Jeżeli A=(x_A, y_A), B=(x_B, y_B), to

$\overrightarrow{AB}=\left[x_B-x_A,y_B-y_A\right]$  

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$\text{a)}A=\left(1,3\right),B=\left(4,5\right)$  

$x_A=1,y_A=3$  

$x_B=4,y_B=5$  

$\overrightarrow{AB}=\left[x_B-x_A,y_B-y_A\right]=\left[4-1,5-3\right]=\left[3,2\right]$  

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$\text{b)}A=\left(-5,-2\right),B=\left(-1,-1\right)$  

$x_A=-5,y_A=-2$  

$x_B=-1,y_B=-1$  

$\overrightarrow{AB}=\left[x_B-x_A,y_B-y_A\right]=\left[-1-\left(-5\right),-1-\left(-2\right)\right]=\left[-1+5,-1+2\right]=\left[4,1\right]$  

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$\text{c)}A=\left(-1,5\right),B=\left(-3,5\right)$  

$x_A=-1,y_A=5$  

$x_B=-3,y_B=5$  

$\overrightarrow{AB}=\left[x_B-x_A,y_B-y_A\right]=\left[-3-\left(-1\right),5-5\right]=\left[-3+1,0\right]=\left[-2,0\right]$  

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$\text{d)}A=\left(6,-3\right),B=\left(6,1\right)$  

$x_A=6,y_A=-3$  

$x_B=6,y_B=1$  

$\overrightarrow{AB}=\left[x_B-x_A,y_B-y_A\right]=\left[6-6,1-\left(-3\right)\right]=\left[0,1+3\right]=\left[0,4\right]$