
coordinate geometry
1)find in terms of t,the equation of the line through the point (t,0) with gradient t.
2)find the equation of the line with gradient $\displaystyle \frac {1}{t}$,(t does not equal to 0)which passes through the point ($\displaystyle t^2$,2t).I fthis line passes through the point (2,1) find the possible values of t.

1) well from the linear line general equation of y= mx +c, where m is your gradient and c is your yintercept;
substituting the coordinates of the point (t,0) → xcoordinate = t, ycoordinate = 0 and also substituting the gradient into the general equation we get:
0 = t(t) + c
c = t^2
therefore the equation for the line with gradient t and passing through the point (t,0) is
y = tx + t^2
2)
using the same method above with gradient now 1/t, xcoordinate = t^2 and ycoordinate = 2t,
2t = (1/t)(t^2) + c
c = t
therefore the equation of the line is:
y = (1/t)x + t
when the line passes through (2,1) we substitute xcoordinate = 2 and ycoordinate = 1 into the equation to get:
1 = (1/t)(2) + t
t = 2 + t^2
t^2  2  t = 0
(t +2)(t1) = 0
t=2,1