Gradient??

Im a bit confused





Work out the gradient of the line joining


A (-2,3)


and


B (4,1)


Does the point (0,2) lie on this line?





Please help?


And please show your working show i can understand how you get the answer


Thanks =DGradient??
Eqn of the line,


{y-3}/(x-(-2)} = (3-1)/(-2-4)


or, (y-3)/(x+2) = -1/3


or, 9 - 3y = x + 2


or, y = 7/3 - x/3 . .. . .....(1)


ie, x + 3y = 7


From (1) you get, the slope of the line is


m = -1/3





Putting x = 0 %26amp; y = 2 in the eqn of the line.


0 + 3*2 = 6 is not equal to 7





Hence (0,2) do not lie on the line.





I have given you two answers, now you give me one answer.


What does ';Thanks=D'; mean? I have seen a number of people using it.Gradient??
Ok first you need slope-intercept y=mx+b where m=slope=y2-y1/x2-x1= -2/6= -1/3 then plug in either point A or B to solve for b in the above equation. You will get 7/3=b, then check to see if the test point fits. plug 0 in for x, because -1/3*0=0 and 0+2 1/3 + 2 1/3 it dosen't =2, so no the point is not on the line
let x1= -2 y1= 3


let x2= 4 y2= 1


then gradient=(y2-y1)/(x2-x1)


=(1-3)/(4-(-2))


= -2/6


=-1/3


now d straight line joining these 2 points is


y-y1=gradient(x-x1)


y-3=-1/3(x+2)


3y-9=-x-2


x+3y=7 this is d straight line.


now to know if (0,2) lies on dis line,simply put this point on dis line i.e. replace x by 0 and y by 2.


we can c that LHS=6 which is not equal to 7.


hence (0,2) doesnt lie on dis line.


i hope dis will suffice.
difference in y = 3-1 = 2


difference in x = -2-4 = -6





gradient = dy / dx = 2 / -6 = -1/3





no (0,2) not on line


halfway between 3 %26amp; 1 = 2


halfway between -2 %26amp; 4 = 1


so (1,2) is on the line