Question

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

Answer 1

From the given equation of lines $x=3,x=5$ and $2x-y-4=0$.

Step 1: Finding the table of the equations.

Table for line $2x-y-4=0$ is tabulated below

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

On plotting the graph, we obtain the graph as shown below

Step 2: Finding the value of BC from the graph.

From the graph, we get,

$$\begin{gathered} AB=OB-OA=5-3=2 \\ AD=2 \\ BC=6 \end{gathered}$$

Thus, quadrilateral ABCD is a trapezium.

Step 3: Finding the Area of Quadrilateral

Hence,

Area of Quadrilateral, ABCD =

$$\begin{gathered} =\frac{1}{2}\times (dis\tan cebetweenprallellines)\times (Sumoflengthofparallellines) \\ =\frac{1}{2}\times \left(AB\right)\times (AD+BC) \\ =\frac{1}{2}\times 2\times (2+6) \\ =8squnits \end{gathered}$$

Hence, area of Quadrilateral, $ABCD=8squnits$.