Question

A system of line is given as $y=m_{i}x+c_i,$ where $m_i$ can take any value out of 0, 1, -1 and when $m_i$ is positive then $c_i$ can be 1 or -1 when $m_i$ equal 0, $c_i$ can be 0 or 1 and when $m_i$ equals -1, $c_i$ can take 0 or 2. Then the area enclosed by all these straight lines is

• A. sq. units
• B. sq. units
• C. sq. untis
• D. sq. units

Answer 1

The correct option is C sq. untis

From the given question, the set of lines are:
y= x+1; y=x-1; y=0; y=1; y=-x; y=-x+2
Plotting then on graph

The area enclosed by the six lines is shaded above. It forms a hexagon ABCDEF. So, area of polygon using stair’s Method.
$A=\frac{1}{2}\left|\begin{array}{cc}0& 1\\ 1& 1\\ \frac{3}{2}& \frac{1}{2}\\ 1& 0\\ 0& 0\\ -\frac{1}{2}& \frac{1}{2}\\ 0& 1\end{array}\right|=\frac{1}{2}|(\frac{1}{2}+\left(\frac{-1}{2}\right))-(1+\frac{3}{2}+\frac{1}{2})|$
$=\frac{3}{2}sq.units.$
Also, in this figure one could see that the hexagon ABCDEF comprises of a square BCEF of side 1 unit and 2 triangles AFB & CED of base 1 unit and ht $\frac{1}{2}$ unit.
$\therefore$ Area=Area of square BCEF+2(area of $\Delta$ AFB)
$=1\times 1+2(\frac{1}{2}-\frac{1}{2}=1)$
$=\frac{3}{2}sq.units.$