To find the area bounded by
Y = 4x + 5 (Say AB) …(i)
Y = 5 – x (Say BC) ...(ii)
4y = x + 5 (Say AC) ...(iii)
By solving equation (i) and (ii), points of intersection is B (0, 5)
By solving equation (ii) and (iii), points of intersection is C (3, 2)
By solving equation (i) and (iii), we get points of intersection is A (– 1, 1)
These are shown in the graph below:
Required area = area of (ΔABD) + area of (ΔBDC) …(1)
Area of (ΔABD) = \(\int^0_{-1}(y_1-y_3)dx\)
Using equation (1), (2) and (3)
Required area = Area (ΔABD) + Area (ΔBDC)
Required bounded area of (\(\Delta ABC\)) = Area of (ΔABD) + Area of (ΔBDC) = \(\frac{15}{2}\) sq. units.