+357 The coordinates of three vertices of a parallelogram are $(-3,1)$, $(5,-2)$, and $(0,0)$. Find the sum of the coordinates of the fourth vertex which is in the third quadrant.
+14917 Find the sum of the coordinates of the fourth vertex.
A(-3,-1)
C(5,-2)
D(0,0)
\(m_{da}=\frac{y_d-y_a}{x_d-x_a}=\frac{0-(-1)}{0-(-3)}=\frac{1}{3}=m_{cb}\\ m_{dc=}\frac{y_d-y_c}{x_d-x_c}=\frac{0-(-2)}{0-5}=-\frac{2}{5}=m_{ab}\)
\(f_{ab}(x)=m_{ab}(x-x_a)+y_a=-\frac{2}{5}(x-(-3))+(-1)=-\frac{2}{5}(x+3)-1\\ \color{blue}f_{ab}(x)=-\frac{2}{5}x-\frac{11}{5}\\ f_{cb}(x)=m_{cb}(x-x_c)+y_c=\frac{1}{3}(x-5)+(-2)=\frac{1}{3}x-\frac{11}{3}\\ \color{blue}f_{cb}(x)=\frac{1}{3}x-\frac{11}{3}\)
\(-\frac{2}{5}x-\frac{11}{5}=\frac{1}{3}x-\frac{11}{3}\\ (\frac{1}{3}+\frac{2}{5})x=\frac{11}{3}-\frac{11}{5}\\ x=\frac{\frac{11}{3}-\frac{11}{5}}{\frac{1}{3}+\frac{2}{5}}\\ \color{blue}x_b=2\)
\(f_{cb}(x)=y_b=\frac{1}{3}x-\frac{11}{3}=\frac{1}{3}\cdot 2-\frac{11}{3}\\ \color{blue}y_b=-3\)
-3 + 2 = -1
The sum of the coordinates of the fourth vertex is -1.
!
Sorry, I wrote down the coordinates of my point A incorrectly. 
+128475 Just a slight mistake by asinus
Slope of AD = -1/3
We need to solve this to find the x coordinate of the missing vertex
(-2/5)(x + 3) + 1 = (-1/3)(x -5) - 2
-6 ( x + 3) + 15 = -5 (x - 5) - 30
-6x - 3 = -5x -5
x = 2
y = (-2/5)(2 + 3) + 1 = -1
The 4th vertex = (2. -1) { This is in the 4th Q , not the 3rd Q }
Sum of coordinates = 1
