**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.

Started by Rocky, Aug 25 2005 03:18 PM

9 replies to this topic

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.

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

then for b) i can't figure that one out yet its been a while

Posted 25 August 2005 - 03:47 PM

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

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

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

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.

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 - 04:08 PM

You've started it again!

going after my answers with a better one

going after my answers with a better one

Posted 25 August 2005 - 05:52 PM

Dammit lol, reli silly error... thanks

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

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

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