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Math Help - Parallel and Perpendicular lines

  1. #1
    Newbie Temperamental's Avatar
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    Parallel and Perpendicular lines

    Please help me with these. I can't understand them at all! I don't need just answers, I want to understand how to solve each one of them. Please help!


    Write an equation in slope-intercept form of the line in parallel to the graph of each equation and passes through the given point.

    1. y=4x+5; (2,-3)

    2. y=-1/2x-3; (0,0)

    3. y=4; (2, 5)

    4. 3x+y=3; (3, 5)

    _________________________________________________
    Write an equation in slope-intercept form of the line that is perpendicular to the graph of each equation and passes through the given point.

    1. y=2x+3; (3, -4)

    2. y=-4x+5; (1, 1)

    3. 2x-5y=3; (-2,7)

    4. x=6; (4,2)
    _________________________________________________

    Determine whether AB or CD are parallel, perpendicular, or neither.

    1. A(4, 3), B(5, 2), C(8, 3), D(9,2)
    2. A(-2, 3), B(4,0), C(2, 5), D(-1, 11)
    _________________________________________________

    Wirte an equation in slop intercept form of each line.

    • passes through the point at (4, -2) and is parallel to the graph of 5x-2y=6
    • perpendicular to the line through points at (1,2) and (8,6) and passes through the point at (0, 5)
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  2. #2
    Jen
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    Consider the slope of each line, for a parallel line they will have the same slope. Parallel line have the same slope because they have to change at the same rate or else they would cross each other eventually right?

    Perpendicular lines have slopes that are negative reciprocals of eachother i.e.

    the negative reciprocal of \frac{1}{2} is \frac{-2}{1} If we multiplied these together we would get -1 that would be the case with all perpendicular lines, if you multiply the slopes together you should get -1
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  3. #3
    Jen
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    So since we are worried about slopes putting them in the standard slope intercept form would be a good place to start. y=mx+b
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by Temperamental View Post
    Write an equation in slope-intercept form of the line in parallel to the graph of each equation and passes through the given point. Note the parallel part.

    1. y=4x+5; (2,-3)

    2. y=-1/2x-3; (0,0)

    3. y=4; (2, 5)

    4. 3x+y=3; (3, 5)
    Because it's parallel, the equation of the second line will have the same slope.

    1) y_2 = 4x + c

    Substitute x = 2 ; y = -3 (Given point)

    (-3) = (4)(2) + c

    Solve for c.

    c = -11

    Thus: y_2 = 4x - 11

    =======

    Same story. Gradient is the same, but we don't know c.

    2) y_2 = - \frac{1}{2} x + c

    Substitute x = 0 ; y = 0 (Given point)

    0 = 0 + c

    Solve for c.

    c = 0

    Thus: y_2 = - \frac{1}{2} x + c

    =======

    3) y=4

    Try to envision this line. It's a horizontal line.

    Basically any y = c line would be parallel to this, but they specifically ask for the one which goes through the point where y = 5.

    So y_2 = 5

    =======

    Try number 4 for yourself.
    _________________________________________________
    Quote Originally Posted by Temperamental View Post
    Write an equation in slope-intercept form of the line that is perpendicular to the graph of each equation and passes through the given point.

    1. y=2x+3; (3, -4)

    2. y=-4x+5; (1, 1)

    3. 2x-5y=3; (-2,7)

    4. x=6; (4,2)
    Same as the first ones we did. Except here the gradients are not equal, but their products equal -1. Always. If the product of lines' gradients equal -1 then they are perpendicular.

    I'll do number 1.

    1.) y=2x+3

    Now the gradient of y_2 should be the negative reciprocal of y.

    m_{y_{2}} = - \frac{1}{2}

    y_2 = - \frac{1}{2} x + c

    You know the drill, substitute (3, -4) and solve for c.
    _________________________________________________

    Quote Originally Posted by Temperamental View Post
    Determine whether AB or CD are parallel, perpendicular, or neither.

    1. A(4, 3), B(5, 2), C(8, 3), D(9,2)
    2. A(-2, 3), B(4,0), C(2, 5), D(-1, 11)
    You know now that when gradients are equal they're parallel and when their product is -1 they're perpendicular. They could also be neither.

    _________________________________________________

    Quote Originally Posted by Temperamental View Post
    Wirte an equation in slop intercept form of each line.

    • passes through the point at (4, -2) and is parallel to the graph of 5x-2y=6
    • perpendicular to the line through points at (1,2) and (8,6) and passes through the point at (0, 5)
    Same drill as all the other stuff.
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  5. #5
    Newbie Temperamental's Avatar
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    Quote Originally Posted by janvdl View Post
    Because it's parallel, the equation of the second line will have the same slope.

    1) y_2 = 4x + c

    Substitute x = 2 ; y = -3 (Given point)

    (-3) = (4)(2) + c

    Solve for c.

    c = -11

    Thus: y_2 = 4x - 11
    What happened to the 5? why did you replace it with c?
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  6. #6
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    Quote Originally Posted by janvdl View Post
    ...
    2) y_2 = - \frac{1}{2} x + c

    Substitute x = 0 ; y = 0 (Given point)

    0 = 0 + c

    Solve for c.

    c = 0

    Thus: \boxed{\color{blue}y_2 = - \frac{1}{2} x }
    Quote Originally Posted by Temperamental View Post
    What happened to the 5? why did you replace it with c?
    The equation of the given line is: y = 4x+5

    A parallel to this line has the same slope (here m = 4) but a different constant summand. This summand is unknown to us and therefore janvdl labeled it c. And you were asked to calculate the value of c using the fact that the parallel has to pass through the point P(2, -3), that means the coordinates of the point must satisfy the equation of the parallel:

    To satisfy an equation means: If you plug in the x- and y-values the equation must be true. And that's the reason why you can use it to calculate the value of c.
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