The midpoint coordinate is equal to half of the sum of the coordinates at both ends.
The midpoint of AB is d (8/2+4/2,5/2-2/2), that is, d (6,3/2).
Similarly, BC midpoint E (- 1, 1/2) and AC midpoint F (1, 4).
The slope of the straight line DE is (3/2-1/2)/(6+1) =1/7, and the equation can be set as: y=x/7+b, which is obtained by substituting d (6 6,3/2).
In the same way, the EF equation is: y = 7x/4+9/4; The DF equation is: y=-x/2+9/2.
1(2)
The middle line on the side of BC is the connecting line between BC midpoint D and point A. If the coordinates of B and C are known, the coordinates of BC midpoint D (-5/2, 1) can be obtained.
As in the previous question, we can get the AD equation: y=8x/5+5.
2
4x 2-4x √ 3+3 = 0, but the formula: (2x-√ 3) 2 = 0, 2x-√3=0, x=√3/2.
Let the dip angle be θ ∈ [0, 180] and x=sinθ, that is, sinθ=√3/2, so θ = 60 or 120.
When θ = 60, slope = tangent of inclination angle = tan θ = tan 60 = √ 3.
When θ = 120, the slope = tan θ = tan 120 =-tan 60 =-√ 3.
So choose d
three
Let the inclination of the straight line L be θ ∈ [0, 180] and the slope be k, then k=tanθ.
Similarly, according to A (- 1, -5) and B(3, -2), the slope of straight line AB can be found.
[(-2)-(-5)]/[3-(- 1)]=3/4
According to the topic, the inclination of the straight line AB is twice that of the straight line L, that is, 2θ, so the slope is equal to the sine value of the inclination, as follows: tan2θ=3/4.
tan2θ=sin2θ/cos2θ=2sinθcosθ/[(cosθ)^2-(sinθ)^2]=2tanθ/[( 1-(tanθ)^2]
Namely: 2tanθ/[( 1-(tanθ) 2] = 3/4, remove the denominator and move the term to simplify:
3 (tan θ) 2+8 tan θ-3 = 0, substitute k=tanθ:
3k^2+8k-3=0
This is a quadratic equation about k, which can be solved as follows: k= 1/3, or k=-3.
Given that the inclination of straight line AB is 2θ, then 2θ ∈ [0, 180], so θ ∈ [0,90], then k = tan θ > 0, so k=-3 should be omitted.
Therefore, the slope of the straight line L is k= 1/3.
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