Draw the graph of 2x-y-2=0,4x+3y-24=0 and …

Given equations, 
{tex}2 x - y - 2 = 0{/tex}
{tex}4x + 3y - 24 = 0{/tex}
{tex}y + 4 = 0{/tex}

We have, {tex}2x - y - 2 = 0{/tex} or {tex}x = {{y+2}\over2} {/tex}
When y = 0, we have {tex}x = {{0+2}\over2} = 1{/tex}
When x = 0, we have y = -2.
Thus, we obtain the following table giving coordinates of two point on the line represented by the equation {tex}2x - y - 2 = 0{/tex} and its graph is shown below.

Now we have, {tex}\begin{array}{l}4x+3y-24=0\Rightarrow y=\frac{24-4x}3\\\end{array}{/tex}
When y = 0, we have x = 6
When x = 0, we have y = 8
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation {tex}4x + 3y - 24 = 0{/tex} and its graph is shown below.

Also we have {tex}y + 4 = 0{/tex}
Clearly, y = - 4 for every value of x.
So, let E(2, -4) and F(0, -4) be two points on the line represented by y + 4 = 0. Plotting these points on the same graph and drawing a line passing through them, we obtain the graph of the line represented by the equation y + 4 = 0 as shown in Figure.



From Fig. we have {tex}\begin{array}{l}\bigtriangleup PQR\\\end{array}{/tex} having vertices P(3,4), Q(-1,-4) and R(9, -4). 

Also, PM = 8 and QR = 10.
{tex}\therefore \quad \text { Area of } \triangle P Q R = \frac { 1 } { 2 } ( \text { Base } \times \text { Height } ){/tex}
{tex}\Rightarrow \quad \text { Area of } \triangle P Q R = \frac { 1 } { 2 } ( Q R \times P M ) = \frac { 1 } { 2 } ( 10 \times 8 )sq.\ units{/tex}
{tex}\Rightarrow \quad \text { Area of } \triangle P Q R = 40 \mathrm { sq } .units.{/tex}