If we know the value of a (2,3), B (5,4), C (7,10), if the vector AP = vector AB + λ vector AC (λ∈ R), then it is in the third quadrant

If we know the value of a (2,3), B (5,4), C (7,10), if the vector AP = vector AB + λ vector AC (λ∈ R), then it is in the third quadrant

Your question should be: when λ is what, P is in the third quadrant?
If so, let P (x, y), then AP (X-2, Y-3), ab (3,1), AC (5,7), and,
Vector AP = vector AB + λ vector AC,
So X-2 = 3 + 5 λ, Y-3 = 1 + 7 λ
If P is in the third quadrant, X and y are both less than 0,
That is 5 + 5 λ

As shown in the figure, ad is the angle bisector of triangle ABC, De is parallel to AB and intersects at point E, AE ratio EC = 3:5, then AB ratio EC =?, de ratio AB =?

(1) De ‖ AB, AD are bisectors,
∴∠BAD=∠CAD=∠ADE
The ade is an isosceles triangle,
∴AE=DE
The results show that AE: EC = de: EC = 3:5, and ab: de = 8:5
∴AB:EC=8:3
(2)DE:AB=EC:AC=EC:(AE+EC)=5:8

Is it possible to factorize (the second power of X + the second power of Y) times (the fourth power of X + the fourth power of Y)

It can be decomposed
(x²+y²)(x^4+y^4)
=(x²+y²)[(x²+y²)²-2x²y²]
=(x²+y²)(x²+y²-√2xy)(x²+y²+√2xy)

Point P is a point on the ellipse x ^ 2 / 25 + y ^ 2 / 16 = 1. The distance from point P to the left focus is equal to twice the distance from point P to the right focus?

F1P+F2P=2a=10
F1P=2F2P
That is: F2P = 10 / 3
The distance from P to right focus is 10 / 3

In the triangle ABC, CD and AE are the heights on the sides of AB and BC respectively, and the angle B = 60 degrees. The proof is: de = 1 / 2Ac

Certification:
∵∠AEC=∠CDA=90°
The four points a, D, e and C are in the same circle,
∴∠BED=∠BAC.
And ∵ ∠ B is the common angle
∴△BED∽△BAC,
∴DE/AC=BD/BC.
In RT △ BDC, ∠ B = 60 °
∴BD=1/2BC,
∴DE/AC=1/2,
∴DE=1/2AC
If you agree with my answer, please click "adopt as a satisfactory answer". I wish you progress in your study!

Given the imaginary number (X-2) + Yi, where x and y are real numbers, when the imaginary number module is 1, find the value range of X / y,

Module is 1, that is, (X-2) ^ 2 + y ^ 2 = 1, Y / X is when the line y = KX is tangent to the circle, K obtains the critical value, that is, the slope is calculated to draw the picture
The radius is 1, the distance from the center of the circle to the origin is 2, so it's 30 degrees, so - root 3 / 3

The known function f (x) = 2Sin ω x × cos ω x (ω > 0, X ∈ R)
(1) Find the range of F (x) (2) if the minimum positive period of F (x) is 4 π, find the value of ω
(the process should not be too concise)

（1）
f(x)=2sinωx×cosωx=sin2wx,
Since - 1 ≤ sin2wx ≤ 1,
So the range of F (x) is [- 1,1];
（2）
The minimum positive period of F (x) is t = 2 π / 2W = π / w = 4 π,
w=π/4π=1/4.

The length of two right angles of a right triangle is 6cm and 8cm respectively. After rotating one circle along the hypotenuse, a body of revolution is obtained and the volume of the body of revolution is calculated

V=1/3∏R²H=1/3*3.14*(6.4)²*10=428.71

What function is cos (SiNx)?

This is a composite function composed of y = COSU and u = SiNx

The known set a = {(x, y) | (Y-3) / (X-2)
Given the set a = {(x, y) ∣ (Y-3) / (X-2) = a + 1}, B = {(x, y) ∣ (a ^ 2-1) x + (A-1) Y-15 = 0}, if a ∩ B = Φ, when a takes what value, a ∩ B = Φ

Analysis, set a = {(x, y) ∣ (Y-3) / (X-2) = a + 1} and set B = {(x, y) ∣ (a ^ 2-1) x + (A-1) Y-15 = 0}, then when two lines a and B are parallel, a ∩ B = Φ
There are two general solutions
1. A ∩ B = Φ, i.e. equations
(y-3)/(x-2)=a+1
(a ^ 2-1) x + (A-1) Y-15 = 0, no solution trouble
2. Classified discussion a ∩ B = Φ,
1) A and B are not empty sets, and the line (Y-3) / (X-2) = a + 1 is parallel to the line (a ^ 2-1) x + (A-1) Y-15 = 0
If the slopes are equal, we get - (a ^ 2-1) / (A-1) = a + 1, and the solution is a = - 1
When a = - 1, the two lines are y = 3 and y = - 15 / 2 respectively, which do not coincide
When a = - 1, a ∩ B = Φ
2) When a is an empty set, because (2,3) is not ∈ a, when (2,3) ∈ B, a ∩ B = φ is substituted into B to get the solution
In this case, a = 5 / 2 or - 4
3) If B is an empty set, then a = 1
So when a = 1, - 1,5 / 2, - 4, a ∩ B = Φ