General Help

[crazygoat]

Hi,

I have a few problems which i dont know how to sort out,

Firstly how can i find the equation of line AB:
A(0,7) and B is (4,9)

Also how do i find the gradient of y=3x-2 of a perpendicular line

and make t the subject in:
v=u+at

and

y=2t-1 OVER t+3

I know that there is alot of problems but i am struggling how to do them as we have not gone over them in class

[KatieB]

first find the gradient of AB often given by the letter m in the equation of a line y=mx+c.
gradient = difference in y/difference in x
=(9-7)/(4-0)
=2/4
=1/2
the c in y=mx+c is where the line grosses the y-axis which in this case is (0,4) so c=4
therefore the equation of the line is y=1/2x+4



The gradient of y=3x-2 is given by m in y=mx+c so the gradient is 3
a perpendicular gradient is given by -1/m so it will be -1/3



In v=u+at

v-u=at
so t=(v-u)/a




y=(2t-1)/(t+3)
so
y(t+3)=2t-1
yt+3y=2t-1
3y+1=2t-yt
3y+1=t(2-y)
t=(3y+1)/(2-y)

[crazygoat]

Thank you so much for your explanation