Posted 25 August 2005 - 03:18 PM
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.
Posted 25 August 2005 - 03:30 PM
then for b) i can't figure that one out yet its been a while
Posted 25 August 2005 - 03:54 PM
Posted 25 August 2005 - 03:56 PM
Posted 25 August 2005 - 04:06 PM
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.
Posted 25 August 2005 - 04:07 PM
Whoops.. got there before me
Posted 25 August 2005 - 04:08 PM
going after my answers with a better one
Posted 25 August 2005 - 05:26 PM
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.
Posted 25 August 2005 - 05:52 PM
Just a slight difference in factorisation in the end though. Cheers.
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