① When k is greater than 0, substitute (-3, -8) and (4, 13).

? Equations: -3k+8+b=0.

4k- 13+b=0

? Find k=3? ，b=- 1？ ; The linear equation is 3x-y- 1=0.

② When k is less than 0, substitute (-3, 13), (4,8),?

? Equations: -3k- 13+b=0.

4k+8+b=0

? Find k=-3 and b = 4；; The linear equation is: -3x-y+4=0?

So the linear equation is 3x-y- 1=0? Or -3x-y+4=0?

2.( 1)AC⊥BC' diagram is not clear, so it should be.

Syndrome: ∵ the side is perpendicular to the bottom, ∴CC'⊥ surface ABC.

? ∵AC？ Facing ABC, ∴AC⊥CC'

? Calculating AC⊥BC from Pythagorean Theorem

∴AC⊥ surface BCC'B'

BC? Surface BCC'B'

∴AC⊥BC'

(2) From (1), we know that the required angle of the AC⊥ plane bcc'b' and ∴ is ∠ AC 'c. 。

tan∠AC'C=AC/CC '

? AC = 3，CC'=AA'=4

? ∴tan∠AC'C=3/4

? So the tangent of the angle formed by the straight line AC' and the plane BCC'B' is 3/4.

(3) The intersection of B' c and BC' is E, and E is the midpoint of BC'.

? Connect the neutral line with DE △ABC'

? ∴DE∥AC'

? ∵ Germany? "CDB surface"

? ∴AC'∥ Ground CDB'