This is METH-ness! (H/w help trend).

[Momo.05]

Urgh screw this. Ahaha, better get back on to finishing the question. --

[the.watchman]

Good luck learning latex!

[Momo.05]

[Momo.05]

OHHH :D:D IT WORKED ! IT WORKED ! IT WORKED !
Thank you brightsky and yeah ahaha lesson learnt the.watchman

Plus i also finish this chapter WOOT ! Two wins !

[brightsky]

Ok, back to the question.

When (remember that t is an x-value), the area is a triangle.

Draw any line , where t is between 0 and 1 inclusive.

The "base" of the triangle would have length whilst the "height" of the triangle would have length as the line has a positive 3 gradient.

Hence, when t is between 0 and 1 (inclusive), the area under the function would be .

Now, when the hybrid function progresses into , the area would be a triangle plus a rectangle.

Draw any line , where t greater than 1.

In this region, the function has already covered the "whole triangle", which has a total area of . So we know that the area in this second region would be .

To find , we need to look back at our graph. In the rectangular portion, the "base" of the rectangle would have a length of , whilst the "height" of the rectangle would have a length of . Hence, .

Hence the whole area under the function when would be equal to:

[the.watchman]

Ahhh, smart!

I was using a trapezium for .
So the area is .
I prefer your way, even though it's longer.

[Momo.05]

 .. And the domain and range of this function is both equal to [0, infinity) =D (Y)

[Momo.05]

Long time no assistance, but once again im stuck on another question

Three points have coordinates A(1,7) , B(7,5) and C(0,-2). Find:

(a) the equation of the perpendicular bisector of AB
(b) the point of intersection of this perpendicular bisector and BC .

Thank you in advance for the help  

[brightsky]

a) Gradient of AB =

Equation of AB =



The point at half of AB is

Let the gradient of perpendicular bisector =



The equation of the perpendicular bisector of AB is .

Substitute in the coordinates :







Hence the equation is . Is that right?

[brightsky]

b) The gradient of the line BC =

Equation of BC =

We've gathered that the equation of the perpendicular bisector of AB =

So we have two equations:

.....(1)

......(2)

Substitute equation (1) into equation (2):









Substituting that back into equation (1):



So the point of intersection is at (2,0).

[Momo.05]

Yup, the answers are right
Thank you for the help, its very clear

[brightsky]

An alternate move to the second step of each, i.e. when using the point-gradient formula is to substitute in the two x and y coordinates into the equation in the first one and for the second one and solving simultaneously to find .  

[Momo.05]

Another question ,

In the rectangle ABCD, A and B are the points (4,2) and (2,8) respectively. Given that the equation of AC is y = x - 2, find ;

(a) the equation of BC

Thank you

[cipherpol]

Another question ,

In the rectangle ABCD, A and B are the points (4,2) and (2,8) respectively. Given that the equation of AC is y = x - 2, find ;

(a) the equation of BC

Thank you

Gradient of AB = -3

BC is perpendicular to AB, so gradient of BC is

[Momo.05]

Thank you