2,0), (2,4)$ and $(1,0)$. 

We know that,

Area of a triangle$=\frac{1}{2}bh$

In the graph, the height of the triangle is the distance between point B and x-axis.

Height of the triangle$=4$ units.

Base of the triangle$=$Distance between the points A and C.

Base of the triangle$=1+2=3$ units.

Area of the triangle formed by the given lines and x-axis$=\frac{1}{2}\times4\times3$

$=6$ sq. units. 

The area of the triangle formed by the given lines and the x-axis is $6$ sq. units.