# Discrete Math: Equations of Line in a Plane

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I'm stuck on these questions. Can you please show me a step by step solution?

20) For each of the following, find a normal vector, a direction vector, and a point of each line.

Note, normal vector means perpendicular to the direction vector.

a) 3x - 6y = 14

normal vector = (1 , -2)
direction vector = (2 , 1)
point on the line = [(14/3) , 0]

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21) Find vector, parametric, and symmetric equations of the following lines.

a) -4x + 6y + 9 = 0

Vector Equation -> r = [0 , (-3/2)] + t(3 , 2)
Parametric Equation -> x = 3t and y = 2t - (3/2)
Symmetric Equation -> (x/3) = [y + (3/2)] / 2

• Discrete Math: Equations of Line in a Plane -

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

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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