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Coordinate geometry HELP

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    I don't know why but I can't seem to do this question, any help would be appreciated...
    Find the equation of the line through point (2,6) which is perpendicular to the line PQ whose equation is 2y-x=5
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    (Original post by Wordnerd2)
    I don't know why but I can't seem to do this question, any help would be appreciated...
    Find the equation of the line through point (2,6) which is perpendicular to the line PQ whose equation is 2y-x=5
    Okay so you are given a point (2,6) and you should immediately be thinking of y-y_1=m(x-x_1) to construct your line.

    Next is figuring out the gradient, m. You are told that this line perpendicular to 2y-x=5, therefore you know that the gradient's on these two lines must be negative reciprocal's of one another. Take this line and rearrange it into the form y=Mx+c. Then take the negative reciprocal of M and this will be your perpendicular gradient, m.
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    (Original post by jsk800)
    This question does not match to the coordinate, as the line 2y-x=5 does not go through (2,6), which would otherwise be the point at which both lines meet.
    lol wat


    They don't intersect at (2,6) and they don't have to.
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    (Original post by rdkgames)
    okay so you are given a point (2,6) and you should immediately be thinking of y-y_1=m(x-x_1) to construct your line.

    Next is figuring out the gradient, m. You are told that this line perpendicular to 2y-x=5, therefore you know that the gradient's on these two lines must be negative reciprocal's of one another. Take this line and rearrange it into the form y=mx+c. Then take the negative reciprocal of m and this will be your perpendicular gradient. m.
    thank you !!!
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    (Original post by Wordnerd2)
    thank you !!!
    Glad to have helped.
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    (Original post by Wordnerd2)
    I don't know why but I can't seem to do this question, any help would be appreciated...
    Find the equation of the line through point (2,6) which is perpendicular to the line PQ whose equation is 2y-x=5
    When a line A is perpendicular to Line B, the gradient of line A multiplied by the gradient of line B is -1.

    So rearrange the equation 2y-x = 5 to find the gradient, then try going from there.

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Updated: September 18, 2016
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