13. The distance between two parallel lines 3x + 4y-12 = O and ax + 8y + 11 = o

13. The distance between two parallel lines 3x + 4y-12 = O and ax + 8y + 11 = o

Because it is parallel, so: A / 3 = 8 / 4
A=6
The equation is: 6x + 8y + 11 = 0
3x+4y+5.5=0
The distance is: D = | - 12-5.5 | / radical (3 ^ 2 + 4 ^ 2) = 3.5
In parallel, a = 6
The distance is 23 / 10
370116 - the distance of level 18 is wrong??
First, 6x + 8y-24 = O and 6x + 8y + 11 = o,
The distance is 23 / 10
① (T + 1) (T-3) = - t (3-3t) ② 2 (X & # 178; - x-1) + 4x = 5x use formula method to solve the equation
t²-2t-3=-3t+3t²
2t²-t+3=0
a=2,b=-1,c=3
So △ = B & # 178; - 4ac = - 23
1) Simplifying the original equation
t^2-3t+t-3=-3t+3t^2
It is concluded that 2T ^ 2-T + 3 = 0
a=2,b=-1,c=3
∆=b^2-4ac=(-1)^2-4*2*3=-23
Solution equation: (5x-1) &# 178; = 19-5x
Two
25X +1-10X=19-5X
Transfer and merge
Then two formulas (- B b2-4ac / 2a) are used
If the images of the functions y = ax + 2 and y = bx-3 intersect at the same point on the X axis, then ab=___ .
In the function y = ax + 2, let y = 0, the solution is x = - 2A; in the function y = bx-3, let y = 0, the solution is x = 3B; since two functions intersect at the same point of X axis, then - 2A = 3B. That is: ab = - 23
2 (y + 1) - 6y-5 ≥ 1 solution inequality
2(y+1)-6y-5≥1
2(y+1)-6y-5≥1 -4y-3>=1 -4y>=4 y
Given that X and y are real numbers, X & # 178; + XY + Y & # 178; = 1, find the value range of X & # 178; - XY + Y & # 178
x²+xy+y²=1
1=x²+y²+xy≥2|xy|+xy
If XY ≥ 0, then 1 ≥ 3xy, then XY ≤ 1 / 3
XY
solution
It can be set that
x=a+b, y=a-b, (a,b∈R)
Let's set the conditional equation,
3a²+b²=1
z=x²-xy+y²=a²+3b²
The problem can be reduced to, when 3A & # 178; + B & # 178; = 1,
Find the range of Z = A & # 178; + 3B & # 178.
3Z = 3A and 178;... Expansion
solution
It can be set that
x=a+b, y=a-b, (a,b∈R)
Let's set the conditional equation,
3a²+b²=1
z=x²-xy+y²=a²+3b²
The problem can be reduced to, when 3A & # 178; + B & # 178; = 1,
Find the range of Z = A & # 178; + 3B & # 178.
3z=3a²+9b²=1+8b²≥1
∴z≥1/3
z=a²+3(1-3a²)=3-8a²≤3
1 / 3 ≤ Z ≤ 3
Only if x ^ y ≥ 2 (x ^ y + 2) / y is equal
∴3(x^2+y^2)/2≧x^2+xy+y^2=1
1 = x ^ 2 + XY + y ^ 2 ≥ 3xy (take the equal sign if and only if x = y)
∴x^2+y^2≧2/3 xy≦1/3 ∴-xy≧-1/3
Ψ x ^ 2-xy + y ^ 2 ≥ 2 / 3... Expansion
∵ x ^ 2 + y ^ 2 ≥ 2XY ∵ (x ^ 2 + y ^ 2) / 2 ≥ XY (take the equal sign if and only if x = y)
∴3(x^2+y^2)/2≧x^2+xy+y^2=1
1 = x ^ 2 + XY + y ^ 2 ≥ 3xy (take the equal sign if and only if x = y)
∴x^2+y^2≧2/3 xy≦1/3 ∴-xy≧-1/3
Ψ x ^ 2-xy + y ^ 2 ≥ 2 / 3-1 / 3 = 1 / 3 (take the equal sign if and only if x = y)
So the value range of x ^ 2-xy + y ^ 2 is [1 / 3, + ∞)
It is known that P is the moving point on the straight line 3x + 4Y + 8 = 0, PA and Pb are the two tangent lines of the circle x2 + y2-2x-2y + 1 = 0, a and B are the tangent points, C is the center of the circle, then the minimum area of PACB of quadrilateral is______ .
∵ the equation of circle is: x2 + y2-2x-2y + 1 = 0 ∵ center C (1,1), radius R is: 1. According to the meaning, if the area of quadrilateral is the smallest, when the distance between the center of circle and point P is the smallest, and the distance between the center of circle and straight line is the distance between the center of circle and straight line, the tangent length PA, Pb is d = 3 ∵ PA | = | Pb | = D2 − R2 = 22 ∵ spacb = 2 × 12 | PA | r = 22, so the answer is: 22
As shown in the figure, given that the intersection point of graph of functions y = x + B and y = ax + 3 is p, then the solution set of inequality x + b > ax + 3 is ()
A. x＜1B. x＞1C. x≥1D. x≤1
The image intersection point of the functions y = x + B and y = ax + 3 is p, and the abscissa of P point is 1. According to the image, when x > 1, the image of the function y = x + B is on the top of the image of the function y = ax + 3, then the value of the function y = x + B is greater than the value of the function y = ax + 3, that is, the solution set x > 1 of the inequality x + b > ax + 3
Solving inequality - 6y ^ 2 + 7Y + 5 > 0
-6y²+7y+5>0
6y²-7y-5
-6y^2+7y+5>0
6y^2-7y-5
Let X and y be real numbers, and the square of X + XY + y = 1, find the range of the square of X - XY + y
ditto
Let z = x ^ 2-xy + y ^ 2. And 1 = x ^ 2 + XY + y ^ 2. Add and subtract the two formulas respectively, and get x ^ 2 + y ^ 2 = (Z + 1) / 2. Z-1 = - 2XY. Add and subtract the two formulas again, and get (3z-1) / 2 = (X-Y) ^ 2 ≥ 0. (3-z) / 2 = (x + y) ^ 2 ≥ 0. = = = = = > Z ≥ 1 / 3. Z ≤ 3.. = = = = = > 1 / 3 ≤ Z ≤ 3
[－1，1]