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LoginNeed workings for this sec 3NA maths question
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Hi all, currently teaching a sec 3 NA student ,but really dun quite understand how to derive the answer .I need help to get the detailed workings .Thanks
There are 2 points P(-2,1) and Q(3,4) in the coordinate plane .A line with equation y=3x+7 passes through P . R is a point on the line such that gradient of RQ = [-1/3]
Find
(a) the coordinates of R answer : ([-3/5],[5/1/5])
(b)the equation of QR answer : (y=(-1/3)x+5)
Edited by syncopation_music 17 Jul `08, 12:11PM
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Let R = (a,b)
4 - b / 3 - a = -1/3
sub a, b into y = 3x+ 7
(b - 7 )/3 = a
simul.
12 - 3b = (a - 3)
15 - 3b = (b-7/3)
45 - 9b = b - 7
10b = 52
b = 5.2 = 5/1/5
a = (5/1/5 - 7) / 3
= -3/5
R = (-3/5 ; 5/1/5)Gradient QR: (4- 5/1/5) / 3- (-3/5) =
equation = y = (gradient)x + c
then just sub 3/4 into it.
.should know if ur a teacher
maybe theres a shorter method, havent touched maths for one year, but this is what i would doEdited by 77th Cloud 17 Jul `08, 2:48PM
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b)
Gradient of QR = -1/3
Using coords of Q, eqn of QR is
y - 4 = -1/3 (x - 3)
y = (-1/3)x + 5a)
since the lines
y = 3x + 7 and y = (-1/3)x + 5
intersect at R, use simultaneous eqn to get the coords of R
3x + 7 = (-1/3)x + 5
(10/3)x = -2
x = -3/5Sub x = -3/5 into any of the 2 eqn
y = 3(-3/5) + 7
y = 26/5so R = ( -3/5 , 26/5 )
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