In the plane rectangular coordinate system, the coordinates of the three vertices of square ABCD are a (0,0), B (- 2,0), D (0,2), then the coordinates of point C are Then the coordinates of the midpoint of the square BC are 2. If point P (a, 3-2a) is on the bisector of the second and fourth quadrants, then the value of a is

C is (- 2,2)
Because (2 + 0) / 2 = 1
So the midpoint of BC is (- 2,1)
24 bisector, then a and 3-2a are opposite numbers
So a + 3-2a = 0
a=3

As shown in the figure, in the plane rectangular coordinate system xoy, the vertex a of the square ABCD is on the positive half axis of the Y axis, the vertex B is on the positive half axis of the X axis, and the vertices C and D are in the first quadrant. We know a (0,4), B (m, 0). (1) find the coordinates of the vertices C and D; (2) when the point B moves, the point C moves on a straight line. Please write out the analytical formula of the straight line

(1) Make CE ⊥ X axis intersect X axis at e point, make DF ⊥ Y axis intersect Y axis at f point, the coordinates of ≁ AOB ≌ CBE, ≌ C point are: (M + 4, m); ≌ AOB ≌ ADF, ≌ D point are (4, M + 4); (2) B (m, 0) and C (M + 4, m). The analytical formula of straight line BC is: y − 0m − 0 = x − mm + 4 − m, sorted out: y = x − M4 M

It is known that the two vertices of square ABCD are on the parabola y = x ^ 2 + C, and the other two points c and D are on the X axis. The area of square ABCD is equal to 4. Find the analytical formula of parabola!

The square has an area of 4
So BC = 2, OC = 1
So the coordinates of point B are (1, - 2)
Substituting (1, - 2) into
C = - 3
So the analytic formula is y = x & sup2; - 3

The vertex A and B of square ABCD are on the parabola y = X2, C and D are on the straight line y = x-4
The answer is 3 √ 2 or 5 √ 2

Let the linear equation of AB be y = x + K (k > - 4)
So the side length is (K + 4) / √ 2
Together again
Y = x + K and y = x ^ 2
yes:
x^2-x-k=0
Let two be x1, x2
Then the side length is √ 2 | x1-x2 | = √ (2 + 8K)
So there are
√(2+8k)=(k+4)/√2
The solution is k = 2 or 6
So the side length is 3 √ 2 or 5 √ 2

In square ABCD, one edge AB is on the straight line y = x + 4, and the other two vertices C and D are on the parabola y2 = X

Let the equation of the straight line where CD is located be y = x + T, ∵ y = x + Ty2 = X. by eliminating y, X2 + (2t-1) x + T2 = 0, ∵ CD | 2 [(1-2T) 2-4t2] = 2 (1-4t), and the distance between AB and CD is | ad | = | T-4 | 2, ∵ ad | = | CD |, | t = - 2 or - 6, so that the side length is 32 or 52. Area S1 = (32) 2 = 18, S2 = (52) 2 = 50

The two vertices a and B of square ABCD are on the parabola y ^ 2 = x, and the two vertices C and D are on the straight line y = x + 4

C. If D is on the straight line y = x + 4, a and B must be on the straight line y = x + B (undetermined coefficient), the chord length formula can solve the length of line AB, the parallel line distance formula can solve BC, and the simultaneous solution can solve B, then the side length (that is, the distance between two straight lines) can be calculated
Do it yourself To remember the solution)

Vertex a of square ABCD is on line Mn, point O is the intersection of diagonal AC and BD, passing through point o as OE ⊥ Mn at point E, passing through point B as BF ⊥ Mn at point F
(1) As shown in Figure 1, when two points o and B are above the straight line Mn, it is easy to prove that AF + BF = 2oe (no need to prove) (2) when the square ABCD rotates clockwise around point a to the position of Figure 2 and figure 3, what is the relationship among the line segments AF, BF and OE? Please write your conjecture directly and choose a case to prove it

(1) As shown in the figure, BG ⊥ OE is made at point B as BG ⊥ oein g, then the quadrilateral bgef is rectangular, and {EF = BG, BF = Ge, in the square ABCD, in the square ABCD, OA = ob, AOB = 90, \⊥ o pointb is BG ⊥ oein g, then the quadrilateral bgef is rectangle, and {EF = BG, BF = Ge, in the square ABCD, in the square ABCD, in the square ABCD, in the square, in the square ABCD, in the square ABCD, OA = ob, and in the square ABCD, OA = ob, and △ AOE and △ Obg, in △ AOE and △ Obg, in △ AOE \\\\\\\\\\\\bg, ∵ af-ef = AE, EF = BG = OE, AE = og = oe-ge = oe-bf, ∵ af-oe = oe-bf, ∵ AF + BF = 2oe; (2) conclusion of Figure 2: AF-BF = 2oe, conclusion of Figure 3: bf-af = 2oe. For Figure 2, it is proved that if the extension line of BG ⊥ OE intersection OE is at g, then the quadrilateral bgef is rectangular, ∵ EF = BG, BF = Ge, in square ABCD, OA = ob, ∵ AOB = 90 °, ∵ BG ⊥ OE, ∵ Obg + ∠ BOE = 90 °, and ∵ AOE + ∠ BOE = 90 ° AOE = Obg, in △ AOE and △ Obg, \\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\﹥ goe = 90 °, ﹥ AOE + ﹥ AOG = 9 In the square ABCD, in the square ABCD, in the square ABCD, OA = ob, AOB = 90, AOB = 90, AOG + bog = 90, AOE = bog, and AOE = bog \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

As shown in the figure, in the square ABCD with side length of 2, the coordinates of vertex a are (0, 2). When the image l of the first-order function y = x + T changes with different values of T, the area of the figure surrounded by L and the edge of the square is s (shadow). (1) when t is what value, s = 3; (2) in the plane rectangular coordinate system, draw the function image of S and t

As shown in the figure. (1) let l intersect the edges AD and CD of the square at M and N, and it is easy to prove that RT △ DMN is an isosceles triangle. Only when MD = 2, the area of △ DMN is 1, and T = 4-2 is obtained. It is easy to verify that s = 3. When t = 4-2, s = 3; (2) when 0 ≤ T < 2, s = 12t2; when 2 ≤ T < 4, s = - 12 (4-T) 2 + 4; when t > 4, s = 4. According to the above analytical formula, draw the figure

As shown in the figure, the side length of square ABCD is 10, and the symmetry centers of four congruent small squares are respectively on the vertex of square ABCD, and their sides are parallel or perpendicular to each side of square ABCD

Fourth, I can't see clearly \First of all, the area of a shadow (x / 2) ^ 2 / x0d has a total of shadows, so y = (x / 2) ^ 2 * 4 = x ^ 2 / x0d, so it is a quadratic function image, and 0 & lt; X & lt; = 10,0 & lt; Y & lt; = 100

Two long strips of paper of equal width are obliquely overlapped, trying to explain that the overlapping part ABCD is a diamond

∵AB‖CD
The quadrilateral ABCD is a parallelogram
∵ s quadrilateral ABCD = CD * H1 (one strip width)
=BC * H2 (another note width)
And ∵ H1 = H2
∴CD=BC
The quadrilateral ABCD is a diamond