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

TEXTBOOK ANSWER:
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

TEXTBOOK ANSWER:
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
    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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