If the line ax + y + 1 = 0 and the line 4x + 2Y + B = 0 are symmetric with respect to the point (2, - 1), find the values of a and B

If the line ax + y + 1 = 0 and the line 4x + 2Y + B = 0 are symmetric with respect to the point (2, - 1), find the values of a and B

Let the point (x, y) be on the line ax + y + 1 = 0
∵ the line ax + y + 1 = 0 and the line 4x + 2Y + B = 0 are symmetric about point (2, - 1)
The symmetric point (4-x, - 2-y) of point (x, y) with respect to point (2, - 1) is on the line 4x + 2Y + B = 0
∴4（4-x）+2（-2-y)+b=0
That is, 4x + 2y-b-12 = 0 and line 2aX + 2Y + 2 = 0 are the same line
∴4=2a,-b-12=2
∴a=2,b=-14

If the lines ax + (1-B) y + 5 = 0 and (1 + a) x = y + B are parallel to the line x-2y + 3 = 0 at the same time, find the values of a and B

Parallel slopes equal
X-2y + 3 = 0, the slope is 1 / 2
Ax + (1-B) y + 5 = 0, the slope is - A / (1-B)
The slope of (1 + a) x = y + B is 1 + a
So 1 / 2 = - A / (1-B) = 1 + a
So a = 1 / 2-1 = - 1 / 2
Substituting 1 / 2 = - A / (1-B)
1-b=1,b=0
So a = - 1 / 2, B = 0

If points P (3, a) and Q (- 1, b) are on the image of quadratic function y = - (x-1) ^ 2 + 2, then the length of segment PQ is_____

Substituting P (3, - 2) Q (- 1, - 2), the length of PQ is 4
If you don't understand, you can continue to ask me (^ - ^)

For points a (- 2,1) and B (3,2), it is known that the line L: ax + y + 2 = 0 always intersects with the extension line of line AB, and the value range of real number a is obtained

Let the linear equation be y = KX + B
Passing a and B
∴1=-2k+b
2=3k+b
The solution is: k = 1 / 5
b=7/5
∴y=1/5x+7/5
As long as they are not parallel, there must be an intersection
ax+y+2=0
y=-ax-2
k=-a
Only: - a ≠ 1 / 5
a≠-1/5 a∈R

If the line 2x-5y + 20 = 0 and mx-2y-10 = 0 have a circumscribed circle with the quadrilateral surrounded by two coordinate axes, find the value of M

A circumscribed circle complements each other diagonally
The angle between the axes is a right angle
Two lines are perpendicular to each other
Y = 2x / 5 + 4, the slope is 2 / 5
So y = MX / 2-5, and the slope is - 5 / 2
So m / 2 = - 5 / 2
m=-5

If two straight lines l1:2x-5y + 20 = 0, L2: mx-2y-10 = 0 and the quadrilateral enclosed by two coordinate axes have circumscribed circle, then M = - 5, does the quadrilateral have circumscribed circle?
The coordinate of the center of the circle is () or the radius is ()

When m = - 5, the slope of the line L2 is - 5 / 2, the slope of the line L1 is 2 / 5, and the product of the two is - 1, so the two lines are perpendicular to each other, which makes the diagonal complementation of the quadrilateral (two coordinate axes intersect at right angles), so the quadrilateral has circumscribed circle
For L1, let x = 0 find the intersection of L1 and Y axis as a (0,4). For L2, let x = 0 find the intersection of L2 and X axis as B (- 2,0). The midpoint of AB is the center of the circumscribed circle; half of AB is the radius of the circumscribed circle. (according to the theorem: "the chord to which the circular angle of 90 ° is the diameter")

If the line 2x-5y + 20 = 0 and mx-2y-10 = 0 have a circumscribed circle with the quadrilateral enclosed by two coordinates, find the real number M

That is, four points are in a circle, so they complement each other diagonally
The two axes are perpendicular
So the two lines are perpendicular
2x-5y + 20 = 0, slope = 2 / 5
Mx-2y-10 = 0, the slope is m / 2
For vertical, 2 / 5 * m / 2 = - 1
m=-5

If the line l1:2x-5y + 20 = 0 and the line L2: MX + 2y-10 = 0 have a circumscribed circle with the quadrilateral enclosed by two coordinate axes, then the value of real number m is ()
A. 5B. - 5C. ± 5D

The inscribed quadrilaterals of a circle complement each other diagonally. Because the x-axis is perpendicular to the y-axis, the line L1 is perpendicular to the line L2. The line a1x + b1y + C1 = 0 is perpendicular to the line a2x + b2y + C2 = 0 if and only if A1A2 + b1b2 = 0 is solved by 2 × m + (- 5) × 2 = 0 and M = 5 is obtained, so a is selected

If the line L1 2x-5y + 20 = 0 L2 mx-2y-10 = 0 has a circumscribed circle with the quadrilateral enclosed by two coordinate axes, then the real number m is

If the line L1: 2x-5y + 20 = 0, L2: mx-2y-10 = 0 has a circumscribed circle with the quadrilateral surrounded by two coordinate axes,
Then L1 and L2 are vertical,
2/5×m/2=-1
m=-5

Two lines L1: ax-2y = 2a-4, L2: 2x + A2 = 2A2 + 4 (0

The intersection of L1 and y-axis is (0,2-a), the intersection of L2 and x-axis is (2 + A * a, 0), and the intersection of L1 and L2 is (2,2). Through cutting, we can know that the enclosed quadrilateral can be divided into a trapezoid and a triangle. The area of trapezoid is (upper bottom + lower bottom) * high / 2 = (2-A + 2) * 2 / 2 = 4-A, and the area of triangle is bottom * high / 2 = (2 + A * A-2) * 2 / 2 = a * a, so the area of quadrilateral is a * A-A + 4 = (A-1 / 2) 2 + 15 / 4, So when a = 1 / 2, the minimum area of quadrilateral is 15 / 4