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Maths Q. - HSN forum

# Maths Q.

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### #1Rocky

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Posted 25 August 2005 - 03:18 PM

I've just started Straight Lines in Higher and I can't do this Question, so can any of you help. Well this is the Question:

Q- A line passes through the points (1/2, a/2) and (b, ab).
(i) Show that the Gradient of this line is 'a'.
(ii) Hence show tha no matter what values 'a' or 'b' take this line will always pass through the Origin.

### #2dondon

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Posted 25 August 2005 - 03:30 PM

for a) you use the normal way of working out gradients m=y -y over x -x
then for b) i can't figure that one out yet its been a while

### #3dfx

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Posted 25 August 2005 - 03:47 PM

Just expanding on dondon's point for (i):

### #4Rocky

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Posted 25 August 2005 - 03:54 PM

Thanks dondon and dfx, I now will somehow work out what the second part is all about lol

### #5dondon

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Posted 25 August 2005 - 03:56 PM

I'm thinking you've to use y-b=m(x+a) have you done that yet?? anyway the answer turns out to be y=ax, and it says hence which usually means a lot but i'm not sure how they are connected

### #6dfx

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Posted 25 August 2005 - 04:06 PM

Got it:

Consider the point (b,ab)

Plug it into the line formula:

y - b = m ( x - a )

So y - ab = a ( x - b)

y - ab = ax -ab

y - ab - ax + ab = 0

y - ax = 0

y = ax

Thus you now know that if you write it out as y = mx + c, your "c" (which is your y-intercept) is 0, because the line is just of the form
y = mx. Thus as it has no y-intercept, it always passes through the origin. How to express that mathematically:

y = ax

Consider general equation: y = mx + c

for y = ax, c = 0 line always passes through origin.

### #7dfx

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Posted 25 August 2005 - 04:07 PM

QUOTE(dondon @ Aug 25 2005, 04:56 PM)
I'm thinking you've to use y-b=m(x+a) have you done that yet?? anyway the answer turns out to be y=ax, and it says hence which usually means a lot but i'm not sure how they are connected

Whoops.. got there before me

### #8dondon

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Posted 25 August 2005 - 04:08 PM

You've started it again!
going after my answers with a better one

### #9Shaun

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Posted 25 August 2005 - 05:26 PM

QUOTE(dfx @ Aug 25 2005, 04:47 PM)
Just expanding on dondon's point for (i):

Sorry that is wrong.

You have wrote b/2 when in actual fact it is a/2.

ab - a/2 [over] b - 1/2

= a (b - 1/2) [over] (b- 1/2).

The top and bottom (b - 1/2) cancel out, leaving you with a.

### #10dfx

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Posted 25 August 2005 - 05:52 PM

Dammit lol, reli silly error... thanks

Just a slight difference in factorisation in the end though. Cheers.

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