Given the point a (- 4,0) B (2,0), if the point is on the image of the linear function y = - 2 / 1X + 2 and the triangle ABC is a right triangle, several conditions of point C are satisfied

Given the point a (- 4,0) B (2,0), if the point is on the image of the linear function y = - 2 / 1X + 2 and the triangle ABC is a right triangle, several conditions of point C are satisfied

There are four:
Taking a as a right angle, the intersection point of the vertical line of X axis and the original straight line through a is C;
Similarly, C can be found by taking B as a right angle;
If there are two points of intersection between the circle and the original line C, take the two points of the circle as the intersection point
To sum up, there are 4 C satisfying the condition

Given that the images of the first order functions y = 2x + B and y = - x + a pass through a (0, - 4) and intersect B and C with X axis respectively, then the area of △ ABC should be () A. 13 B. 14 C. 11 D. 12

Put a (0, - 4) into the first order function y = 2x + B and y = - x + A, respectively
When y = 0, the two lines intersect with X axis,
2x-4=0，x=2；
-x-4=0，x=-4；
Therefore, if the coordinates of B and C are B (2,0) and C (- 4,0), then BC = | - 4-2 | = 6,
OA = | - 4 | = 4, then the area of △ ABC should be 1
2×BC×OA=1
2×6×4=12．
Therefore, D

Given that the images of the first order functions y = 2x + B and y = - x + a pass through a (0, - 4) and intersect B and C with X axis respectively, then the area of △ ABC should be () A. 13 B. 14 C. 11 D. 12

Substituting a (0, - 4) into the first order function y = 2x + B and y = - x + A, we get: B = - 4, a = - 4, when y = 0, two straight lines intersect the X axis, 2X-4 = 0, x = 2; - x-4 = 0, x = - 4; therefore, the coordinates of B and C are B (2, 0), C (- 4, 0), then BC = | - 4-2 = 6, OA = | - 4 = 4, then the area of △ ABC should be 12

As shown in the figure, in the plane rectangular coordinate system, it is the coordinate origin, the quadrilateral oabc is a rectangle, and the coordinates of points a and B are a (- 4,0), B (- 4,2)

c（0,2）

It is known, as shown in the figure: in the plane rectangular coordinate system, O is the coordinate origin, the quadrilateral oabc is a rectangle, the coordinates of points a and C are a (10, 0), C (0, 4), respectively; point D is the midpoint of OA, and point P moves on the edge of BC; when △ ODP is an isosceles triangle with waist length of 5, the coordinates of point P are______ .

(1) When od is the bottom of an isosceles triangle, P is the intersection of the vertical bisector of OD and CB, and op = PD ≠ 5;
(2) When od is one waist of isosceles triangle:
① If point O is the vertex of the vertex angle, point P is the intersection point of the arc with point o as the center of the circle and radius of 5 with CB,
In right angle △ OPC, CP=
OP2-OC2=
If 52-42 = 3, then the coordinates of P are (3, 4)
② If D is the vertex of the vertex angle, point P is the intersection point of the arc with point D as the center of the circle and radius of 5 with CB,
D as DM ⊥ BC at point M,
In right angle △ PDM, PM=
PD2-DM2=3，
When p is on the left of M, CP = 5-3 = 2, then the coordinates of P are (2,4);
When p is on the right side of M, CP = 5 + 3 = 8, then the coordinates of P are (8, 4)
So the coordinates of P are: (3,4) or (2,4) or (8,4)
So the answer is: (3,4) or (2,4) or (8,4)

As shown in the figure, in the plane rectangular coordinate system xoy, the coordinates of the vertex B of the rectangular oabc are (6, 4), and the straight line y = - x + B exactly divides the rectangular oabc into two parts with equal area, then B=______ .

∵ the straight line y = - x + B exactly divides the rectangular oabc into two parts of equal area
The line y = - x + B passes through the center of the rectangle
∵ the center of the rectangle is (3,2)
 substituting point (3,2) into y = - x + B, the solution is: B = 5

As shown in the figure, in the plane rectangular coordinate system, the vertices of square oabc are o (0, 0), a (1, 0), B (1, 1), C (0, 1) (1) Judge the straight line y = - 2x + 1 3. Whether there is an intersection with the square oabc, and explain the reasons; (2) Now y = - 2x + 1 3. After translation, oabc can be divided into two parts with equal area

(1) Because the line y = - 2x + 1
3. Submit to (0, 1) with OC
3) (1)
6，0)，
So a line intersects a square
(2) Let the analytic formula of the straight line after translation be y = - 2x + B, which should pass through the intersection of AC and Bo (1
2，1
2) B = 3
2，
Then the analytic expression of the straight line is y = - 2x + 3
2．

