The distance from point (1,1) to line L: 3x + 4Y + 3 = 0 is______ ．

The distance from point (1,1) to line L: 3x + 4Y + 3 = 0 is______ ．

According to the formula of distance from point to line, we can get the distance from (1,1) to line L: 3x + 4Y + 3 = 0 d = | 3 × 1 + 4 × 1 + 3| 32 + 42 = 105 = 2

What is the distance from the origin to the line 3x + 4Y = 5

The distance between point P (m, n) and straight line ax + by + C = 0 is d = | am + BN + C | / √ (A & # 178; + B & # 178;)
In this problem, that is to find the distance from the point (0,0) to the straight line 3x + 4y-5 = 0, where a = 3, B = 4, C = 5, directly replace the formula

The result of simplifying [(xx-2xy + YY) (xx-2xy + YY) - xxxxx + yyyy] / (XX YY) (XX YY) is

The original formula = [(X-Y) & sup2; (X-Y) & sup2; - (X & sup2; + Y & sup2;)] / (X & sup2; - Y & sup2;) & sup2; = [(X & sup2; - 2XY + Y & sup2;) - (X & sup2; + Y & sup2;)] / (X & sup2; - Y & sup2;) = - 2XY / (X & sup2; - Y & sup2;)

2 / 3Y square = 4Y + 1 with matching method!

2 / 3Y & # 178; = 4Y + 1
3y²/2-4y=1
9y²-24y=6
9y²-24y+16=6+16
(3y+4)²=22
3y+4=±√22
y=(-4±√22)/3

1/4y^2-y+1=0
Factorization or direct square root process

1/4y^2-y+1=0
y^2-4y+4=0
(y-2)^2=0
therefore
y=2

Evaluation, to speed.. 1-4y ^ 2 > = 0

Divide both sides by - 4
y²-1/4

Formula method for 4Y ^ 2 - (8 + √ 2) y √ 2 = 0

4y^2-(8+√2)y √2=0
(4y-8√2-2)y=0
(2y-4√2-1)y=0
Y1 = 0 or y2 = (4 √ 2 + 1) / 2

Using the collocation method to solve the equation: 3Y ^ 2 + 1 = 2 √ 3Y

Upstairs friends are familiar with the formula to solve problems, rather than with the method
The steps of the method are as follows:
1. By using the same solution equation, the constant term is moved to the right of the equation, and then the coefficient of the quadratic term is changed to 1;
2. Add the square of half of the coefficient of the first term to both sides of the equation;
3. Let the left side be written as the square of binomial. If the right side of the equation is nonnegative, the equation can be solved by the open method
Example: using the collocation method to solve the equation: 3Y ^ 2 + 1 = 2 √ 3Y
The original formula is rewritten as: y ^ 2-2 √ 3 / 3 * y = - 1 / 3
On both sides of the equation, add the square of half of the coefficient of the first term: y ^ 2-2 √ 3 / 3 * y + (√ 3 / 3) ^ 2 = - 1 / 3 + (√ 3 / 3) ^ 2
Square of binomial on the left: (Y - √ 3 / 3) ^ 2 = 0
Using the open method to solve the equation: Y1 = y2 = √ 3 / 3

Solving equation 3y-4y = - 25-20

-y=-45
y=45

Solving equation 1 / 4y-y + 1 = 0 by factorization

Is 1 / 4Y & # 178; - y + 1 = 0
(1/2y-1)²=0
1/2y-1=0
y=2