Parallelogram

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. 

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