數學問題 point slope equation

Let the point of x-intercept and y-intercept be (a,0) and (0,a) respectively.

The equation of L:
y - a = (a/-a)(x)
y = -x + a
As it passes through (2,1),
1 = -2 + a
a = 3
So, equation of L: y = -x + 3

2011-02-13 17:15:53 補充：
Slope of L = -1
Slope of tangents = 1
Let the equations of tangents be y = x + c.
Sub. the equations of tangents into the circle.
x^2 + (x+c)^2 - 6x - 2(x+c) - 8 = 0
x^2 + x^2 + 2xc + c^2 - 6x - 2x - 2c - 8 = 0
2x^2 + (2c - 8)x + (c^2 - 2c - 8) = 0

2011-02-13 17:15:56 補充：
Discriminant = 0
(2c - 8)^2 - 4(2)(c^2 - 2c - 8) = 0
4c^2 - 32c + 64 - 8c^2 + 16c + 64 = 0
-4c^2 - 16c + 128 = 0
c^2 + 4c - 32 = 0
(c - 4)(c + 8) = 0
c = 4 or c = -8
So equations of tangents are y = x + 4 and y = x - 8 respectively.