1) well from the linear line general equation of y= mx +c, where m is your gradient and c is your y-intercept;

substituting the coordinates of the point (t,0) → x-coordinate = t, y-coordinate = 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, x-coordinate = t^2 and y-coordinate = 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 x-coordinate = -2 and y-coordinate = 1 into the equation to get:

1 = (1/t)(-2) + t

t = -2 + t^2

t^2 - 2 - t = 0

(t +2)(t-1) = 0

t=-2,1