It is known that the polar coordinates of the two ends of the line segment passing through the pole are P1 (ρ 1, θ 1) P (ρ 2, θ 2 + π), and the midpoint m of p1p2 is proved（ It is known that the polar coordinates of the two ends of the line segment passing through the pole are P1 (ρ 1, θ 1) P (ρ 2, θ 2 + π). It is proved that the midpoint of p1p2 is m (ρ 1 - ρ 2, θ 1)

It is known that the polar coordinates of the two ends of the line segment passing through the pole are P1 (ρ 1, θ 1) P (ρ 2, θ 2 + π), and the midpoint m of p1p2 is proved（ It is known that the polar coordinates of the two ends of the line segment passing through the pole are P1 (ρ 1, θ 1) P (ρ 2, θ 2 + π). It is proved that the midpoint of p1p2 is m (ρ 1 - ρ 2, θ 1)

Your questions are wrong. They should all be in the form of θ 1
The polar coordinates of the two ends of the line passing through the pole are P1 (ρ 1, θ 1) P (ρ 2, θ 1 + π),
Draw a graph, and it's obvious that the midpoint of p1p2 is m (ρ 1 - ρ 2, θ 1)
If you want a process,
Can be converted to rectangular coordinates
P1（ρ1cosθ1,ρ1sinθ1)
P2 (ρ 2cos (θ 1 + π), ρ 1sin (θ 1 + π)), namely P2 (- ρ 2cos θ 1, - ρ 1sin θ 1)
The Cartesian coordinates of the midpoint of p1p2 are m ((ρ 1 - ρ 2) cos θ 1, (ρ 1 - ρ 2) sin θ 1)
The polar coordinates are m (ρ 1 - ρ 2, θ 1)

[mathematics] in the polar coordinate system, we know that the polar coordinates of the three vertices of △ ABC are a (2,10 °), B (- 4220 °), C (3,100 °),
There are two days and four hours to the end of the problem
1) Find the area of △ ABC; 2) find the height of AB side of △ ABC

1) ABC = 3 (3) ^ 1 / 2 + 1 is 3 root sign 3 plus 1. If you want to get a good image, you can reduce the area of three small triangles with a large area
I am also a high school student

Two polar coordinates a (3, π / 2) B (3, π / 6) are known
Find (1) the distance between two points a and B, find (2) the area of s triangle ABC, and (3) the angle between straight lines a and B and the positive direction of polar axis

If two polar coordinates a (3, π / 2), B (3, π / 6), then r = 3, the included angle is π / 3, (1) △ OAB is an equilateral triangle, ab = 3. (2) s = (1 / 2) * 3 ^ 2 * sin (π / 3) = 9 √ 3 / 4 (3) the line a, B and the y-axis are negatively angled

As shown in the figure, the vertex of △ OAB is O (0,0), a (2,1), B (10,1), the straight line CD ⊥ X axis, and the area of △ 0ab is bisected. If the coordinate of point D is (x, 0), find the value of X

Let y = KX (K ≠ 0), ∵ B (10,1), ∵ 1 = 10K for OB, and the solution is k = 110. Let y = 110x, ∵ D (x, 0), ∵ f (x, X10), ∵ EF = 1-x10, EB = 10-x, ab = 10-2 = 8, ∵ s △ bef = 12 × 10 − X10 × (10-x) = (10 − x) 220, ∵ s △ AOB = 12 × 8 × 1 = 2 × (10 − x) 220, and the solution is x = 10-210