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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.

 Mar 18, 2024
 #1
avatar+14903 
+1

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.

laugh !

Sorry, I wrote down the coordinates of my point A incorrectly. indecision

 Mar 18, 2024
edited by asinus  Mar 18, 2024
edited by asinus  Mar 18, 2024
 #2
avatar+128406 
+1

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

 

cool cool cool

 Mar 18, 2024

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