100 questions on solving binary linear equations

Formula of distance between two parallel lines

The line with slope k intersects with curve y = LNX at a (x1, Y1), B (X2, Y2) (x1 < x2) to prove x1

Let f (x) be continuous in [a, b] and differentiable in (a, b), then there exists x belonging to (a, b), such that [f (b) - f (a)] / [B-A] = f '(x). The proof is as follows: Let f (x) = LNX, x > 0, obviously it is continuous and differentiable in x > 0

How is the mathematical slope formula (y2-y1) / (x2-x1)? What is the slope formula (y1-y2) / (x1-x2)?

The two formulas are the same principle. For the same problem, the slope calculated by the two formulas is the same. For example, the two points are (3,5), (7,6). When X1 = 3, Y1 = 5, X2 = 7, y2 = 6, then the slope k = (y1-y2) / (x1-x2) = (- 1) / (- 4) = 1 / 4 = (y2-y1) / (x2-x1), so the two formulas are the same

Can the slope formula k = y2-y1 / x2-x1 use k = y1-y2 / x1-x2? Why do the gods help

Because the first point and the second point can form a straight line, the slope formula can only be obtained by subtracting the left side of the second point from the coordinates of the first point, which can be y2-y1 / x2-x1 or y1-y2 / x1-x2

The formula of quadratic function for | y1-y2 | or | x1-x2 | without slope?
(y1-y2) ^ 2 = (A / b) ^ 2-4c / A is there such a formula?

│x1-x2│=√[（x1+x2）²-4x1x2]
=√[（b/a）²-4c/a]

How to find the common chord of two intersecting circles, for example, r = root 50, R2 = root 10, d = 2 times root 5

The two circular equations are multiplied by appropriate multiples respectively, and then subtracted to eliminate the quadratic terms of X and Y. the remaining quadratic terms are the common chords

The characteristic equation R ^ 2-6r-9 = 0, R1 = 3-3 radical 2, R2 = 3 + 3 radical 2. How can we get this?
And R ^ 2 + 4 = 0, R1 = 2I, R2 = - 2I
And R ^ 2-4r + 13 = 0, R1 = 2-3i, R2 = 2 + 3I, how are these equations obtained?

Solution: R ^ 2-6r-9 = 0, R ^ 2-6r = 9 (transposition) R ^ 2-6r + 9 = 18 (complete square) (R-3) ^ 2 = 18, R-3 = plus or minus 3 times radical 2 + 3, so R1 = 3-3 radical 2, R2 = 3 + 3 radical 2 solution: R ^ 2 + 4 = 0, R ^ 2 = - 4 (transposition) r = plus or minus 2I

Root 3 belongs to {x | x = root 2 + A, root 3, a belongs to R}
Is that right? Why

Yeah
x=√2+a√3=√3
Then as long as a = (√ 3 - √ 2) / √ 3
A is a real number
So that's right

|z1+z2+z3+.+zn|

Maybe I can only explain, not strictly prove
First of all, I want to say what your Z1 and Z2 are. I first understand real numbers, then through observing the number axis, there is: if Z1, Z2 have the same number or one is 0, then | Z1 + Z1 + Z2

100 questions on solving binary linear equations