Symmetry formula of point about line

Symmetry formula of point about line

I'm copying it for you. I hope it works
The formula of linear symmetry is as follows:
1. The relation between point (a, b) and line y = KX + m (k = 1 or - 1)
The symmetry point is: (B / K-M / K, Ka + m), which is actually the exchange of the values of X and Y in the expression, because the linear equation y = KX + m has x = Y / K-M / K and y = KX + m, this method is only applicable to K = 1 or - 1
It can also be extended to the symmetric curve f (x, y) = 0 with respect to the line y = KX + M
f（y/k-m/k,kx+m）=0.
2. When k is not equal to 1 or - 1, the symmetric point of point (a, b) with respect to the line ax + by + C = 0 is
(a - (2a * (AA + BB + C)) / (a * a + b * b), B - (2B * (AA + BB + C)) / (a * a + b * b)), can also be extended to the symmetry of the curve with respect to the straight line, where f (x, y) = 0, the symmetry curve with respect to the straight line ax + by + C = 0 is f (x - (2a * (AX + by + C)) / (a * a + b * b), Y - (2B * (AX + by + C)) / (a * a + b * b)) = 0
The above includes all the cases of linear symmetry
By the way, the point about point symmetry is also written here, convenient for everyone to use
The point (x, y) is symmetric with respect to point (a, b), and the point is (2a-x, 2b-y);
The curve f (x, y) = 0 is symmetric with respect to point (a, b), and the curve f (2a-x, 2b-y) = 0

For children to practice

I bought it at home and practiced it every day

The number of roots of equation E ^ - x = log2 ^ x is
To the specific process ah, painting image is OK

Let f (x) = e ^ (- x) - log2 (x)
Because f '(x) = - e ^ (- x) - 1 / (x * LN2), when x > 0, there is obviously f' (x) 0, f (2) = 1 / e ^ 2-1 = (1-e ^ 2) / e ^ 2

As shown in the figure, find the area of the inscribed square ABCD in the ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0)

Guo Dunyong replied:
Let the area of the inscribed square ABCD of the ellipse x2 / A2 + Y2 / B2 = 1 (a > b > 0) be sabcd, then
0.7071b²＜SABCD＜b².

Solving 100 exercises of binary linear equations
As long as the calculation, do not choose and apply!

1.2x+9y=81 3x+y=34 2.9x+4y=35 8x+3y=30 3.7x+2y=52 7x+4y=62 4.4x+6y=54 9x+2y=87 5.2x+y=7 2x+5y=19 6.x+2y=21 3x+5y=56 7.5x+7y=52 5x+2y=22 8.5x+5y=65 7x+7y=203 9.8x+4y=56 x+4y=2110.5x+7...

Given that the function y = f (x) is a decreasing function on R and f (B + 1) > F (2b-3), the value range of real number B is obtained

B + 1 > 2b-3 is 4 > b because R is simple decreasing

Find the eccentricity of major axis and minor axis of ellipse 25X + y = 25

25X quadratic + y quadratic = 25
The formation standard equation is as follows
y²/25+x²=1
The focus is on the y-axis
a²=25,b²=1,c²=a²-b²=24
∴a=5,b=1,c=2√6
The length of long axis is 2A = 10
The length of minor axis is 2B = 2
The focal coordinates are (0, - 2 √ 6), (0,2 √ 6)
Vertex A1 (0, - 5), A2 (0,5)
B1(-1,0),B2(1,0)
Centrifugation e = C / a = 2 √ 6 / 5

A few simple calculation topics!
(1) . 1 - half + one sixth + one twelfth + one twentieth + one thirtieth
(2) 1-quarter-eighth-sixteenth-32nd-64th
(3) 5.34x (multiply) three fifths + 0.6x4.28-0.62x60%
(4) . 1 divided by (1.3 divided by 2.9) divided by (2.9 divided by 9) divided by (9 divided by 2.6)

(1)1-1/2+1/6+1/12+1/20+1/30=1-1/2+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+(1/5-1/6)=1-1/6=5/6

F (x, y) = the negative XY power of Xe, find the derivative FX (1,0), find the derivative FY (1,0)

f(x,y)=x*e^(-xy)
that
fx(x,y)=e^(-xy) +x*(-y) *e^(-xy)=(1-xy)*e^(-xy)
So FX (1,0) = e ^ 0 = 1
and
fy(x,y)=x *e^(-xy) *(-x)= -x² *e^(-xy)
so
fy(1,0)= -1 *e^0= -1

Given the straight line L1: (M + 3) x + 4Y = 5-3m, L2: 2x + (M + 5) y = 8 and L1 parallel L2, find the value of M

L1: from the known formula: 4Y = 5-3m - (M + 3) x, y = 1 / 4 (5-3m - (M + 3) x), K1 = - (M + 3) / 4; L2: from the known formula: (M + 5) y = - 2x + 8, y = (- 2x + 8) / (M + 5), K2 = - 2 / (M + 5); because two straight lines are parallel, K1 = K2, so (M + 3) / 4 = 2 / (M + 5); from the above formula: (M + 1) (M + 7) = 0, so