If the absolute value of M × n is ≥ 2 √ 3, what is the range of K?

If the absolute value of M × n is ≥ 2 √ 3, what is the range of K?

The center of the circle (x-3) ^ 2 + (Y-2) ^ 2 = 4 is (3,2), the radius is equal to 2, the distance from the center of the circle to the straight line y = KX + 3 is equal to d = |3k-2 + 3|k2 + 1 = |3k + 1|k2 + 1, from the chord length formula, Mn = 24 - (|3k + 1|k2 + 1) 2 ≥ 23, ■ (|3k + 1|k2 + 1) 2 ≤ 1, that is 2K (4K + 3) ≤ 0, the solution is - 34 ≤ K ≤ 0

The line L passing through the point P (- 1 1) intersects with the circle x ^ 2 + y ^ 2-4x = 0 at two points a and B. when the absolute value of AB is the minimum, how to find the equation of line l?

When point P is tangent to circle O, two points a and B coincide, and the shortest linear slope of AB is k, then the linear equation is: Y-1 = K (x + 1) kx-y + K + 1 = 0, circle x ^ 2 + y ^ 2-4x = 0, which is rewritten as: (X-2) ^ 2 + y ^ 2 = 2 ^ 2  2k-0 + K + 1 / √ (1 + K ^ 2) = 29K ^ 2 + 6K + 1 = 4K ^ 2 + 4 5K ^ 2 + 6k-3 = 0, K1 = (- 6 + 4 √ 6) / 10 = (- 3 + 2 √ 6) / 5 K2 = (- 6-4 √ 6) / 10 = (- 3-2 √ 6) / 5, then the slope can be brought in

The minimum value of | ab | and the linear equation when crossing point a (3,1) to make a straight line L intersection circle x ^ 2 + y ^ 2-4x-12 = 0 and two points a and B

The solution is from the circle x ^ 2 + y ^ 2-4x-12 = 0
The circle m (X-2) ^ 2 + y ^ 2 = 4 is obtained
The center of the circle is m (2,0) and the radius is 4
The distance from point P (3,1) to m (2,0) is √ 2 < 4
So the point P is inside the circle (X-2) ^ 2 + y ^ 2 = 4
So point P (3,1) makes a straight line L intersecting circle x ^ 2 + y ^ 2-4x-12 = 0 intersecting with two points a and B,
When / AB / is minimum, the line AB is perpendicular to MP
From KMP = (1-0) / (3-2) = 1
So KAB = - 1
So the equation of line AB is
y-1=-（x-3）
That is y = - x + 4

If the line ax + 2BY-2 = 0 (a > 0, b > 0) bisects the circle x2 + y2-4x-2y-6 = 0, then the minimum value of 1 / A + 2 / B is 0________ .

Bisecting a circle indicates the center (2,1) of a straight line passing through a circle. Substituting x = 2, y = 1 into ax + 2BY-2 = 0, a + B-1 = 0 is obtained
So 1 / A + 2 / b = 1 / A + 2 / (1-A), using x > 0, Y > 0, x + Y > = 2 * (radical (x * y)) if and only if x = y,
So the minimum value of 1 / A + 2 / B is 6

If x + 2Y + 3 ≥ 0, then the minimum value of (x + 1) 2 + (y + 2) 2 is______ ．

First, the feasible region is drawn according to the constraint conditions, z = (x + 1) 2 + (y + 2) 2, which means the square of the distance between the point and B (- 1, - 2) in the feasible region. When Z is the square of the distance between the point B and the line x + 2Y + 3 = 0, Z is the minimum, and the minimum value is D2 = (|− 1 − 2 × 2 + 3 | 1 + 4) 2 = 45, so the answer is: 45

If point P is a moving point on the square of circle (x-3) + (y + 4) = 8, then the minimum distance from point P to line 2x-2y + 1 = 0 is

Center C (3, - 4), r = 2 √ 2
The distance from C to the straight line is d = | 6 + 8 + 1 | / √ (2 & # 178; + 2 & # 178;) = 15 √ 2 / 4
So the minimum is D-R = 7 √ 2 / 4

If a, B ∈ {1, 2, 3, 4, 5, 6}, L1: x-2y-1 = 0, L2: ax + BY-1 = 0, then the probability of L1 ⊥ L2 is 0___ ．

Let event a be "line L1 ⊥ L2", ∵ a, B ∈ {1, 2, 3, 4, 5, 6} of the total number of events is (1, 1), (1, 2) ，（1，6），（2，1），（2，2），… ，（2，6），… ，（5，6），… (6, 6) there are 36 kinds, and L1: x-2y-1 = 0, L2: ax + BY-1 = 0, L1 ⊥ L2 ⇔ 1 · a-2b = 0, when a = 2, B = 1; when a = 4, B = 2; when a = 6, B = 3; there are three kinds of cases.. P (a) = 336 = 112.. the probability of line L1 ⊥ L2 is: 112

If a belongs to R, then a = 1 is the condition that the line L1: ax + 2Y = 0 is parallel to the line L2: x + (a + 1) y + 4 = 0

When a = 1, the two lines are x + 2Y = 0 and X + 2Y + 4 = 0
On the contrary, if two straight lines are parallel, then a (a + 1) - 2 = 0, then a = 1 or a = - 2
It should be: [sufficient and unnecessary conditions]

If the line L1: x + y + a = 0, L2: x + ay + 1 = 0, L3: ax + y + 1 = 0 can form a triangle, find the value range of A

The line L1: x + y + a = 0, L2: x + ay + 1 = 0, L3: ax + y + 1 = 0 can form a triangle. That is to say, every two lines intersect, but the three lines are not in common. The slope K1 of L1 = - 1, intercept B1 = - A, L2. K2 = - 1 / A,.. B2 = - 1 / A, L3. K3 = - A,. B3 = - 1, K1 ≠ K2 ≠ K3, a ≠± 1, the intersection point of L1 and L2 (- 1-A, 1) is not on L3, a (- 1

If the distance between L1 and L2 is D, then the value range of D is_______ .

Obviously, if d > 0, the maximum value of D is required
When a and B are perpendicular to L1 and L2, D is the largest
In this case, d = √ (3 & sup2; + 4 & sup2;) = 5
So the value range of D is (0,5]