Vector coordinate calculation

The triangle ABC is given in the plane. A (-3,5), B (2,3), C (-1, -2) write the coordinates of the vectors u, v, w if u = AB, v = AC, and w = BC. Enter the coordinates of the centers of the lines SAB (..), SAC (...), SBC (. ..)

Final Answer:

u₀ =  5
u₁ =  -2
v₀ =  2
v₁ =  -7
w₀ =  -3
w₁ =  -5
x₀ =  -0.5
y₀ =  4
x₁ =  -2
y₁ =  1.5
x₂ =  0.5
y₂ =  0.5

Step-by-step explanation:

$$\begin{gathered} A_0=-3;A_1=5 \\ {B}_0=2;B_1=3 \\ {C}_0=-1;C_1=-2 \\ u=(u_0,u_1)=AB=B-A \\ {u}_0=B_0-A_0=2-(-3)=5 \end{gathered}$$

$u_1=B_1-A_1=3-5=-2$

$$\begin{gathered} v=(v_0,v_1)=AC=C-A \\ {v}_0=C_0-A_0=(-1)-(-3)=2 \end{gathered}$$

$v_1=C_1-A_1=(-2)-5=-7$

$$\begin{gathered} w=(w_0,w_1)=BC=C-B \\ {w}_0=C_0-B_0=(-1)-2=-3 \end{gathered}$$

$w_1=C_1-B_1=(-2)-3=-5$

$$\begin{gathered} S_0=(x_0,y_0)=AB/2 \\ {x}_0=(A_0+B_0)/2=((-3)+2)/2=-0.5 \end{gathered}$$

$y_0=(A_1+B_1)/2=(5+3)/2=4$

$$\begin{gathered} S_1=(x_1,y_1)=AC/2 \\ {x}_1=(A_0+C_0)/2=((-3)+(-1))/2=-2 \end{gathered}$$

$y_1=(A_1+C_1)/2=(5+(-2))/2=1.5$

$$\begin{gathered} S_2=(x_2,y_2)=BC/2 \\ {x}_2=(B_0+C_0)/2=(2+(-1))/2=0.5 \end{gathered}$$

$y_2=(B_1+C_1)/2=(3+(-2))/2=0.5$

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