Helpp maths (higher) +9 rep

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Originally Posted by Cheekykarl

That kids got it :) +REP

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Kid? :o

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Originally Posted by Ostinato

Kid? :o

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Kid/ Adult i dno what you are :P

Ontopic - Well done at working it out :)

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Originally Posted by Cheekykarl

Kid/ Adult i dno what you are :P

Ontopic - Well done at working it out :)

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Lol tar :)

um, i dont understand how k = 0 when from the picture, it clearly doesnt lol. unless that picture isnt right,

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Originally Posted by PTER

um, i dont understand how k = 0 when from the picture, it clearly doesnt lol. unless that picture isnt right,

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Pmsl hmmm...

what i can find is

QP m = 2/3

y = 2/3x + b
7 = 2/3(-3) + b
7 = -2 + b
b = 7 + 2
b = 9

y = 2/3x + 9


QR since it is perpendicular with QP, its slope would be m = -1.5

y = -1.5x + b
7 = -1.5(-3) + b
7 = 4.5 + b
b = 2.5

y = -1.5x + 2.5

substituting the given value, 5 into the second equation, you'll have

-1.5(5) + 2.5 = k
-7.5 + 2.5 = k
k = -5


and i dont know gradients, sorry

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Originally Posted by Ostinato

I'VE GOT IT :)

Lol right...

First of all you find the gadient of QP, which = 1/2 and as PQR is 90degrees you can state that the gradient of QR is therefore -2 (perpendicular).

You then have the gradient of the line QR and you have the point Q so you just find the equation of that line - which is y= -2x + 10.

You then substitute the point R into the equation of the line so you have:-

k = -2(5) + 10
k = -10 + 10
k = 0 :)

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Lol that can't be right - the point R isn't on 0 :)
I think that PTER's right.

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Originally Posted by PTER

what i can find is

QP m = 2/3

y = 2/3x + b
7 = 2/3(-3) + b
7 = -2 + b
b = 7 + 2
b = 9

y = 2/3x + 9


QR since it is perpendicular with QP, its slope would be m = -1.5

y = -1.5x + b
7 = -1.5(-3) + b
7 = 4.5 + b
b = 2.5

y = -1.5x + 2.5

substituting the given value, 5 into the second equation, you'll have

-1.5(5) + 2.5 = k
-7.5 + 2.5 = k
k = -5


and i dont know gradients, sorry

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k is -5 :P

woteva gurlfriend

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Originally Posted by Ostinato

woteva gurlfriend

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Don't give in!!