Let p be the interior point of the directed line p1p2, where P1 (2,3) P2 (8,4) is known and p1p / PP2 = PP2 / p1p2, try to find the coordinates of P Vector is the title, thank you, please write the process

Vector p1p2 = (6,1), since P is the interior point of the directed line p1p2, it can be set that
Vector p1p = a * vector p1p2 = (6a, a)
Length p1p = ap1p2, PP2 = (1-A) p1p2
So a / (1-A) = (1-A) / 1
So a = (3+_ Root 5) / 2
Since P is an interior point, a

Let p be a point on the line P1 (1, - 2), P2 (- 3,5), and vector p1p = (- 1 / 2) vector PP2, then what is the coordinate of point P?

Let P coordinate be (x, y), then vector p1p = (x-1, y + 2), vector PP2 = (- 3-x, 5-y)
From the vector p1p = (- 1 / 2) vector PP2, we get: X-1 = (- 1 / 2) (- 3-x), y + 2 = (- 1 / 2) (5-y)
The solution is: x = 5, y = - 9, so the coordinates of point P are (5, - 9)

P1 (2, - 1), P2 (0, 5), and P is on the extension line of p1p2, so that | p1p | = 2 | P & nbsp; P2 |, then point P is ()
A. （2，11）B. （34，3）C. （23，3）D. （2，-7）

Let P (x, y), then (- 2,6) = (x, Y-5), ｜ x = - 2y-5 = 6, ｜ x = - 2Y = 11, and the coordinates of point P are (- 2,11)

Given P1 (2, - 1), P2 (0, 5) and point P is on the extension line of p1p2, | p1p | = 2 | PP2 |, then the coordinate of point P is ()
A. （2，11）B. (43，3)C. (23，3)D. （-2，11）

∵ point P is on the extension line of p1p2, | p1p | = 2 | PP2 |, and | point P2 is the midpoint of the line p1p.. 0 = 2 + xp2, 5 = − 1 + Yp2. The solution is XP = - 2, YP = 11.. P (- 2, 11). So D

Given that the moving point P of two points m (4.0) n (1.0) on the plane satisfies the equation PN = 2pm (1) to find the trajectory C of the moving point P (2) if the point Q (a, 0) is a point in the trajectory C, any line L intersects the trajectory C through Q at two points AB, so that the value of vector QA multiplied by vector QB is only related to a; Let f (a) = vector QA multiplied by vector QB, find the value range of F (a)?

(1) Let point P coordinate (x, y), then vector PM + (4-x, - y), vector PN + (1-x, - y), the absolute value of vector PM is equal to (4-x) &# 178; + Y & # 178;) under the root sign, by the absolute value of vector PM is equal to 2, the absolute value of vector PN is equal to (4-x) &# 178; + Y & # 178; = 2 under the root sign (1-x & # 178;) + Y & # 178;, sort out

It is known that the center of a circle is the coordinate origin, and the radius is 2. From any point on the circle p to the X axis, make a vertical line PP ', and find the locus of the midpoint m of the line PP'

Let the coordinates of point m be (x, y) and point p be (x0, Y0), ∵ m be the midpoint of line segment PP ′, ∵ x = x0, y = Y02 from the midpoint coordinate formula, that is, x0 = x, Y0 = 2Y. ∵ P (x0, Y0) is on the circle x2 + y2 = 4, ∵ X02 + Y02 = 4

If the center of a circle is the origin of coordinates and the radius is 2, and a vertical line Pb is made from any point on the circle to the x-axis, then the trajectory of the midpoint of the line Pb is a circle

The center of a circle is the origin of coordinates, and its radius is 2. The equation x & sup2; + Y & sup2; = 2 & sup2; (1)
Make a vertical line Pb from any point on the circle to the x-axis, then the midpoint of the line Pb 2 * Y1 = y is substituted into (1)
x²+(2*y1)²=2² x²+4y1²=2² x²/2²+y²=1
Its trajectory is an ellipse whose major axis is 2 and minor axis is 1

From a point Q on the hyperbola x2-y2 = 1, the perpendicular of the straight line x + y = 2 is drawn, and the perpendicular foot is n. The trajectory equation of the midpoint P of the line QN is obtained

Let the coordinates of the moving point p be (x, y), and the coordinates of the point Q be (x1, Y1), then n (2x-x1, 2y-y1) is substituted by X + y = 2, and 2x-x1 + 2y-y1 = 2 & nbsp; & nbsp; & nbsp; & nbsp; ① and PQ is perpendicular to the straight line x + y = 2, so y − y1x − X1 = 1, that is, X-Y + y1-x1 = 0 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ② is solved by ① and ②

From a point Q on the hyperbola x2-y2 = 1, the perpendicular of the straight line x + y = 2 is drawn. If the perpendicular foot is n, then the trajectory equation of the midpoint P of the line QN is______ ．

Let P (x, y), q (x1, Y1), then n (2x-x1, 2y-y1), ∵ n is on the straight line x + y = 2, ∵ 2x-x1 + 2y-y1 = 2, and ∵ PQ is perpendicular to the straight line x + y = 2, ∵ y − y1x − x1 = 1, that is, X-Y + y1-x1 = 0

From a point Q on the hyperbola x2-y2 = 1, the perpendicular of the straight line x + y = 2 is drawn. If the perpendicular foot is n, then the trajectory equation of the midpoint P of the line QN is______ ．

Let p be the interior point of the directed line p1p2, where P1 (2,3) P2 (8,4) is known and p1p / PP2 = PP2 / p1p2, try to find the coordinates of P Vector is the title, thank you, please write the process