Draw the graphs of the equations $ x=3, x=5 $ and $ 2 x-y-4=0 $. Also find the area of the quadrilateral formed by the lines and the $ x $-axis.

Given:

The given equations are:

$x=3, x=5$ and $2x-y-4=0$

To do:

We have to find the area of the quadrilateral formed by the lines and the x-axis.

Solution:

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

Every point on the line $x=3$ will have $x$ coordinate as 3.

Therefore, 

$x$	$3$	$3$
$y$	$0$	$3$

Every point on the line $x=5$ will have $x$ coordinate as 5.

Therefore, 

$x$	$5$	$5$
$y$	$0$	$6$

For equation $2x-y-4=0$,

$y=2x-4$

If $x=2$ then $y=2(2)-4=0$

If $x=3$ then $y=2(3)-4=6-4=2$

$x$	$2$	$3$
$y$	$0$	$2$

The equation of x-axis is $y=0$.

The above situation can be plotted graphically as below:

The lines AB, CD and EF represent the equations $x=3$, $x=5$ and $2x-y-4=0$ respectively.

As we can see, the points of intersection of the lines AB, CD, EF and x-axis taken in pairs are the vertices of the required quadrilateral.

Hence, the vertices of the quadrilateral are $(3,0), (5,0), (5,6)$ and $(3,2)$. 

We know that,

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

In the graph, the area of the required quadrilateral is the difference of the areas of triangles ECD and EAF.

In triangle ECD,

The height of the triangle is the distance between point D and EC.

Height of the triangle$=6$ units.

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

Base of the triangle$=5-2=3$ units.

Area of the triangle ECD$=\frac{1}{2}\times6\times3$

$=9$ sq. units. 

In triangle EAF,

The height of the triangle is the distance between point F and EA.

Height of the triangle$=2$ units.

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

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

Area of the triangle EAF$=\frac{1}{2}\times2\times1$

$=1$ sq. units.

Area of the quadrilateral ACDF$=(9-1)=8$ sq. units.

The area of the quadrilateral formed by the given lines and the x-axis is 8 sq. units.