Find the area of the quadrilateral formed with vertices at (1,2) (5,8) (8,6), (4,1)
The easiest way to solve this is with something known as "Pick's Theorem"
Definition : Lattice point = point with integer coordinates
Area =
Number of interior lattice points in the figure + ( number of lattice points on the border of the figure ) / 2 - 1
So we have
21 + 5/2 - 1 =
21 + 2.5 - 1 =
22.5 = Area
See the following graph :
Find the area of the quadrilateral formed with vertices at (1,2) (5,8) (8,6), (4,1)
Hello Guest!
\(m_1=\frac{8-2}{5-1}=\frac{3}{2}\\ m_2=\frac{6-8}{8-5}=-\frac{2}{3}\\ m_3=\frac{1-6}{4-8}=\frac{5}{4}\\ m_4=\frac{2-1}{1-4}=-\frac{1}{3}\)
\(f(x)=m(x-x_P)+y_P\)
\(f_1(x)=\frac{3}{2}(x-1)+2=\frac{2}{3}x-\frac{3}{2}+2\\ f_1(x)=\frac{2}{3}x-\frac{1}{2}\\ f_2(x)=-\frac{2}{3}(x-5)+8=-\frac{2}{3}x+\frac{10}{3}+8\\ f_2(x)= -\frac{2}{3}x+\frac{34}{3} \)
\(f_3(x)=\frac{5}{4}(x-8)+6=\frac{5}{4}x-10+6\\ f_3(x)=\frac{5}{4}x-4\\ f_4(x)=-\frac{1}{3}(x-4)+1=-\frac{1}{3}x+\frac{4}{3}+1\\ f_4(x)=-\frac{1}{3}x+\frac{7}{3}\)
\(A=\int_{a}^{b} \! f_1(x) \, dx+\int_{a}^{b} \! f_2(x) \, dx -\int_{a}^{b} \! f_3(x) \, dx -\int_{a}^{b} \! f_4(x) \, dx \)
\(\int_{1}^{5} \! f_1(x) \, dx=\int_{1}^{5} \! (\frac{2}{3}x-\frac{1}{2}) \, dx=\left[\frac{2}{3}\cdot \frac{x^2}{2}-\frac{1}{2}x\right]_{5}^{1} \\ =\frac{1}{3}\cdot 5^2-\frac{5}{2}\)
\(\int_{5}^{8} \! f_2(x) \, dx=\int_{5}^{8} \! ( -\frac{2}{3}x+\frac{34}{3} ) \, dx=\left[-\frac{2}{3}\cdot \frac{x^2}{2} +\frac{34}{3}x\right]_{5}^{8} \)
\(\int_{4}^{8} \! f_3(x) \, dx=\int_{1}^{5} \! (\frac{5}{4}x-4) \, dx=\left[ \frac{5}{4}\cdot \frac{x^2}{2}-4x\right]_{4}^{8} \)
\(\int_{1}^{5} \! f_1(x) \, dx=\int_{1}^{5} \! (\frac{2}{3}x-\frac{1}{2}) \, dx=\left[\frac{2}{3}\cdot \frac{x^2}{2}-\frac{1}{2}x\right]_{5}^{1} \)
and so on
to be continued
!