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Math Help - Co-ordinate Geometry Help Please

  1. #1
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    Co-ordinate Geometry Help Please

    Is anyone able to help me with these questions ? they are most likely easy but im in a hurry as it is due tomorrow.
    Cheers

    1)
    Find the equation of the line in the form ax + by + c = 0 that
    a) passes through (-2,5) and has a gradient of 2/3
    b) passes through (-2,6) and (-3,-7)

    2)
    Find the equation of the line that passes through (4,-7) and is parallel to y=3x-2

    3) Find the equation of the line that passes through (-3,8) and is perpindicular to the line 3x-4y+2=0

    Any help will be highly appreciated thanks.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Brownhash View Post
    Is anyone able to help me with these questions ? they are most likely easy but im in a hurry as it is due tomorrow.
    Cheers

    1)
    Find the equation of the line in the form ax + by + c = 0 that
    a) passes through (-2,5) and has a gradient of 2/3 When you say gradient, are you referring to the slope?

    b) passes through (-2,6) and (-3,-7)

    2)
    Find the equation of the line that passes through (4,-7) and is parallel to y=3x-2

    3) Find the equation of the line that passes through (-3,8) and is perpindicular to the line 3x-4y+2=0

    Any help will be highly appreciated thanks.
    1b) If the line passes through 2 points, we can determine the slope of the line. Our two points are (-2,6) \text{ and }(-3,-7)

    Thus m=\frac{y_2-y_1}{x_2-x_1}=\frac{-7-6}{-3-(-2)}=\frac{-13}{-1}=13

    Thus, the line will have the form y=13x+b.

    To solve for b, pick one of the points that the line passes through. I'll choose (-2,6)

    6=13(-2)+b\implies b=32

    Thus our line is y=13x+32. Our line could also be written in the form \color{red}\boxed{13x-y+32=0}

    2)

    This time, we are to find a line that is \parallel to y=3x-2. It also passes through the point (4,-7).

    If lines are parallel, they must have the same slope.

    Thus our parallel line has the form y=3x+b. To find b, plug in the point:

    -7=3(4)+b\implies b=-19

    Thus, our parallel line is {\color{red}\boxed{y=3x-19}}=\color{red}\boxed{3x-y-19=0}

    3)

    The line we are looking for is \bot to 3x-4y+2=0 and it passes through the point (-3,8)

    First, rewrite the line in slope intercept form:

    3x-4y+2=0\implies y=\frac{3}{4}x+\frac{1}{2}.

    The slope of the perpendicular line is the negative reciprocal of this line : \therefore m=-\frac{4}{3}

    Thus the perpendicular line has the form of y=-\frac{4}{3}x+b

    Plug in the point to find b:

    8=-\frac{4}{3}(-3)+b\implies b=4

    Thus, the line is {\color{red}\boxed{y=-\frac{4}{3}x+4}}=\color{red}\boxed{4x+3y-12=0}

    Does this make sense?

    --Chris
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  3. #3
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    Thankyou

    Yes that helps alot thanks for the help
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