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'