Megan_90
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Hi,
I have been trying to answer this question since today morning I still have no clue, how to do this..

The line l1 passes through the point P(-1,2) and Q(11,8). Find an equation for l1 in the form y=mx+C, where m and c are constants.
a) line l2 passes through the point R(10,0) and is perpendicular to l1. The lines l1 and l2 intersect at point S.

b) calculate the coordinates of S.
c) show that the length of RS is 3root5
d) hence,of otherwise, find the exact area of the triangle PQR

PLEASE,PLEASE HELP ME OUT... I really need help with this...

Thanks a lot,
Megan. (once again please help me)
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bigshiv2
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(Original post by Megan_90)
Hi,
I have been trying to answer this question since today morning I still have no clue, how to do this..

The line l1 passes through the point P(-1,2) and Q(11,8). Find an equation for l1 in the form y=mx+C, where m and c are constants.
a) line l2 passes through the point R(10,0) and is perpendicular to l1. The lines l1 and l2 intersect at point S.

b) calculate the coordinates of S.
c) show that the length of RS is 3root5
d) hence,of otherwise, find the exact area of the triangle PQR

PLEASE,PLEASE HELP ME OUT... I really need help with this...

Thanks a lot,
Megan. (once again please help me)
Do you just want a solution?
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samsama
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First you must find the gradient (m_{l1}) of the line l1. This is found using:\frac{y_2 - y_1}{x_2 - x_1}.
Then substitute m_{l1}, and the coordinates for P (or Q), into the general equation: (y-y_1)=m(x-x_1)
Then simplify and rearrange the equation into the form y=mx+c.

Find the gradient of the line l2. Since it is normal to l1, m_{l2} = -\frac{1}{m_{l1}}.
Then substitute m_{l2}, and the coordinates for R, into the general equation: (y-y_1)=m(x-x_1).
Then simplify and rearrange the equation into the form y=mx+c.

You now have 2 equations. Put them equal to each other to solve them simultaneously to find the coordinate for S.

To find the length of RS, use Pythagora's theorem, which is: a^2=b^2+c^2
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Megan_90
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(Original post by bigshiv2)
Do you just want a solution?
I wanted to know how to work out the answer to this question. yup solution
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Megan_90
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also can someone please help me with this question too? please?

Find the set of values for 6x-7<2x+3 and 2x^2-11x+5<0

I will really appreciate if you could help me.
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bigshiv2
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Image You should be able to do the last part since it is pretty self explanatory.
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Megan_90
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(Original post by bigshiv2)
you should be able to do the last part since it is pretty self explanatory.

thank you sooooooo much for helping!!! it is well explained too....Image
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bigshiv2
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(Original post by Megan_90)
thank you sooooooo much for helping!!! it is well explained too....Image
No probs
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samsama
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(Original post by Megan_90)
also can someone please help me with this question too? please?

Find the set of values for 6x-7<2x+3 and 2x^2-11x+5<0

I will really appreciate if you could help me.
6x-7 < 2x+3
6x-2x < 3+7
4x < 10
x < 10/4
x < 5/2




2x^2-11x+5\textless 0

2x^2-10x-x+5\textless 0

2x(x-5)-1(x-5)\textless 0

(2x-1)(x-5)\textless 0

critical values: x=\tfrac{1}{2} or x=5

*imagine graph*

\tfrac{1}{2}\textless x\textless 5
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Megan_90
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(Original post by samsama)
6x-7 < 2x+3
6x-2x < 3+7
4x < 10
x < 10/4
x < 5/2




2x^2-11x+5\textless 0

2x^2-10x-x+5\textless 0

2x(x-5)-1(x-5)\textless 0

(2x-1)(x-5)\textless 0

critical values: x=\tfrac{1}{2} or x=5

*imagine graph*

\tfrac{1}{2}\textless x\textless 5
Thank you very much!
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