Posted 26 October 2005 - 03:58 PM
(a) Find the gradient of AB
(b) Hence show that AB is perpendicular to only one of these 2 lines.
*Help will be appreciated, thanks.
Posted 26 October 2005 - 04:37 PM
Posted 26 October 2005 - 04:42 PM
b)then work out the gradients of both the lines and then see if they are perpendicular.
eg. if the gradient of AB was 3, the gradient of the perpindicular line would be -1/3
Posted 26 October 2005 - 04:57 PM
Posted 26 October 2005 - 04:59 PM
Posted 26 October 2005 - 05:08 PM
y2-y1 OVER x2-x1? But for the B co-ords or Aco-ords?
Posted 26 October 2005 - 05:11 PM
Posted 26 October 2005 - 05:16 PM
Ok I get M=7, so now what do I do with the 7 for part (b). As you saying I need to see what line (out of the 2) will be perpendicular, so do I sub X=-7 into the 2 functions?
Posted 26 October 2005 - 05:19 PM
Posted 26 October 2005 - 05:23 PM
Posted 26 October 2005 - 05:29 PM
you take the the gradient of each line say the first one was -2, for AB to be perpindicular it would need to be 1/2.
Because you turn the gradient round.
Posted 26 October 2005 - 05:33 PM
Posted 26 October 2005 - 05:42 PM
If not i think maybe youve maybe copied it down wrong.
Posted 26 October 2005 - 05:44 PM
Posted 26 October 2005 - 06:18 PM
x + 3y + 1 = 0 ... (i)
2x + 5y = 0 ... (ii)
x = -1 - 3y
2(-1 - 3y) + 5y = 0
-2 - y = 0 [=>0 y = -2
2x + 5(-2) = 0 2x = 10 then x = 5
so B is (5,-2)
M AB (the gradient of AB) is calculated to be 3
(b) Let's get the gradients of the two lines.
Let L1 be x+3y+1=0 3y = -x - 1 m = -1/3 . Let's call this m1.
Let L2 be 2x+5y=0 5y = -2x m = -2/5 . Let's call this m2.
One of these lines (L1 or L2) is perpendicular to AB. Only way to find out is to compare the gradient of AB ( m ab) with those of L1 and L2 (m1 and m2), using the formula:
mPmQ = -1 (for perpendicular lines) ... we commonly know this as "flip the gradient and change the sign for perpendicular lines." (where mP and mQ are the two gradients)
Anyway, test using that and you see that only L1 is perpendicular to AB, and not L2.
Posted 27 October 2005 - 03:14 PM
The Q was that write x2-10x+27 in the form (x+b)2 +c.
Which is completing the square, and I got (x-5)2 +2.
So all good, not exactly, as it then PART B says
"Hence show that g(x)=1/3(x-cubed) -5(x-squared)+27x-2 is always increasing"
Well I differentiated it kk and when I am trying to get the x values, it does not factorise!! So anyone can help me on this problem please.
Posted 27 October 2005 - 03:19 PM
From that you can get the turning points - look it up in your textbook.
Posted 27 October 2005 - 03:36 PM
g(x)= 1/3(x-cubed) -5(x-squared)+27x-2
g-1(x)= 2/3(x-squared) -10x+27 (NOW I MULTIPLY BY 3)
2(x- squared) -30x+81=0
*Yeah it is very similar to my "completing the square" Eqn.
Plus it doesn't even bloody factorise properly.
Posted 27 October 2005 - 04:29 PM
You've differentiated wrong.
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users