On the point of intersection of the symmetric axis of parabola and the axis of parabola 1. If the point q is on the symmetric axis of the parabola, the point P is on the parabola, and the quadrilateral with points c, C ', P and Q as its vertex is a parallelogram, the coordinates of point Q and point P can be obtained (which can be expressed by an algebraic formula containing m) 2. Find out the perimeter of parallelogram

First of all, we need to consider how many kinds of parallelogram with points c, C ', P, Q as vertices? Or what conditions can be satisfied to show that a parallelogram can be formed with points c, C', P, Q as vertices. From the viewpoint of vector, a parallelogram can be formed with points c, C ', P, Q as vertices if and only if the vector CP = c'q (CC' is an edge
One
The solution of the equation: 3 (3x-2) = 10-0.5 (x + 3.5) x + 5Y = 21 9x + 2Y = 15 X-Y = 3 3x + 4Y = 10 3x + 12.5 = 5x + 7.7 4 (x + 5) = 80-x
1.2.5(x-2)-x+2
2.3（3X-2）=10-0.5（X+3.5）
3.{ X+5Y=21
{ x-y=3
4.{9x+2y=15
{3x+4y=10
5.3x+4y=23
6.3x+12.5 =5x+7.7
7.4（x+5）=80-x
3（3X-2）=10-0.5（X+3.5）X+5Y=21x=21-5Y 9x+2y=15 4.3x+12.5 =5x+7.7 5.4（x+5）=80-x
The parabola y = (m ^ 2 + 4m + 3) x ^ 2 - (3m ^ 2 + m-2) x + 1. When m =, the parabola takes the Y axis as the symmetry axis
When the parabola takes Y-axis as symmetry axis, (3m ^ 2 + m-2) = 0
The solution is m = - 1 or 2 / 3
And because m ^ 2 + 4m + 3 is not equal to 0, m cannot be equal to - 1
That is, when m = 2 / 3, the parabola takes the y-axis as the symmetry axis
{(2x + 5Y) 2 = 36 {5 (3x + 2Y) = 80 solution equation
｛（2x+5y)2=36
｛5（3x+2y）=80
（2x+5y)2=36
5（3x+2y）=80
Simplification: 2x + 5Y = 18 (1)
3x+2y=16 （2）
(1) 3 - (2) × 2
15y-4y=54-32
11y=22
Y=2
Substituting y = 2 into (1) yields:
2x+5×2=18
X=4
∴ x=4
Y=2
That is 4x + 10Y = 36 (1)
15x+10y=80 (2)
(2)-(1)
15x-4x=80-36
11x=44
X=4
y=(36-4x)/10=2
X=4，Y=2
It is known that the parabola y = x2 + 2x + m has and only has one common point [1] with the x-axis to find the value of M and the symmetry axis of the parabola,
【2】 If P [n, Y1], Q [2, Y2] is a point on the parabola, and Y1 is greater than Y2, the value range of real number n is obtained
(1) Y = x2 + 2x + M = (x + 1) 2 + M-1, the axis of symmetry is a straight line, x = - 1, ∵ and X-axis have only one common point, the ordinate of the vertex is 0, and the vertex coordinate of the vertex is (- 1,0); (2) let the functional expression of C2 be y = (x + 1) 2 + K, substituting a (- 3,0) into the above expression, we can get (- 3 + 1) 2 + k = 0, k = - 4, ∵ C2's
Solve the binary linear equations: 3x + 5Y = 82x − y = 1
3x + 5Y = 8 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① 2x − y = 1 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ②, ② × 5 + ① is: 13X = 13, X = 1, substituting x = 1 into ② is: 2 × 1-y = 1, y = 1, so the solution of the equations is: x = 1y = 1
Parabola y = (M-10) x ^ 2-2mx-3m-1, please prove that when m takes different values, the parabola will pass through two fixed points, and calculate the two points
y=(m-10)x^2-2mx-3m-1=mx^2-10x^2-2mx-3m-1=m(x^2-2x-3)-10x^2-1=m(x+1)(x-3)-10x^2-1,
When x = - 1 or x = 3, no matter what real value m takes, y = - 11 or - 91,
When two parabolas pass through a fixed point, they will be different
Let this formula be equal to zero - Open - extract m to get: m (x2-2x-3) - 10x2-1 = 0, then your formula seems to be wrong
m-1>0,m>1.....1)
y=(m-1)x2+2mx+3m-2
=(m-1)[x+m/(m-1)]^2+(2m^2-5m+2)/(m-1)
(2m^2-5m+2)/(m-1)=0
(2m-1)(m-2)=0
M = 1 / 2 or M = 2... 2)
Total 1), 2)
The value of M is m = 2
y=(m-10)x^2-2mx-3m-1
=m(x^2-2x-3)-10x^2-1
That is, the coefficient of M should be 0
x^2-2x-3=0
x1=-1,x2=3
Find y,, and the vertex is
(3,-91)
(-1,-11)
Let x = 3 have y = 9 (M-10) - 6m-3m-1 = - 91, so it passes through the fixed point (3, - 91)
Let x = - 1 have y = (M-10) + 2m-3m-1 = - 11, so it passes the fixed point (- 1, - 11)
So after two fixed points (3, - 91) (- 1, - 11)
By sorting out M, we can get
y=(x^2-2x-3)m-10x^2-1
It can be seen that when x ^ 2-2x-3 = 0, no matter what value m takes, it has no effect on y. therefore, the two vertices must be the points where x ^ 2-2x-3 = 0. By solving this equation, the roots of X are - 1 and 3, that is, the fixed points are (- 1, - 11) and (3, - 91).
Solve the equations 1.3x-2y = 1 2x-5y = - 25
2.3x-2y=4,2x-3y=-5
(1) If 3x-2y = 1 × 5, then (3) 15x-10y = 5
(2) 2x-5y = - 25 × 2 to get (4) 4x-10y = - 50 minus (3)
11x=55
Substituting x = 5 into (1)
2y=3x-1=14
Y=7
(1) If 3x-2y = 4 × 3, then (3) 9x-6y = 12
(2) 2x-3y = - 5 × 2 to get (4) 4x-6y = - 10 minus (3)
5x=22
X = 22 / 5 to (1)
2y=3x-4=3*22/5-4=46/5
y=23/5
The axis of symmetry of the parabola y = (M2-2) x2 + 2mx + 1 with downward opening passes through the point (- 1,3), and the value of M is obtained
∵ the symmetric axis of the parabola y = (M2-2) x2 + 2mx + 1 with downward opening passes through the point (- 1,3), ∵ - 2M2 (M2-2) = - 1, M2-2 < 0, and the solution is: M1 = - 1, M2 = 2 (not suitable for the problem), and∵ M = - 1
Use the substitution elimination method to solve the following binary linear equations: 2x + 3Y = 5, 4x-y = 3 tonight!
2x+3y=5①
4x-y=3 ②
Change ② to y = 3-4x ③
③ Substitute ① for 2x + 3 (3-4x) = 5
2x+9-12x=5
-10x=-14
x=1.4
The second formula moves to y = 4x-3, brings in 2x + 12x-9 = 5, x = 1, y = 1

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