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Find the distance between the parallel lines 3x-2y-5 = 0 and 6x-4y + 3 = 0
l1:ax+by+c1=0
l2:ax+by+c2=0
The distance is the absolute value of (C1-C2) divided by (a square plus b square) under the root sign
According to the formula, we first convert a and B of the two formulas into the same, that is
6x-4y-10=0
6x-4y+3=0
The absolute value of (- 10-3) is divided by the root sign (the square of 6 and the square of 4)
That's 13 divided by root 52
Find the distance between two parallel lines: 3x-2y + 4 = 0 and 6x-4y + 5 = 0 Ask every day and do good every day
It is converted to 3x-2y + 4 = 0 and 3x-2y + 5 / 2 = 0
So the distance is | 4-5 / 2 | / (under the root sign (3 ^ 2 + 2 ^ 2)) = 3 / (13 under the root sign)
Find the distance between two parallel lines 3x-2y-1 = 0 and 6x-4y + 2
So simple, let me give you a formula directly, two parallel lines ax + by + C1 = 0, ax + by + C2 = 0, then the distance between them is ABS (C1-C2) / root sign (a ^ 2 + B ^ 2)
The chord length of the line 3x-4y-4 = 0 cut by circle (x-3) 2 + y2 = 9 is () A. 2 Two B. 4 C. 4 Two D. 2
According to the equation of the circle, the center of the circle is (3,0), and the radius is 3
Then the distance from the center of the circle to the straight line is | 9 − 4|
9+16=1
The chord length is 2 ×
9−1=4
Two
Therefore, C is selected
The length of the chord AB cut by the line L: 3x-y-6 = 0 by the circle C: x2 + y2-2x-4y = 0 is () A. 10 B. 5 C. Ten D. Ten Two
The circle equation x2 + y2-2x-4y = 0 is transformed into the standard equation, and (x-1) 2 + (Y-2) 2 = 5 is obtained
The coordinates of the center of the circle are (1, 2), and the radius is r=
5．
| the distance from the center of the circle to the straight line d = | 3 − 2 − 6|
12+32=
Ten
2．
The length of string AB is 2
5−5
2=
10．
Therefore, C
Find the length of the chord that the line 3x-4y + 10 = 0 is cut by the square of circle x + the square of y-6x-2y-15 = 0
Circle x ^ 2 + y ^ 2-6x-2y-15 = 0
(x-3)^2+(y-1)^2=25
Radius of center (3,1) = 5
Distance d from center of circle to straight line
d=|9-4+10|/5=3
Chord length / 2 = 4
Chord length = 8
The length of chord AB cut by the line 3x + 4y-15 = 0 by circle x2 + y2 = 25 is______ .
The center coordinate of x2 + y2 = 25 is (0, 0) and the radius is: 5, so the distance from the center of the circle to the straight line is: D = | - 15|
32+42＝3，
So 1
2|AB|=
52−32=4，
So | ab | = 8
So the answer is: 8
The chord length of the line 3x-4y + 1 = 0 cut by circle (x-3) 2 + y2 = 9 is () A. Five B. 4 C. 2 Five D. 2
According to the equation of the circle, the center of the circle is (3,0), and the radius is 3
Then the distance from the center of the circle to the straight line is | 9 + 1|
9+16=2
The chord length is 2 ×
9-4=2
Five
Therefore, C
M is the linear equation of the shortest chord in the circle x ^ 2 + y ^ 2-6x-4y + 5 = 0 The coordinates of M are (2, 0)
The standard equation of a circle is: (x-3) 2 + (Y-2) 2 = 8. Therefore, if the chord passing through M (2,0) is the shortest, then the chord is perpendicular to MCK (MC) = 2. Therefore, the slope of the line where the shortest chord is located k = - 1 / 2, and passing through the point m (2,0), the linear equation is: y-0 = (- 1 / 2) (X-2)
Given that the circle C: x2 + y2-6x-4y + 4 = 0, the midpoint of the chord of the line L1 cut by the circle is p (5, 3) (1) Find the equation of line L1; (2) If the line L 2: x + y + B = 0 intersects the circle C at two different points, the value range of B is calculated
(1) From the circle C: x2 + y2-6x-4y + 4 = 0, we get (x-3) 2 + (Y-2) 2 = 9, ᚉ the center of the circle C (3,2), and the radius is 3. According to the vertical diameter theorem, it is known that the slope of the straight line L1 ⊥ the line CP, ∵ the slope of the straight line CP is KCP = 3-25-3 = 12,
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