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Coordinate geometry HELP

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
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)(2,6) and you should immediately be thinking of y−y1=m(x−x1)y-y_1=m(x-x_1) to construct your line.

Next is figuring out the gradient, mm. You are told that this line perpendicular to 2y−x=52y-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+cy=Mx+c. Then take the negative reciprocal of MM and this will be your perpendicular gradient, mm.
(edited 7 years ago)
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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Wordnerd2
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Original post by rdkgames
okay so you are given a point (2,6)(2,6) and you should immediately be thinking of y−y1=m(x−x1)y-y_1=m(x-x_1) to construct your line.

Next is figuring out the gradient, mm. You are told that this line perpendicular to 2y−x=52y-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+cy=mx+c. Then take the negative reciprocal of mm and this will be your perpendicular gradient. mm.


thank you !!!
Original post by Wordnerd2
thank you !!!


Glad to have helped. :smile:
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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