Can Any Mathematical Geniuses help????

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PhRose
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I was doing this question: "the points A and B have coordinates (-4,6) and (2,8) respectively. A line p is drawn perpendicular to AB to meet the y axis at the point C. Find an equation of the line p. Determine the coordinates of C.
The answers are y=3x+14 and (0,14). I got the y=3x part but couldnt get the y intercept to be 14; but this book is known to have incorrect answers sometimes (rarely though) so I was wondering if it is me or the book.
2. Also need help on this one:
The line l has equation 2x-y-1=0 the line m passes though the point (0,4) and is perpendicular to the line l. Find an equation of m and show that the lines l and m intersect at the point p(2,3)
b) the line n passes through the point (3,0) and is parallel to the line m. Find an equation of n and hence find the coordinates of the point Q where the lines l and n intersect.
the answers are y=-1/2x +4 and (b) y=-1/2x+3/2 (1,1)
Source; Edexcel Maths AS Core 1 book
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PhRose
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its ok I have managed to do question 2 but I still need help on question 1 PLEASE.
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zed963
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You have the coordinates for A and B, from this you can work out the gradient of that line. y2-y1/x2-x1

To find the perpendicular gradient all you do is -1/(gradient)

You find coordinate C.

Then use the formula y-y1=m(x-x1)

And the answer should come out, have a go and see if you get it now.
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PhRose
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I dont get it; this is what I am doing
I used y-y1/y2-y1=x-x1/x2-x1 to find the equation of the first line AB; this has a gradient of 1/3
because that is 1/3 a line perpendicualr to it ie p must have gradient -3 which I got
but then I use y-y1=m(x-x1) to find the equation of the new line using -3 as m and (2,8) as the coordinates because p also passes through B which would give an equation of y=-3x+32 not y=-3x+14
please help
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PhRose
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that does help a tonne
but I dont understand how to get the equation in the first place; its not the coordinates I am confused with but the equation- please read my last comment regarding this.
Thank you very much
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Noble.
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(Original post by PhRose)
I dont get it; this is what I am doing
I used y-y1/y2-y1=x-x1/x2-x1 to find the equation of the first line AB; this has a gradient of 1/3
because that is 1/3 a line perpendicualr to it ie p must have gradient -3 which I got
but then I use y-y1=m(x-x1) to find the equation of the new line using -3 as m and (2,8) as the coordinates because p also passes through B which would give an equation of y=-3x+32 not y=-3x+14
please help
No it doesn't. If you've got a line with gradient -3 and you want it to pass through the point (2,8) then:

y = -3x + c where c is the constant to be determined

8 = -3(2) + c

8 = -6 + c

 c = 14
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PhRose
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oh, thank you very much I guess I probably made a stupid mistake somewhere :confused:
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