If the line L1 passes through the point (3,0), the line L2 passes through the point (0,4), and L1 / / L2, if D represents the distance between L1 and L2, then the value range of D? RT

If the line L1 passes through the point (3,0), the line L2 passes through the point (0,4), and L1 / / L2, if D represents the distance between L1 and L2, then the value range of D? RT

L3 is the line passing (3.0) (0.4)
When L1L2 is perpendicular to L3, the maximum d = 5
When L1L2 is close to L3, it is minimum but not zero, so (0,5]
Suppose that the line L3 = 5 in two parallel lines L1L2
The angle between parallel line and line segment is a
If the distance between parallel lines is 5sina, it is easy to draw

If three straight lines L1: X-Y = 0, L2: x + Y-2 = 0: and L3: 5x-ky-15 = 0 do not form a triangle, then the sum of all values of the real number k is______ ．

If L1 ∥ L3, then k = 5, if L2 ∥ L3, then k = - 5, from X − y = 0x + y − 2 = 0, then x = 1y = 1, if (1,1) is on L3, then k = - 10. The sum of all values of the qualified real number k is 5 + (- 5) + (- 10) = - 10, so the answer is: - 10

Given that the line L passes through the point P (- 1,1), it is cut by two parallel lines L1: x + 2y-1 = 0, L2: x + 2y-3 = 0, and the midpoint m of line m1m2 is on the line L3: x-y-1 = 0, the equation of line L is obtained,

2X+7Y-5=0

Given that the line L passes through the point (2,4), and the midpoint of the line cut by the parallel line L1: X-Y + 1 = 0 and the line L2: x-y-2 = 0 is on the line L3: x + 2y-3 = 0, the equation of the line L is obtained

The equation of line L
y-4=k(x-2)
kx-y+4-2k=0
x-y+1=0
x1=(2k-3)/(k-1)
y1=(3k-4)/(k-1)
kx-y+4-2k=0
x-y-2=0
x2=(2k-6)/(k-1)
y2=-4/(k-1)
Midpoint of line segment (a, b)
a=(x1+x2)/2=(4k-9)/2(k-1)
b=(y1+y2)/2=(3k-8)/2(k-1)
On the line L3: x + 2y-3 = 0
a+2b-3=0
k=19/4
The equation of line L
y-4=19(x-2)/4

Given that the line L1: (K-3) x + (4-K) y + 1 = 0 is parallel to l2:2 (K-3) x-2y + 3 = 0, then the value of K is______ ．

When k = 3, the two lines are parallel, when k ≠ 3, there is 2 = − 24 − K ≠ 3 & nbsp; so & nbsp; & nbsp; k = 5, so the answer is: 3 or 5

Find K parallel to line L1: (K-3) x + (4-K) y + 1 = 0 and line l2:2 (K-3) x-2y + 3 = 0

K = 4, x + 1 = 0 and 2x + 2Y + 3 = 0
Nonparallel
k≠4
Then L1 slope is - (K-3) / (4-K)
The slope of L2 is K-3
So - (K-3) / (4-K) = K-3
Then K-3 = 0 or - 1 / (4-K) = 1
k=3,k=5
So k = 3, k = 5

As shown in the figure, AB is the chord of ⊙ o, M is the point on AB, if AB = 20cm, MB = 8cm, OM = 10cm, find the radius of ⊙ o

Let the radius of the circle be r, then cm = r-om = r-10cm, DM = R + om = R + 10cm. ∵ AB = 20cm, MB = 8cm, ∵ am = ab-mb = 20-8 = 12cm. ∵ DM · cm = am · MB, ∵ (R + 10) (r-10) = 12 × 8 = 96, that is, R2 = 196, ∵ r = 14cm

The longest chord of a point m in the circle 0 is 10cm and the shortest chord is 8cm. Find the radius of circle O and the length of OM

Because the longest chord passing through a point m in circle 0 is 10cm, we know that the chord is diameter, so the radius r of ⊙ o is 5cm
Because the shortest chord through M should be vertically bisected by the diameter, so om ^ 2 = 5 ^ 2-4 ^ 2 = 9, so the length of OM is 3cm

If the circle x2 + y2 = 4 is known and the secant ABC of the circle is made through a (4,0), then the trajectory equation of the midpoint of the chord BC is ()
A. （x-2）2+y2=4B. （x-2）2+y2=4（0≤x＜1）C. （x-1）2+y2=4D. （x-1）2+y2=4（0≤x＜1）

Let the midpoint (x, y) of the chord BC, the slope of the straight line passing through a be K, and the equation of secant ABC be y = K (x-4); make the secant ABC of the circle, so the line between the midpoint and the center of the circle is perpendicular to the secant ABC, and the equation is x + KY = 0; because the intersection point is the midpoint of the chord and it is on these two straight lines, the trajectory equation of the point in the chord BC is x2 + y2-4x = 0, as shown in the figure, so select B

If we know that the square of circle x + the square of y = 4 and make the secant ABC of circle through a (40), then the trajectory equation of the midpoint of chord BC is

Let the origin of the coordinate, that is, the center of the circle be o, and the midpoint of BC be d
Because D is the midpoint of chord BC, OD is perpendicular to secant ABC, and AO distance is always 4
Therefore, the trajectory of point D is a circle with diameter Ao
Its center is (2,0) and its radius is 2
The trajectory equation is: (X-2) ^ 2 + y ^ 2 = 4