It is known that the distance between a and B is 5cm. Find a point C in the plane so that the sum of the distances between point C and a and B is 5cm, then point C is in the plane? 1-4.2 exercises

It is known that the distance between a and B is 5cm. Find a point C in the plane so that the sum of the distances between point C and a and B is 5cm, then point C is in the plane? 1-4.2 exercises

Point C is on line AB, that is, the sum of the distances from point C to a and B is equal to 5cm

Given that the length of line AB is 2, and the sum of squares of the distances between moving point m and two points a and B is 10, the trajectory equation of point m is obtained
Please explain in detail

A (- 1,0), B (1,0)
Let m (x, y)
Then (x + 1) ^ 2 + y ^ 2 + (x-1) ^ 2 + y ^ 2 = 10,
We get x ^ 2 + y ^ 2 = 4,
This is the trajectory equation of point M

Given that the length of line AB is 10 and the sum of squares of the distances from the moving point P to a and B is 122, the trajectory equation satisfied by the moving point P is obtained

I forgot. I'll try
Establish rectangular coordinate system X-Y, set AB line segment, a point coordinates at the origin (0,0), B point (10,0), P point (x, y)
According to the square sum of the distances from the moving point P to a and B is 122, the equation can be formulated as follows:
x2+y2+（x-10）2+y2=122
It is reduced to: x2 + y2-10x = 11

If the distance from the midpoint m of AB to plane α is 4cm, and the distance from point a to plane α is 6cm, then the distance from point B to plane α is___ cm．

(1) If a and B are on the same side of plane α, and pass through the midpoint C of a, B and AB, make vertical lines to plane α, and the vertical feet are A1, B1 and C1 respectively, then Aa1 ‖ CC1 ‖ BB1, and | CC1 | = | Aa1 | + | BB1 | 2, | BB1 | = 2 (2) if a and B are on both sides of plane α, and the midpoint C and B are on the same side, pass through the midpoint C of a, B and ab respectively

Given that AB is the diameter of circle O, PD tangents circle O at point C, the extension line of intersection AB at point D, and co equals CD, then the angle PCA =?

Connect to OC
Then OC ⊥ CD
∵CO=CD
∴∠COD=∠D=45°
∴∠A=22.5°
∴∠PCA=∠D+∠A=45+22.5=67.5°

As shown in the figure, it is known that OA is the diameter of ⊙ o, B is the center of the circle, OA = 20, DP and ⊙ B are tangent to point D, DP ⊥ PA, the perpendicular foot is p, PA and ⊙ B intersect at point C, PD = 8

PD = 8.1) find the length of AC. 2) establish a plane rectangular coordinate system with o as the origin and OA as the x-axis, and find the analytic formula of AD. 1) make the vertical line of PA from B intersect Pa at point E, with diameter OA = 20, radius AB = DB = BC = 10, ∵ PD tangent circle B at point D, ∵ d = ∵ P = 90 degrees, AP ∥ BD, and be ⊥ AP, ∥ dbep are rectangular, PE = DB = 10, be = DP = 8, AE =

AB is the diameter of circle O, point D is a point on the extension line of AB, PD is the tangent line of circle O, P is the tangent point, ∠ d = 30 ° to prove: PA = PD

Connect OP, QP
∵ P is the tangent point of circle o
∴∠OPD=90°
∴∠POD=60°
∴∠AOP=120°
And ∵ OA = OP = R
∴∠OAP=∠APO=30°
∴∠A=∠D=30°
∴PA=PD

As shown in the figure, AB is the diameter of circle O, point P is on the extension line of AB, PD and circle O are tangent to point D, point C is on circle O, and PC = PD

So OC = OD, because PC = PD, and the triangle pod and the triangle POC share the same side Po, we can see that the three sides of the triangle pod and the triangle POC are equal, OC = OD, PC = PD, Po = Po, and the two triangles are congruent

How to change the distance between the mark line and the drawing in CAD

1. Once and for all method: in the format menu, select the currently used dimension style, click Modify, and in the straight line item, increase the "beyond dimension line" and "start offset". 2. Just change one or two dimensions: select the dimension, open the properties window, and in the straight line and

It is known that, as shown in the figure, the diameter ab of semicircle o is 12cm, and points c and D are the three equal points of this semicircle. Calculate the degree of ∠ CAD and the area s of the figure (shaded part) enclosed by strings AC, ad and CD

Connecting Co, OD, CD, ∵ C, D are the triangles of this semicircle, ∵ CD ∥ AB, ∵ CDO = 60 °, the degree of ∵ CAD is 30 °, ∵ OC = OD, ∵ OCD is equilateral triangle, CD = OC = 12ab = 6, ∵ OCD and △ CDA are equal base and equal height triangles, ∵ s shadow = s sector, OCD = 16 π × 62 = 6 π cm2