Draw the graphs of the equations 5x − y = 5 and 3x − y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and y-axis. Calculate the area of the triangle so formed.
The given equations are:
$5 x-y=5$$. .(i)$
$3 x-y=3$....(ii)
Putting $x=0$ in equation (i) we get:
$\Rightarrow 5 \times 0-y=5$
$\Rightarrow y=-5$
$x=0, \quad y=-5$
Putting $y=0$ in equation (i) we get:
$\Rightarrow 5 x-0=5$
$\Rightarrow x=1$
$x=1, \quad y=0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $A(0,-5), B(1,0)$ from table
$3 x-y=3$
Putting $x=0$ in equation $(i i)$ we get:
$\Rightarrow 3 \times 0-y=3$
$\Rightarrow y=-3$
$x=0, \quad y=-3$
Putting $y=0$ in equation $(i i)$, we get:
$\Rightarrow 3 x-0=3$
$\Rightarrow x=1$
$x=1, \quad y=0$
Use the following table to draw the graph.
Draw the graph by plotting the two points $C(0,-3), D(1,0)$ from table.
Hence the vertices of the required triangle are $B(1,0), C(0,-3)$ and $A(0,-5)$.
Now,
$\Rightarrow$ Required area $=$ Area of $\mathrm{PCA}$
$\Rightarrow$ Required area $=1 / 2($ base $\times$ height $)$
$\Rightarrow$ Required area $=1 / 2(2 \times 1)$ sq. units
Hence the required area is 1 sq.units
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