Let points a (2, - 3), B (- 3, - 2) and a straight line L pass through point P (1,1) and intersect with line AB, then the slope k of L is? Why is k greater than or equal to three-quarters or K less than or equal to - 4, rather than - 4 greater than or equal to K less than or equal to three-quarters

Let points a (2, - 3), B (- 3, - 2) and a straight line L pass through point P (1,1) and intersect with line AB, then the slope k of L is? Why is k greater than or equal to three-quarters or K less than or equal to - 4, rather than - 4 greater than or equal to K less than or equal to three-quarters

The closer the slope is to the y-axis, the closer it is to infinity, and the closer it is to the x-axis, the closer it is to 0

If the line L passes through point P (1,1) and intersects line AB, then the slope k of line L is ()
A. K ≥ 34B. 34 ≤ K ≤ 2C. K ≥ 2 or K ≤ 34d. K ≤ 2

The slope of the line PA is k = 3 − 12 − 1 = 2, and the slope of the line Pb is k ′ = − 2 − 1 − 3 − 1 = 34. Combined with the image, the slope of the line L is k ≥ 2 or K ≤ 34

Given that the line L passes through the point P (- 1,2) and intersects the line segment with a (- 2, - 3), B (3,0) as the end point, the value range of the slope of the line L is obtained

The slope of the straight line AP is k = - 3 − 2 − 2 + 1 = 5, and the slope of the straight line BP is k = 0 − 23 + 1 = - 12. Let L and line AB intersect at point m, and M moves from a to B, and the slope becomes larger and larger. At a certain point, am will be parallel to the Y axis, and there is no slope. That is, K ≥ 5. After this point, the slope increases from - ∞ to - 12 of the straight line BP

It is known that the eccentricity of ellipse C: x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (a > b > 0) is root sign 3 / 2, and the line L passing through the origin o of coordinate and with a slope of 1 / 2 intersects with C, a, B. | ab | = 2 root sign 10

The slope is 1 / 2 and the linear equation is y = x / 2
OA=√10
The coordinates of point a are (2 √ 2, √ 2)
e=√3/2
a=2b
8/4b^2+2/b^2=1
b^2=4
a^2=16
:x^2/16+y^2/4=1
a=4,b=2

The triangle ABC is isosceles triangle, ab = AC, BD is the midline on the side of AC, BD divides the circumference of ABC into 18cm and 21cm
Finding the length of each side of a triangle

Let AB be a and BC be B
(a 1/2a) (1/2a b)=24 18=42
｜a-b｜=24-18=6
① If a > b,
a-b=6;2a b=42
The solution is as follows
A = 18, B = 12; a ^ 2, a ^ 2 > b ^ 2 satisfy the basic definition of triangle
So the length of each side of this triangle is: 18, 18, 12
② If

What is the circumference of the triangle ABC when a = 60 degrees and BC = 3 in the triangle ABC

Sin a = opposite / hypotenuse
Sin 60 degree = 3 / bevel
(radical 3) / 2 = 3 / hypotenuse
Hypotenuse = 2 (radical 3)
Cos a = adjacent / hypotenuse
Cos 60 degree = adjacent side / [2 (radical 3)]
1 / 2 = adjacent edge / [2 (radical 3)]
Adjacent edge = root 3
Perimeter = (2 radical 3) + 3 + radical 3
=(3 radical 3) + 3

In ABC, ab = 2, BC = 7, the length of AC is even

According to the sum of the two sides of the triangle is greater than the third side, the difference between the two sides is less than the third side
So: let the length x of the third side of the triangle be 7-2

In triangle ABC, AC = 6, B = 60, perimeter = 16, area =?

AB+BC=10
AC^2=AB^2+BC^2-2AB*BC*cos60°
36=AB^2+BC^2-AB*BC=(AB+BC)^2-3AB*BC=100-3AB*BC
3AB*BC=64
AB*BC=64/3
S=1/2*AB*BC*sin60°=16√3/3

If the lengths of the two sides of the triangle are 2 and 7, then the value range of the third side length C is 0______ When the perimeter is odd, the third side length is______ ．

According to the trilateral relation of a triangle, 7-2 ＜ C ＜ 7 + 2 is obtained, that is, 5 ＜ x ＜ 9. If ∵ the circumference is odd, then C is even, and the length of the third side is 6 or 8. Therefore, the answer is 5 ＜ C ＜ 9, 6 or 8

The lengths of two sides of a triangle are 2 and 7, and the third side is odd. Find the perimeter of the triangle

Third side: less than 2 + 7 = 9, more than 7-2 = 5
The numbers of are: 6, 7, 8 and are odd
So: 7
The side length of triangle is: 2 + 7 + 7 = 16