Area of a quadrilateral 
The perpendicular bisector of the line
  joining $A(0,1)$ and $C(-4,7)$ intersects
  the $x$-axis at $B$ and the $y$-axis at $D$.
  Find the area of the quadrilateral.

Thank you in advance!

The perpendicular bisector will pass through the mid-point: $(-2, 4)$  of the line passing through the points $A(0, 1)$ & $C(-4, 7)$ & have a slope: $\frac{2}{3}$ normal to the line $AC$. 
The equation  of perpendicular bisector is given as $$y-4=\frac{2}{3}(x-(-2))$$ $$\implies y=\frac{2x+4+12}{3}=\frac{2x+16}{3}$$ By substituting $y=0$ & $x=0$ respectively, the points of intersection of perpendicular bisector with the axes are determined as : $B(-8, 0)$ & $D\left(0, \frac{16}{3}\right)$. Now divide the quadrilateral $ABCD$ into two triangles $\Delta ABC$ & $\Delta ACD$ Thus we have $$\text{area of quadrilateral}\space ABCD=\text{area of}\space \Delta ABC \space \text{with vertices}\space (0, 1),(-8, 0)\space \text{&}\space  (-4, 7)+\text{area of }\space \Delta ACD \space \text{with vertices}\space (0, 1),(-4, 7)\space \text{&}\space  \left(0, \frac{16}{3}\right)$$   $$\implies \text{area of quadrilateral}\space ABCD=\frac{1}{2}\left|0(0-7)-8(7-1)-4(1-0)\right|+\frac{1}{2}\left|0\left(7-\frac{16}{3}\right)-4\left(\frac{16}{3}-1\right)+0(1-7)\right|$$ $$=\frac{1}{2}\left|-60\right|+\frac{1}{2}\left|\frac{-52}{3}\right|=30+\frac{26}{3}$$$$=\frac{116}{3}\space \text{sq.unit}$$   
