The coordinates of the three vertices of △ ABC are a (1, - 1), B (4, 1) and C (2, 2). When △ ABC is rotated 180 ° around the origin of the coordinates, the coordinates of each vertex of the triangle are as follows______ ，______ ，______ .

The coordinates of the three vertices of △ ABC are a (1, - 1), B (4, 1) and C (2, 2). When △ ABC is rotated 180 ° around the origin of the coordinates, the coordinates of each vertex of the triangle are as follows______ ，______ ，______ .

∵ △ ABC rotates 180 ° around the coordinate origin,
The new triangle and the original triangle are symmetrical about the origin,
The coordinates of the vertices of the triangle after rotation are as follows: (- 1,1), (- 4, - 1), (- 2, - 2),
So the answer is: (- 1,1); (- 4, - 1); (- 2, - 2)

As shown in the figure, in the plane rectangular coordinate system, the coordinates of the three vertices of △ ABC are a (2, 3), B (2, 1), C (3, 2) (1) Judge the shape of △ ABC; (2) If △ ABC is rotated one circle along the line of edge AC, the volume of the rotating body is obtained

(1) Answer: triangle is isosceles right triangle;
From the coordinates of points a, B and C,
AC=
(2-3)2+(3-2)2=
2，
BC=
(3-2)2+(2-1)2=
2，
AB=3-1=2，
Because（
2）2+（
2) 2 = 4 = 22, namely ac2 + BC2 = AB2, AC = BC,
Therefore, the triangle is an isosceles right triangle;
(2) The volume of the cone is 1
3π•BC2•AC=1
3π×（
2）2×
2=2
Three
2π．

As shown in the figure, O is the origin of rectangular coordinate system, and the coordinates of a and C are (3,0), (0,5) (1) Write the coordinates of point B directly; (2) If the straight line CD passing through point C intersects AB and edge on point D, and the circumference of rectangular oabc is divided into two parts of 1:3, the analytic formula of straight line CD is obtained

(1) The coordinates of point B are (3, 5). (2)

As shown in the figure, the coordinates of the points in the rectangular coordinate system of the parallelogram ABCO are o (0,0) a (4,0) B (6,2) C (2,2). The four points of oabc are changed as follows 1. Add 3 to the abscissa and ordinate respectively, and the points are connected by line segments. What is the change of the pattern compared with the original pattern? 2. When the ordinate remains unchanged and the abscissa becomes twice as much as the original, what is the change of the pattern compared with the original one? 3. The ordinate and abscissa are multiplied by - 1 respectively, and then the obtained points are connected by line segments. What is the change of the pattern compared with the original pattern?

Shape unchanged (position shifts up and right)
The lateral side length is stretched twice, the height remains unchanged, and the shape is slightly flattened (because the angle c0a becomes smaller)
Shape unchanged (position mirrored to third quadrant)

As shown in the figure, in the trapezoidal oabc, O is the origin of the rectangular coordinate system, and the coordinates of a, B and C are (14, 0), (14, 3), (4, 3). Points P and Q start from the origin at the same time and move at a uniform speed, where point P moves along OA to terminal a at a speed of 1 unit per second; point Q moves along OC and CB to the end point B. when one of these two points reaches its own end point, the other point stops Let P move for T seconds from the start (1) If the speed of point q is 2 units per second, ① Try to write out the coordinates of the point Q on OC or CB respectively (expressed in the algebraic formula containing T, the range of T is not required to be written out); ② When calculating the value of T, PQ ∥ OC? (2) If the sum of the distances between point P and point q is exactly half of the perimeter of the trapezoidal oabc, ① Try to express the distance and speed of point Q by using the algebraic formula containing t; ② Question: is it possible for the straight line PQ to divide the area of trapezoidal oabc into two equal parts at the same time? If possible, find out the corresponding value of T and the coordinates of P and Q; if not, please explain the reason

(1) (2) if the velocity of Q is V, the perimeter of the trapezoid is 32, then t + VT = 16, so v = 16 − TT, the point Q passes through the point Q

As shown in the figure, in rectangular oabc, O is the origin of rectangular coordinate system, and the coordinates of a and C are (3,0), (0,5) respectively. (1) if the straight line CD passing through point C intersects AB edge and point D, and the circumference of rectangular oabc is divided into 1:3 parts, the analytic formula of straight line CD is obtained, Make the triangle feeding the vertex C, D, e similar to the triangle with B, C, D as the vertex. If there is a request for the coordinates of e point; if not, please explain the reason

1. Let D (3, y) be CO + 0A + ad = 3 (BD + BC), so 5 + 3 + x = 3 [3 + (5-x)] x = 4, then d (3,4)  CD is y = (- 1 x / 3) + 52. Draw it by yourself. When ① angle ECD ② angle CED ③ angle EDC is right angle, ① does not hold, because CE ⊥ CD then when CD is y = KX + B1, CE is y = (- 1x / k) + B2, that is, CE is y = 3x + B

(2007 Lianyungang) as shown in the figure, in the rectangular coordinate system, the vertex o of the rectangular oabc coincides with the coordinate origin, and the vertices a and C are on the coordinate axis, OA = 60cm, OC = 80cm. Starting from point O, the moving point P moves uniformly along the x-axis at the speed of 5cm / s, and stops when it reaches point C (1) Pass through the point P as the vertical line of the diagonal ob, and the perpendicular foot is the point T. find the functional relationship between the length y of Pt and time t, and write out the value range of the independent variable t; (2) At this time, when the motion point of OBO 'falls on the diagonal point of the straight line exactly, when it falls on the diagonal point of the straight line; (3) To explore whether the area of △ apt with a, P, t as the vertex can reach 14% of the area of rectangular oabc Please state the reasons