As shown in the figure, in the plane rectangular coordinate system xoy, the coordinates of the vertex a of the rectangular AOCD are (0,4). There are two moving points P, Q, which start from point O and move to point C at a speed of 2 unit length per second along line OC (excluding end points O, c), and point Q starts from point C and moves towards point d at a speed of 1 unit length per second along line CD (excluding end points c and D), Suppose the movement time is t (seconds), when t = 2 (seconds), PQ = 2, as shown in the figure. In the plane rectangular coordinate system xoy, the coordinates of the vertex a of the rectangular AOCD are (0,4). The existing two moving points P, Q and P start from point o along the line OC (excluding the end points o, c) at a uniform speed to point C at a speed of 2 units per second, and point Q starts from point C along the line CD (excluding the end point C), D) Point P and Q start at the same time and stop at the same time. Set the movement time as t (s). When t = 2 (s), PQ = 2 √ 5 (1) Find the coordinates of point D and write the range of t directly (2) Connect AQ and extend the cross X axis at point E, fold AE along AD and the extension line of intersection CD at point F, and connect EF, then whether the area s of △ AEF changes with the change of T? If it changes, find the functional relationship between S and T; if not, find the value of S (3) Under the condition of (2), what is the value of T when apqf is trapezoid?

0

As shown in the figure, in the plane rectangular coordinate system xoy, the vertex e coordinates of rectangular oefg are (4,0), and the vertex g coordinates are (0,2). Rotate the rectangular oefg anticlockwise around the point O, so that the point F falls on the point n of the y-axis, and the rectangle omnp, OM and GF intersect at the point (1) Judge whether △ OGA and △ NPO are similar and explain the reasons; (2) The analytic formula of inverse proportional function of point a is obtained; (3) If the image of the inverse scaling function obtained in (2) intersects with EF at point B, please explore whether the line AB is perpendicular to OM and explain the reason

(1) The △ OGA is similar to △ NPO for the following reasons:
∵ the rectangular oefg rotates anticlockwise around point o so that the point F falls at the point n of the y-axis to obtain the rectangular omnp,
∴∠P=∠POM=∠OGF=90°，
∴∠PON+∠PNO=90°，∠GOA+∠PON=90°，
∴∠PNO=∠GOA，
∴△OGA∽△NPO；
(2) ∵ the coordinates of e point and G point are (4,0) and (0,2),
∴OE=4，OG=2，
∴OP=OG=2，PN=GF=OE=4，
∵△OGA∽△NPO，
ν og: NP = GA: OP, i.e. 2:4 = GA: 2,
∴GA=1，
The coordinates of point a are (1, 2),
Let the analytic formula of inverse proportional function of point a be y = K
x，
Substitute a (1,2) into y = K
X is k = 1 × 2 = 2,
The analytic formula of inverse proportional function of passing point a is y = 2
x；
(3) The straight line AB is perpendicular to OM
Replace x = 4 with y = 2
In X, y = 1
2，
The coordinates of point B are (4, 1
2），
∴BF=2-1
2=3
2，
The coordinates of point a are (1, 2),
∴AG=1，AF=4-1=3，
∴OG：AF=2：3，GA：FB=1：3
2=2：3，
∴OG：AF=GA：FB，
And ∠ OGA = ∠ F,
∴△OGA∽△AFB，
∴∠GAO=∠ABF，
∵∠ABF+∠BAF=90°，
∴∠GAO+∠BAF=90°，
∴∠OAB=90°，
The straight line AB is perpendicular to OM

As shown in the figure, in the plane rectangular coordinate system xoy, the vertex e coordinate of rectangular oefg is (4,0), and the vertex g coordinate is (0,2)

(1) It is known that ∠ OGA = M = 90 °, Goa = ∠ Mon, and △ OGA ∷ omn are easily obtained
(2) According to the conclusion of (1), the value of Ag, i.e. the coordinate of a, can be obtained. Suppose the inverse proportional function y = KX, substituting a (1,2) into, we can get k = 2, that is, y = 2x
(3) Let y = MX + N, substituting a (1,2) and B (4,12) into the equation system, the value of Mn can be obtained, and the analytic formula of the straight line AB can be obtained by substituting it into the solution;
(4) Let the symmetry center of rectangular oefg be q and the coordinates of easily obtained point Q be (2,1), and the answer can be determined by substituting it into the analytical formula. (1) △ OGA ∽ △ omn. (1 point)
From the known, ∠ OGA = ∠ M = 90 ° and ∠ Goa = ∠ mon,
ν Δ OGA ∽ omn. (2 points)
(2) From (1), agmn = ogom
∴ AG2=24,AG=1,
A (1,2). (3 points)
Suppose the inverse proportional function y = KX, substituting a (1,2) into, we get k = 2, that is, y = 2x. (4 points)
(3) ∵ the abscissa of point B is 4. Substituting x = 4 into y = 2x, y = 12, i.e. B (4,12). (5 points)
Let y = MX + N, substituting a (1,2), B (4,12) into {m + n = 24m + n = 12, and {M = - 12n = 52
ν y = - 12x + 52. (8 points)
(4) Let the symmetry center of rectangular oefg be q, then the coordinates of point q are (2,1)
Substituting x = 2 into y = 2x gives y = 1
The image of inverse scale function passes through the symmetry center of rectangular oefg. (10 points)