The characteristics of slope formula k = (y1-y2) / (x1-x2) and (x1 ≠ x2) are analyzed. It is known that a (2,4) and B (3,3) points P (a, b) are the moving points on line AB (including the end points), then what is the value range of (B-1) / (A-1)

The characteristics of slope formula k = (y1-y2) / (x1-x2) and (x1 ≠ x2) are analyzed. It is known that a (2,4) and B (3,3) points P (a, b) are the moving points on line AB (including the end points), then what is the value range of (B-1) / (A-1)


B-1 / A-1 can be regarded as the slope between the fixed point P and the fixed point Q (1,1), so the range of the slope can be calculated
kAQ=(4-1)/(2-1)=3
kBQ=(3-1)/(3-1)=1
So the range of B-1 / A-1 is 1



How to determine the slope formula y2-y1 / x2-x1 = K (x1, Y1) and (X2, Y2)
It's not a slope. Let's take an example


1. The first is to draw a triangle
Y2-y1 is the direction of Y axis, which is the opposite side
X2-x1 is the direction of the X axis, which is the adjacent edge
Slope formula = Tan β = (y2-y1) / (x2-x1)
It should be noted that the order of (x1, Y1) and (X2, Y2) should be consistent, that is, which of (y2-y1) and (x2-x1) should be put in front when subtracting



A 120cm long wire is divided into two parts, each part is bent into a square, what is the sum of their areas?
Others list it as follows:
The side length of one of the squares is X
S=x^2+(30-x)^2
Why 30-x? How


Because a 120cm long wire is divided into two parts, each part is bent into a square
So the sum of each of the two squares is 30, and the side length of one of the squares is X
The other square has a side length of 30-x, of course~



2A = m + n deduce 4A ^ 2 = m ^ 2 + n ^ 2 + 2Mn greater than or equal to 2 (m ^ 2 + n ^ 2) why?


Wrong, less than or equal to
2a=m+n
Then (2a) &# 178; = (M + n) &# 178;
That is 4A & # 178; = M & # 178; + 2Mn + n & # 178;
And 2 (M & # 178; + n & # 178;) - (M & # 178; + 2Mn + n & # 178;) = M & # 178; - 2Mn + n & # 178; = (m-n) & # 178; ≥ 0
SO 2 (M & # 178; + n & # 178;) ≥ M & # 178; + 2Mn + n & # 178;



As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 °. D is the midpoint of BC, de ⊥ AB is proved at point E: EB = 3EA


It is proved that: ∵ AB = AC, ∵ BAC = 120 ∵ B = ∵ C = 30 ∵ D is the midpoint of BC ∵ ad ⊥ BC and ad bisects ∵ BAC, ∵ bad = 60 ∵ ADB = 90 ∵ ad = 12ab and