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    yep y=-2x is the equation for l2 so now you need to equate l1 to l2 so 3(-2x) -x + 21 = 0 then find x, sub that value back into either equation to get your y coordinate
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    (Original post by chemkid)
    yep y=-2x is the equation for l2 so now you need to equate l1 to l2 so 3(-2x) -x + 21 = 0 then find x, sub that value back into either equation to get your y coordinate
    Thank you.
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    (Original post by PoliticallyDiverse)
    This is probably extremely simple, but I'm not sure if I know what's missing.

    I have this question:



    To which I have (for (a)):

    y--4 = \frac{1}{3}(x-9)

    3y+12 = x-9

    3y-x+21=0

    Then for (b), I think it's y=-2x, but I'm not entirely sure as to the logic behind it; I'd essentially like to know why.

    Thank you for any assistance.
    If you think about it from the general equation of a line which you know:

    y - y_1 = m(x-x_1)

    What do you know here? Well, x_1 = y_1 = 0 as it passes through the origin.

    So what are you left with?

    (Equally, just consider that a line can be expressed as  y = mx + c: m is -2, as stated, and c will be 0 as that's the y-intercept (0, as crosses at the origin)).
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    (Original post by EEngWillow)
    If you think about it from the general equation of a line which you know:

    y - y_1 = m(x-x_1)

    What do you know here? Well, x_1 = y_1 = 0 as it passes through the origin.

    So what are you left with?

    (Equally, just consider that a line can be expressed as  y = mx + c: m is -2, as stated, and c will be 0 as that's the y-intercept (0, as crosses at the origin)).
    Ah, brilliant. Thank you very much.
 
 
 
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