Perpendicular projection

Determine the distance of a point B[1, -3] from the perpendicular projection of a point A[3, -2] on a straight line 2 x + y + 1 = 0.

Correct result:

d =  0

Solution:

$$\begin{gathered} B_x=1;B_y=-3 \\ {A}_x=3;A_y=-2 \\ p:2x+y+1=0. \\ q\perp p;A\in q \\ q:-x+2y+c=0 \\ -A_x+2\cdot A_y+c=0 \\ -3+2\cdot (-2)+c=0 \\ c=7 \\ q:-x+2y+7=0 \\ X=(x,y)=q\cap p \\ -x+2\cdot y+7=0 \\ 2\cdot x+y+1=0 \\ x-2y=7 \\ 2x+y=-1 \\ x=1 \\ y=-3 \\ d=\sqrt{(B_x-x{)}^2+(B_y-y{)}^2}=\sqrt{(1-1{)}^2+((-3)-(-3){)}^2}=0 \end{gathered}$$

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