Thread: 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 ,(t does not equal to 0)which passes through the point ( ,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 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