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How was AS Math P1?

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The question gives point R (-1.3) and line of 3y + 2x = 33 as the mirror. We are looking for the image of R, (let's name it as P)

Rearranging the equation of the mirror gives us y = -2x/3 + 11

Therefore gradient (m) of mirror is -2/3.

Since we understand that the line PR will be PERPENDICULAR to mirror (properties of reflection), therefore m of PR = 1.5.

Work out the equation of PR from the point (-1,3) and m of 1.5

y - 3 = 1.5 (x+1)

y = 1.5 x + 4.5

Another properties of mirror is that distance from object to mirror = distance from image to mirror. Therefore point of intersection of line PR and the mirror will be the MIDPOINT for line PR.

Working out simultaneous equation on PR and mirror,

y = 1.5 x + 4.5
y = -2x/3 + 11

1.5 x + 4.5 = -2x/3 + 11

9x + 27 = -4x + 66

13x = 39
x = 3

Subs x = 3 to any of the 2 equation above, we will get y = 9

Since we know R, and the midpoint, we can work out P.

(-1 + x)/2 = 3 for the x-coordinate and (3+y)/2 = 9 for the y-coordinate

Therefore x = 7 and y = 15

Wow, and I was only able to work out the gradient- ran out of time :(
 
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Is there any hope that the Math P1 gt will be around 55/56 ?
I mean it can't be 60 or above right ?
This paper was pretty hard compared to Nov. 2012 and June 2012 don't you think ?
 
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it will be less than 60
insha-allah
Is there any hope that the Math P1 gt will be around 55/56 ?
I mean it can't be 60 or above right ?
This paper was pretty hard compared to Nov. 2012 and June 2012 don't you think ?
Less than 60 ? u mean 59 ? !! yeah that is exactly what its gonna be :p
my sir says 58-62 so I expect a 60 to be A ... all the best
 
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In Q 7 Guys point R is the reflection of point (-1,3) on that line giving and we were supposed to find R not that pointtt.....cuz i saw u guys took point R as (-1,3) and ur finding its reflection
 
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In Q 7 Guys point R is the reflection of point (-1,3) on that line giving and we were supposed to find R not that pointtt.....cuz i saw u guys took point R as (-1,3) and ur finding its reflection
the point R doesn't matter, once we get the answer The examiner will ignore that mistake
 
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OKEI GUYS !
for the domain range question
1- the function was increasing, bcuz the gradient is above zero ( dy/dc = positive value)
2-the range is y>=2
domain -2<=x<0
don't ask me how, cuz I asked the same thing to my teacher (HOW??) and his reply was 'don't ask me how' :p
 
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Hi fellow members,

Just want to ask you guys what the answers/solutions are for the quadratic equation in cos^2x (or was it cos^2 theta??)

Thanks!
(Were there four solutions?)
answer to the first part was 10
and second part, we had 2 answers.
 
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OKEI GUYS !
for the domain range question
1- the function was increasing, bcuz the gradient is above zero ( dy/dc = positive value)
2-the range is y>=2
domain -2<=x<0
don't ask me how, cuz I asked the same thing to my teacher (HOW??) and his reply was 'don't ask me how' :p
it was -2.5 instead of -2
 
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guys don't you think that oct/nov 2012 (12) was easier than this paper ? and A was 55/75 . it would be perfect if this paper is also 55/75
 
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