just_sanaa
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the points A (1, 7) B (20, 7) C (p, q) form the vertices of a triangle ABC. the point D (8, 2) is the midpoint of AC.

a) find the value of p and q

the line L passes through D and is perpendicular to AC intersects AB at E.

b) find an equation for L

c) find the exact x coordinate of E

help will be much appreciated! :yep:

Sana x
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refref
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Have you drawn a diagram?
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zdo0o
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(Original post by just_sanaa)
the points A (1, 7) B (20, 7) C (p, q) form the vertices of a triangle ABC. the point D (8, 2) is the midpoint of AC.

a) find the value of p and q

the line L passes through D and is perpendicular to AC intersects AB at E.

b) find an equation for L

c) find the exact x coordinate of E

help will be much appreciated! :yep:

Sana x
did these in my head so not sure if they're right,

15,-3

y=14/10 x + 6.6

x=0.4/1.4

coudl you check the mark scheme pls if theyre right ill write out my method
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just_sanaa
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(Original post by ShortRef)
Have you drawn a diagram?
yehh heres the diagram ....
Attached files
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just_sanaa
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(Original post by zdo0o)
did these in my head so not sure if they're right,

15,-3

y=14/10 x + 6.6

x=0.4/1.4

coudl you check the mark scheme pls if theyre right ill write out my method
i would if i had the markschemes lol. but if u look at the diagram i posted in the previous post, it doesnt seem like 15,-3 cud be the coordinates
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zdo0o
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(Original post by just_sanaa)
i would if i had the markschemes lol. but if u look at the diagram i posted in the previous post, it doesnt seem like 15,-3 cud be the coordinates
but you drew those cooirdinates without knowing p and q?

D is the midpoint of AC

A --> D is, move 7 along x axis, 5 down y axis.
therefore
D --> C must be the same process, since it's the midpoint. 7 along, 5 down, and you reach 15,-3

EDIT: in fact if you look at your diagram, your line AD is longer than DC, if you make DC the same length it fits to 15,-3 exactly xD
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zdo0o
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what paper is it and can you find the mark scheme online?
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Ejit
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a) use the midpoint formula (I got (15,-3) also)
b) use the gradient formula for AC, and know that the gradient multiplied by the perpendicular gradient will equal -1, subsitute values
c) simultaneous equation

I don't know if that helps, but you didn't give much of an impression where exactly you were stuck.
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just_sanaa
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(Original post by zdo0o)
but you drew those cooirdinates without knowing p and q?

D is the midpoint of AC

A --> D is, move 7 along x axis, 5 down y axis.
therefore
D --> C must be the same process, since it's the midpoint. 7 along, 5 down, and you reach 15,-3

EDIT: in fact if you look at your diagram, your line AD is longer than DC, if you make DC the same length it fits to 15,-3 exactly xD
actualli yes i see your point. i mustve misread the coordinates or something but your explanation makes sense. i wish id thought of that haha!

its edexcel jan 05 but its very hard to find the markschemes. our schools gonna give them when we get bk to school.

so how would you do the other ones? assuming ur answers are right?

Thanks for ur help! x
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zdo0o
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ok just did part b, gradient was right bu my C was wrong, correct eqn is y = 14/10 x - 9.2

basicaly, the gradient of AC is deltaY/deltaX = -10/14, and using the rule that a perpendicular line has gradient -1/m (m being the original gradient), you know that the gradient of the perpendicular line must be 14/10 (came from -1 divided by -10/14 if you're unsure)

you also have a point, 8,2 that you know it passes through, and so the equation must be true at this point. y = 14/10x + C, y=2, x=8, so just stick the numbers in and you get C = -9.2

y = 14/10x - 9.2

or to simplify, maybe your question asked in the form ax + by + c = 0

14x - 10y - 92 = 0

divide by 2 to be proper about it

7x - 5y - 46 = 0

EDIT: and ive just checked the mark scheme and thats the right equation
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zdo0o
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(Original post by just_sanaa)
actualli yes i see your point. i mustve misread the coordinates or something but your explanation makes sense. i wish id thought of that haha!

its edexcel jan 05 but its very hard to find the markschemes. our schools gonna give them when we get bk to school.

so how would you do the other ones? assuming ur answers are right?

Thanks for ur help! x
and for part C (which was also wrong in my original post because i used my wrong part B answer )

the line it has to cross is just a straight line y=7

so all you have to do is say, what does the equation look like when y = 7

by subbing it in, you end up with 7x - (5x7) - 46 = 0

7x = 81

x = 81/7

and i've checked that one as well and it's correct if you don't understand just ask me, and ill attach the mark scheme to that paper in a sec
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zdo0o
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FreeExamPapers.com.doc

that's the mark scheme for the jan 05 edexcel c1 paper
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just_sanaa
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(Original post by zdo0o)
FreeExamPapers.com.doc

that's the mark scheme for the jan 05 edexcel c1 paper
lol ok i get that. thankyou soo much for ur help and time!!! i am in debt to you lol! if you need any help, not that you would, but i could always try! loll

also thanks for the markscheme!

xxxx
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Nosh_0306
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Thank you soo much this helped a lot!
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lilianleung
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(Original post by Ejit)
a) use the midpoint formula (I got (15,-3) also)
b) use the gradient formula for AC, and know that the gradient multiplied by the perpendicular gradient will equal -1, subsitute values
c) simultaneous equation

I don't know if that helps, but you didn't give much of an impression where exactly you were stuck.
by using the midpoint formula, how did you get (15,-3)?
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Sir Cumference
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(Original post by lilianleung)
by using the midpoint formula, how did you get (15,-3)?
This is a 10 year old thread so they're obviously not going to reply to you If you have a question, please start a new thread and someone will help.
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