As shown in the figure, in the plane rectangular coordinate system, ⊙ P and X axis are tangent to origin o, and the straight line parallel to y axis intersects ⊙ P at M and N. if the coordinates of point m are (2, - 1), then the coordinates of point n are () A. （2，-4） B. （2，-4.5） C. （2，-5） D. （2，-5.5）

As shown in the figure, in the plane rectangular coordinate system, ⊙ P and X axis are tangent to origin o, and the straight line parallel to y axis intersects ⊙ P at M and N. if the coordinates of point m are (2, - 1), then the coordinates of point n are () A. （2，-4） B. （2，-4.5） C. （2，-5） D. （2，-5.5）

Ma ⊥ OP is used for crossing point m, and a is for perpendicular foot
Let PM = x, PA = X-1, Ma = 2
Then x2 = (x-1) 2 + 4,
The solution is x = 5
2，
∵OP=PM=5
2，PA=5
2-1=3
2，
/ / op + PA = 4, so the coordinates of point n are (2, - 4)
Therefore, a

It is known that O is the origin of the plane rectangular coordinate system. The straight line passing through the point m (- 2,0) and the circle x ^ 2 + y ^ 2 = 1 intersect P and Q If the vector OP * vector OQ = - 1 / 2, find the equation of the line L If the area of the triangle OMP and the triangle OPQ are equal, find the slope of the line L

Because the two points of P and Q are on the circle x2 + y2 = 1, because op →· OQ → = - 12, so op →· OQ → = | op → = | op → = | op →| = | op →| - 12

It is known that O is the origin of the plane rectangular coordinate system. The line passing through the point m (- 2,0) and the circle x ^ 2 + y ^ 2 = 1 intersect P and Q

Because | PQ | = - x0d3, the radius of the circle is 1, and the two points of P and Q are on the circle x2 + y2 = 1, the distance between the center of the circle and the line L is d = 1 - (32) 2 = 12

It is known that O is the origin of the plane rectangular coordinate system, and the line L passing through the point m (- 2,0) and the circle x ^ 2 + y ^ 2 = 1 intersect at two points P and Q (1) If | PQ | = radical 3, find the linear l equation (2) If the vector MP = V vector MQ, find the intersection coordinates of the line L and the circle

Let P (x1, Y1) Q (X2, Y2)
The linear equation y = K (x + 2) and the equation of circle eliminate y to get the quadratic equation of one variable about X
X 1 + x 2 x 1.x 2 is obtained by the relation between root and coefficient
The chord length formula pq| = root sign (1 + K.K) (x1-x2) (x1-x2) can obtain K

In the plane rectangular coordinate system, the line y = x + 2 intersects point a with X axis, and intersects point B with y axis (1) The coordinates of the symmetric point a 'with respect to the origin are______ The coordinates of point B 'symmetric about the origin are______ ; (2) Find the analytic formula of the line y = x + 2 which is symmetric about the origin

（1）∵y=x+2，
When y = 0, x + 2 = 0, x = - 2,
When x = 0, y = 2,
The coordinates of point a are (- 2,0), and the coordinates of point B are (0,2),
The coordinates of point a and point B are (2, 0) and (0, - 2) respectively;
(2) The analytic formula of symmetry of the line y = x + 2 about the origin is y = X-2
So the answer is (2,0), (0, - 2)

It is known that there are two points a (- 1,0) B (0,2) in plane rectangular coordinate system. Point C is symmetric at point a with respect to the origin of coordinate. The moving line L passing through point C intersects with D and intersects with line ab (2) if the angle between the straight line L and the y-axis is 45 °, find the area of △ BCE. ③ if l ⊥ BC, try to find the coordinates of point E

