Who has a more comprehensive formula for the plane Cartesian coordinate system, which can be used in

There are a lot of knowledge points.

1.The basic form of the proportional function is: y=kx (k≠0).

When k>0, the image (straight line) crosses the first triangular limit;

k<0, the image passes the second quadrant.

2.The basic form of a primary function is: y=kx+b(k≠0, k and b are constants) k>0, b>0, the image (straight line) passes the first, two, and third quadrants;

When k>0 and b<0, the image passes the first, third, and fourth quadrants;

When k<0 and b<0, the image passes the second, third, and fourth quadrants;

k<0,b>0, the image crosses the 1st, 24th quadrants.

When k in two primary functions is equal and b is not equal, their images are parallel to each other;

When the product of k in two primary functions is multiplied by -1, their images are perpendicular to each other.

3.The basic expression of the inverse proportional function is: y=k x. (k≠0).

At k>0, the image (hyperbola) is located in the first and third quadrants, respectively.

At k<0, the image is in the second and fourth quadrants, respectively.

4.The basic expression for a quadratic function is: y=ax +bx+c(a≠0, a, b, c are constants).

a>0, the image (parabolic opening up);

a<0, the image opening is downward.

When a and b have the same sign, the axis of symmetry of the image is to the left of the y-axis;

When the a,b sign is different, the axis of symmetry of the image is to the right of the y-axis.

c>0, the image and the y-axis intersect on the positive semi-axis;

At c<0, the image intersects the y-axis on the negative half-axis.

5.If point a is (m,n), point b is (p,q).

The abscissa of the midpoint of the line segment AB is: (m+p) 2, and the ordinate is (n+q) 2

The length of the line segment AB is: [m-p) +n-q) ]

Midpoint formula: (x1+x2) 2,(y1+y2) 2 y=kx+b, find the formula for k: k=(y1-y2) (x1-x2) The distance between two points in the plane:

x1-x2) +y1-y2) length of the upper segment of the x-axis: δa

The formula is dead, learning is alive, slowly accumulating, eating a fat man in one bite? Not that way. Take it easy.

(1) 2 times (0, 4) (1, 2) (2, 0).

3 times (0, 6) (1, 4) (2, 2) (3, 0).

3) Let the point p be translated to the right x times from the point o, each time 1 unit length, then the upward translation is n - x times, each time 2 unit length.

When we reach the point q on the line y=x, we know that the coordinates of the point q are ( x,2(n-x) ) where 0 x n

Since the point q is on the straight line y=x, then there is: x=2(n-x), i.e., 3x=2n, we know that x is an even number.

So the path length of the translation is equal to x+2(n-x)=2x

If the length of the path of the translation is known to be not less than 50 and not more than 56, then:

50≤2x≤56

Solve 25 x 28

Since x is an even number, so:

It can be seen that x = 26 or 28

When x=26, n=39 is obtained from 3x=2n, which is in line with the title, and the coordinates of the point q are (26,26);

When x=28, n=42 is obtained from 3x=2n, which is in line with the title, and the coordinates of the point q are (28, 28).

(1)xy<0

The abscissa ordinate is different from the abscissa ordinate, m is in the , quadrant, and is not on the coordinate axis.

2)x+y=0

i.e. y=-x, m is on the bisector of the angle of the quadrant , .

3)x/y=0

x is always equal to 0 and y is not equal to 0, so it is on the y axis, but not at the coordinate origin.

1. If the bottom edge is BC

then the height is 1bc=8,c(-2,0).

2. If the bottom edge is AC

and let c(1,x).

It can be seen that there is 5(x+1) 2=4

x=c(1, none of which are in line with the topic.

3. The bottom edge is AB

c on the y-axis.

Let c(0,y).

Make a rectangle through a, b, and c.

Then there is an equation. 4=[5(y+1)-(5y+y+1+4)/2]y=2c(0,2)

q(4,-2),r(1,-3), the abscissa of the midpoint is equal to half of the sum of the abscissa of the two endpoints of the corresponding line segment, and the ordinate of the midpoint is equal to half of the sum of the ordinates of the two endpoints of the corresponding line segment.

If the coordinates of the two ends of a line segment are (a, b), (c, d), then the coordinates of the midpoint of the segment are ((a+c) 2,(b+d) 2).

It can be seen that ab is on the x-axis, and the ordinates of cd are equal to 1, so ab cd, which is a trapezoid.

The upper bottom of the trapezoid cd = 2, the lower bottom ab = 6, and the height is 1.

So, the area is.

Thank you for adopting

The big triangle on the left - the small triangle on the bottom + the triangle on the right = [(1*5) 2-1*1 2] + 2*(5-1) 2=6

Find a c(0,a) on the y-axis and see ab as the bottom, can't you use the formula, s=, get a=3

That's his y is 3, x is arbitrary.

According to the idea: (1) ab=4, to make the abc area 6, let the height of the triangle be h, then: 2h=6, that is, h=3

There are two points on the y-axis that meet the requirements: c has coordinates of (0,3) or (0,-3).

2) From the above, as long as the distance from the axis where AB is located is 3, all are satisfied, so there are infinite points that meet the conditions. These points form two straight lines, i.e., y=3 and y=-3

(1) Let c (an infinite number of bars, all on the straight lines y=3 and y=-3.

d is chosen because when the axial is symmetrical with respect to the x-axis, the value of the abscissa does not change, and the value of the ordinate becomes the opposite number, so b(2,2).

When the symmetry about the origin is the abscissa, the ordinate and the ordinate must become opposites, so the symmetry of the point b about the origin is the coordinate of c.

Hope mine is helpful to you.

If the symmetry of point A with respect to the x-axis is B, then B(2,2), only the ordinate is negated.

The symmetry of point b with respect to the origin point is c, then c(-2, -2), and the horizontal and vertical coordinates are inverted d d

The coordinates of point B are (2,2) and the coordinates of point C are...

Haha, kid, think for yourself in your own homework.