I will do the second one, since it looks a bit more difficult than the first
You are clearly working in only 2-D here.
I assume you know that for any straight line
Ax + By + C = 0
[A,B] is a normal to the line which makes
[B,-A] a direction vector
and the slope of the line is -A/B
the vector equation of a line in 2-D is
r = [any point on the line] + t[direction vector of the line], where t is a parameter.
so they started by letting x=0, then y = -9/6 = -3/2
so the point they used is (0,-3/2)
and the direction of the line is [6,4] or [3,2]
so the vector equation is
r = [0,-3/2] + t[3,2] as shown in your text.
if we "expand" that we get
r = [0+3t,-3/2 + 2t]
from that you can see that the
x=3t and the
y=-3/2 + 2t as shown in your book answer
now if we solve these last two equations for t we get
t = x/3 and t = (y+3/2)/2
equating t=t gives us their third answer of
x/3 = (y+3/2)/2
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