(1) Let AB analytic formula be y = KX + B
A(-1,0) B(0,2)
Then 0 = - K + B
B=2
K = 2, B = 2
The analytic formula of AB is y = 2x + 2
(2) Point C is symmetric to point a with respect to the origin of coordinates, C (1,0)
The angle between the straight line L and the Y axis is 45 degrees
① The angle between Y positive half axis and 45
k=-1
The linear equation of the line L is y = - x + B, passing through C (1,0)
Then B = 1
The intersection point of equation y = - x + 1 of line L and line ab
y=-x+1
y=2x+2
E (- 1 / 3,4 / 3) was obtained
Δ BCE area = (1 / 2) * (2 + 4 / 3) * 1 / 3 + 2 * 1 / 2 - (1 + 1 / 3) * (4 / 3) / 2
=5/9+1-8/9
=2/3
② The angle between the negative axle shaft and the shaft is 45
Go home and explain it again

In the plane rectangular coordinate system, the symmetric point of point a about the y-axis is B, and the symmetric point of point a with respect to the origin o is point C. (2) if the coordinate of point a is (a, b) (AB ≠ 0) Please judge the shape of △ ABC

A right triangle! Because a and B are symmetrical about the Y axis, AB is perpendicular to the Y axis and BC is perpendicular to the X axis, so AB is perpendicular to BC
esteem it a favor.

In the plane rectangular coordinate system, the parabola passes through the origin O and intersects with the x-axis at another point a, and its vertex is B. Kong Ming used a 3 cm wide rectangular ruler with scale to measure the parabola as follows: ① OA = 3cm; ② The left side of the ruler is overlapped with the symmetry axis of the parabola, so that the lower left end point of the ruler coincides with the apex of the parabola (as shown in Figure 1). The scale reading of the intersection point C between the parabola and the right side of the ruler is 4.5 Please complete the following questions: (1) Write the symmetry axis of parabola; (2) Find the analytic formula of parabola; (3) Translate the ruler (enough length) in the figure to the right of point a along the horizontal direction to the right of point a (as shown in Fig. 2). Both sides of the ruler intersect X axis at points h and G, and parabola at points E and F. verification: s trapezoid efgh = 1 6（EF2-9）．

(1) Straight line x = 3
2；
(2) Let the analytic formula of parabola be y = ax (x-3),
When x = 3
2, y = - 9
4A, namely B (3
2，−9
4a)；
When x = 9
2, y = 27
4A, namely C (9
2，27
4a)，
27
4a−(−9
4a)＝4.5，
The solution is: a = 1
2，
The analytical formula of parabola is: y = 1
2x2−3
2x；
(3) It is proved that the crossing point E is ed ⊥ FG, and the vertical foot is d,
Let e (x, 1
2x2−3
2x)，
Then f (x + 3,1
2x2+3
2x)，
Results: s trapezoid efgh = 3
2(EH+FG)＝3
2•[(1
2x2−3
2x)+(1
2x2+3
2x)]＝3
2x2，
∵1
6(EF2−9)＝1
6×9x2＝3
2x2，
ν s trapezoidal efgh = 1
6(EF2−9)．

In the plane rectangular coordinate system, if point P is in the first quadrant, ⊙ P is tangent to point Q with X axis, and intersects with y axis at two points m (0, 2), n (0, 8), then the coordinates of point P are______ .

Pass point P as PD ⊥ Mn to D, connect PQ
⊙ P is tangent to the x-axis at point Q, and intersects with the y-axis at M (0, 2), n (0, 8),
∴OM=2，NO=8，
∴NM=6，
∵PD⊥NM，
∴DM=3
∴OD=5，
∴OQ2=OM•ON=2×8=16，OQ=4．
∴PD=4，PQ=OD=3+2=5．
That is, the coordinates of point P are (4, 5)
So the answer is: (4, 5)

In the plane rectangular coordinate system, point P is in the first quadrant, circle P and X axis tangent with point Q, and intersect with y axis at M (0,2), n (0,8), and find the mark of point P

It intersects with Y-axis at M and N points, so the ordinate value of point P is between 2 and 8, that is, 5, so PM = PN = 5. A straight line passing through point P is parallel to X axis, and intersecting Y axis with point is marked as Q (0,5), QM = QN = 3. PQ = 4 is the abscissa of point P, so point P is (4,5)