**Q-**The point A has co-ords (7,4). The straight line with equations x+3y+1=0 & 2x+5y=0 and they intersect at B.

**(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 - 06:18 PM

(a) Let's work out point B:

x + 3y + 1 = 0 ... (i)

2x + 5y = 0 ... (ii)

From (i)

x = -1 - 3y

Then

2(-1 - 3y) + 5y = 0

-2 - y = 0 [=>0** y = -2 **

From (ii)

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.

x + 3y + 1 = 0 ... (i)

2x + 5y = 0 ... (ii)

From (i)

x = -1 - 3y

Then

2(-1 - 3y) + 5y = 0

-2 - y = 0 [=>0

From (ii)

2x + 5(-2) = 0 2x = 10 then

so

(b) Let's get the gradients of the two lines.

Let

Let

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

Thanks but I got the answer just like you did. But can someone help as I am very frustrated with a little part in a Q, I need to do:

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.

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

That's the whole point of getting you to complete the square, don't you see. Once you differentiate g(x) you end up with the previous function they asked you to complete the square with. Now you can rewrite the derivative the form

From that you can get the turning points - look it up in your textbook.

From that you can get the turning points - look it up in your textbook.