How many minutes after the hour hand points to 4 o'clock, the hour hand coincides with the minute hand?

How many minutes after the hour hand points to 4 o'clock, the hour hand coincides with the minute hand?

When pointing to 4 o'clock, the actual angle difference between the hour hand and minute hand is 4 × 30-0 = 120 degrees
It can be regarded as a travel problem
The minute hand moves 360 ﹣ 12 ﹣ 5 = 6 degrees per minute, and the hour hand moves 30 ﹣ 60 = 0.5 degrees per minute
The speed difference is 6-0.5 = 5.5 degrees
120 △ 5.5 = 240 / 11 minutes
After 240 / 11 minutes

L1: ax by + 4 = 0, L2: (A-1) x + y + B = 0, L1 and L2 are parallel, and the distance from the coordinate origin to the two straight lines is equal
Solution: first, the expressions of L1 and L2 are transformed to obtain the
L1：y=(a/b)x+4/b
L2:y=(1-a)x-b
From L1 parallel to L2, we get
A / b = 1-A (slope equal)
If the distance from the origin of the coordinate to the two straight lines is equal, then
-B = 4 / B or B = 4 / b
That is - B ^ 2 (- the square of B) = 4 (this obviously does not hold)
Or B ^ 2 = 4
Then B = 2 or B = - 2
Substituting a / b = 1-A, we get
a=2,b=-2
Or a = 2 / 3, B = 2
Problem. If the distance from the origin of the coordinate to the two straight lines is equal, we can get the following result
-B = 4 / B or B = 4 / b
This step is how to find what formula

That is, when x = 0, Y1 = 4 / b
y2=-b
Because L1 and L2 are parallel, and the distance from the origin to the two straight lines is equal. Drawing a picture, we can get that the values of Y1 and Y2 are opposite to each other, and we can get b = 4 / b
-B = 4 / b means that the two lines are equal in the ordinate, if they are equal, they are the same line

If the moving points a (x1, Y1) and B (X2, Y2) move on the line L1: x + Y-7 = 0 and L2: x + Y-5 = 0 respectively, the minimum distance from the midpoint m of line AB to the origin is ()
A. 23B. 33C. 32D. 42

The trace of point m is a straight line L parallel to the straight lines L1 and L2 with equal distance to L1 and L2, so the equation is x + y-6 = 0, and the minimum distance from m to the origin is d = 62 = 32

Given that a (x1, Y1) and B (X2, Y2) are respectively on the straight line x + Y-7 = 0 and X + Y-5 = 0, the minimum distance from the midpoint m of AB to the origin is obtained

Let the midpoint of AB be (x0, Y0) ∫ x0 = X1 + x22y0 = Y1 + Y22 & nbsp; & nbsp; & nbsp; and ∫ X1 + Y1 − 7 = 0x2 + Y2 − 5 = 0 ∫ X1 + x2) + (Y1 + Y2) = 12 ∫ 2x0 + 2y0 = 12 ∫ x0 + Y0 = 6, that is, x0 + y0-6 = 0, that is, the distance from the origin (0, 0) to x + y-6 = 0 on the straight line x + y-6 = 0, that is, the minimum distance from the midpoint m to the origin is d = | 0 + 0 − 6 | 2 = 32

If the moving points P1 (x1, Y1) and P2 (X2, Y2) move on the line L1: x-y-5 = 0 and L2: x-y-15 = 0 respectively, the minimum distance from the midpoint P of p1p2 to the origin is ()
A. 522B. 52C. 1522D. 152

Since points P1 (x1, Y1), P2 (X2, Y2) move on two parallel lines L1: x-y-5 = 0, L2: x-y-15 = 0 respectively, the linear equation of the midpoint P of p1p2 is x-y-10 = 0, then the minimum distance from the midpoint P of p1p2 to the origin is the distance from the origin o to the line x-y-10 = 0, which is equal to | 0 − 0 − 10 | 2 = 52, so B

If the moving points a (x1, Y1) and B (X2, Y2) move on the line L1: x + √ 2y-10 = 0 and L2: x + √ 2y-8 = 0 respectively, then the minimum distance from the midpoint m of AB to the origin is?
A. 3 root 2
B. 2 radical 3
C. 3 root 3
D. 4 root 2

Choose c.. Because the lines L1 and L2 are parallel lines, so the trajectory of the midpoint m of a and B is x + √ 2y-9 = 0. So the shortest distance from the point to the line can be obtained

Let C: x ^ 2 + y ^ 2 + 4x-6y = 0, if C is symmetric with respect to the line L: a * (x-2y) - (2-A) * (2x + 3y-4) = 0, find the real number a

Circle C: x ^ 2 + y ^ 2 + 4x-6y = 0
That is, (x + 2) ^ 2 + (Y-3) ^ 2 = 13
So the center of the circle is (- 2,3)
Circle C is symmetric with respect to line L: a * (x-2y) - (2-A) * (2x + 3y-4) = 0
That is, the center of the circle is on the line
a*(-2-2*3)-(2-a)*(2*（-2）+3*3-4)=0
-8a-(2-a)=0
a=-2/7

The set of points equidistant from a pair of parallel lines 5x-2x-6 = 0, 10x-4y + 3 = 0 is

Two parallel lines,
10x-4y-12=0
10x-4y+3=0
The set of points equal to their distance is a line parallel to them and sandwiched between them
So the equation is
10x-4y+ (-12+3)/2=0
10x-4y-9/2=0
20x-8y-9=0

Find the distance between the following two parallel lines 3x + 4Y + 12 = 0 and 3x + 4y-3 = 0; 5x-12y + 7 = 0 and 5x-12y-6 = 0

The first group: L12 - (- 3) l / √ 4 * 4 + 3 * 3 = 15 / 5 = 3
The second group L7 - (- 6) l / √ 5 * 5 + 12 * 12 = 13 / 13 = 1

Given the line L1: MX + 8y + n = 0 and L2: 2x + MY-1 = 0, find m, n when the following conditions are satisfied
1. If two straight lines intersect at point P (m, - 1), then,
2. If two lines are parallel,
3. If the distance between point Q (0,1) and line L2 is 1, find the value of M

(1) From m2-8 + n = 0; 2m-m-1 = 0, M = 1, n = 7
(2) When m = 0 is not true; when m is not equal to zero, the slopes of two parallel lines are equal, that is, M2 = 16, M = + - 4; when m = 4, n is not equal to - 2; when m = - 4, n is not equal to 2
(3) M = - 1.5, which can be obtained directly from the formula of distance from point to line