(1) In the rectangular oabc,
Because OA = 60, OC = 80,
So ob = AC = 602 + 802 = 100
Because Pt ⊥ ob,
So RT △ opt ∽ RT △ OBC
Because Pt BC = OP ob, i.e. Pt 60 = 5T 100,
So y = Pt = 3T
When point P moves to point C, it stops moving, and the maximum value of T is 805 = 16
(2) (as shown in Fig. 2), when the symmetry point o 'of point o with respect to the straight line AP is exactly on the diagonal ob, points a, t, P are
In a straight line
So AP ⊥ ob, ∠ 1 = ∠ 2
So RT △ AOP ∽ RT △ OCB,
So OP CB = Ao OC
So OP = 45
So the coordinates of point P are (45,0)
Let the function analytic formula of straight line AP be y = KX + B
The point a (0,60) and point P (45,0) are substituted into the analytic formula,
60 = 0 + b 0 = 45k + B,
By solving the equations, k = - 43 B = 60
So the analytic formula of the straight line AP is y = - 4 3x + 60
(3) It is known from (2) that when t = 45, 5 = 9, points a, t, P are on a straight line, and points a, t, P are not constructed
Form a triangle
Therefore, there are two situations:
1. When 0 < T < 9, the point t is located in the interior of △ AOP (as shown in Figure 1), and AE ⊥ ob is made through point a, and point E is perpendicular to the foot,
From Ao? AB = ob? AE, AE = 48
So s △ apt = s △ aop-s △ ato-s △ OTP = 12 × 60 × 5t-1 2 × 4T × 48-1 2 × 4T × 3T = - 6t2 + 54t
If s △ apt = 1 4 s rectangular oabc,
Then - 6t2 + 54t = 1200, that is, t2-9t + 200 = 0
At this time, △ = (- 9) 2-4 × 1 × 200 ＜ 0,
Therefore, the equation has no real root
Therefore, when 0 ＜ t ＜ 9, the area of △ apt with a, P, t as the vertex cannot reach 14% of the area of rectangular oabc
2. When 9 ＜ t ≤ 16, the point t lies outside △ AOP
At this time, s △ apt = s △ ATO + s △ otp-s △ AOP = 6t2-54t
If s △ apt = 1 4 s moment oabc,
Then 6t2-54t = 1200, that is, t2-9t-200 = 0
T 1 = 9 + 8812, T 2 = 9 - 8812 ＜ 0
Because 881 ＞ 625 = 252,
So t = 9 + 8812 > 9 + 6252 = 17
At this time, 9 < T ≤ 16,
Therefore, t = 9 + 8812 does not conform to the meaning of the question and should be omitted
Therefore, when 9 ＜ t ≤ 16, the area of △ apt with a, P, t as the vertex can not reach 14% of the area of rectangular oabc
To sum up, the area of △ apt with a, P, t as the vertices cannot reach 14% of the area of the rectangular oabc

As shown in the figure, in the plane rectangular coordinate system, the vertex of rectangular oabc coincides with the origin of vertex o coordinate

(1) Let the analytic formula of the straight line de be, ∵ the coordinates of point D and E are (0,3), (6,0), ? the point m is on the edge of AB, B (4,2), while the quadrilateral oabc is a rectangle, and the ordinate of point m is 2. And ∵ point m is on the straight line, ? 2 =. X = 2. ? m (2,2). ? (x > 0

As shown in the figure, in the rectangular coordinate system, point a is a point on the image of the inverse scale function Y1 = K / x, the positive half axis of ab ⊥ X axis is at point B, C is the midpoint of ob, the image of the first order function y2 = ax + B passes through two points a and C, and intersects point d (0, - 2) and s triangle AOD = 4 (1) The analytic formula of inverse proportional function and first order function (2) Observe the image, please point out the value range of X when Y1 > Y2 on the right side of Y axis

(1) Make AE ⊥ Y axis at E,
∵S△AOD=4,OD=2
∴ OD•AE=4
∵ ab ⊥ ob, C is the midpoint of ob,
∴∠DOC=∠ABC=90°,OC=BC,∠OCD=∠BCA
∴Rt△DOC≌Rt△ABC
∴AB=OD=2
ν a (4,2) replace a (4,2) with a (4,2), and K = 8,
The analytic formula of inverse proportional function is:,
Substitute a (4,2) and D (0, - 2) into y2 = ax + B,
The solution is as follows:
The analytic formula of the first order function is: y2 = X-2;
(2) On the right side of y-axis, when Y1 > Y2, 0 < x < 4
Master the method of further solving the function analytic formula by calculating the coordinates of points; when solving the inequality by observing the image, the function value at the top of the function image is large from the intersection point

In the rectangular coordinate system, O is the coordinate origin, and the point P (m, n) is on the image of the inverse scale function y = K / x, if M + n = √ 2, Op = 2, and the inverse scale function y = KX satisfies: when x > 0, y decreases with the increase of X, then k =? This is the problem of inverse proportional function in mathematics 9 of Zhejiang Education Press

The inverse proportional function y = KX satisfies: when x > 0, y decreases with the increase of X, so K is less than 0. Because P (m, n) is on the inverse proportional function y = K / x, M = K / N, because K is greater than 0, so m and n are different signs. Because OP = 2, that is, m ^ 2 + n ^ 2 = 4 (^ 2 represents Square), and because m + n = √ 2, m, n are (√ 2 + √ 6) / 2