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Find the equation of the line with root sign 10 which is perpendicular to the straight line x + 3Y + 5 = 0 and the distance from the point P (- 1.0) is three fifths times
Perpendicular to x + 3Y + 5 = 0, let the equation be 3x-y + B = 0
To p distance d = | B-3 | / under root sign (3 ^ 2 + 1) = | B-3 | / root 10 = three fifths root 10
So | B-3 | = 6
B = 9 or - 3
Then the linear equation is 3x-y + 9 = 0 or 3x-y-3 = 0
Find that x + 3y-5 = 0 perpendicular to the straight line, and the distance from the point P (- 1, 0) is 3 Five The equation of a straight line of 10
∵ the slope of the line x + 3y-5 = 0 is − 1
3，
The slope of the line perpendicular to the line x + 3y-5 = 0 is 3,
Then the equation perpendicular to the line x + 3y-5 = 0 can be set as y = 3x + m, that is, 3x-y + M = 0
From the distance formula of point to line, the distance d from point P (- 1,0) to 3x-y + M = 0 is | - 1 × 3 + M|
10＝3
Five
10，
The solution is: M = - 3 or M = 9
The linear equation obtained is: 3x-y-3 = 0 or 3x-y + 9 = 0
Given that a straight line passes through point P (2,0), and the distance from point Q (- 2,4 / 3, root sign 3) to the line is 4, find the equation of this line
If the slope does not exist
Then it is perpendicular to the X axis
So x = 2
Then the distance from Q to the straight line = | - 2-2 | = 4
Slope exists
y-0=k(x-2)
kx-y-2k=0
Distance = | - 2k-4 √ 3 / 3-2k | / √ (k? 2 + 1) = 4
|k+√3/3|=√(k²+1)
square
k²+2√3k/3+1/3=k²+1
k=√3/3
So X-2 = 0 and √ 3x-y-2 √ 3 = 0
Find the equation of the line parallel to the line x-3y + 2 = 0 and the distance from it is root 10 Please write the process
Because of the parallel relationship, the slope is the same, only the intercept is different. Let the equation of the straight line be: x-3y + C = 0. According to the formula of the distance from the straight line to the straight line: √ 10 = | 2-C | / √ (1 ﹤ 3 √ 10 = | 2-C | 1010 = | 2-C | 10 ﹤ 4-4c + C | 4c-96 = 0 (C-12) (c + 8) = 0C = 12 or C = - 8
Two points a (1, 6) are known 3)，B(0，5 3) If the distance to the straight line L is equal to a, and such a line l can be made into four lines, then the value range of a is______ .
∵ if a and B are on the same side of the line L, two straight lines can be made,
If such a line l can be made into four lines, then when two points a and B are on both sides of the line L, there should be two more
ν 2a is less than the distance between a and B
∵|AB|=
(1−0)2+(6
3−5
3)2=2
∴0＜2a＜2，∴0＜a＜1
So the answer is 0 < a < 1
It is known that the distance from two points a (1,6 ∫ 3) B (0,5 ∫ 3) to the straight line L is equal to a, and four such lines can be made, then the value range of a is
Greater than 0 and less than half of the distance between two points
The line y = KX + 3 and circle (x-3) ^ 2 + (Y-2) = 4 intersect at m, N. if / Mn / > is equal to 2 root sign 3, then the value range of K is
kx-y+3=0
R=2
MN>=√3/2
Then the center distance of the circle=
Given that a, B, C satisfy the equation: 3 root sign A-B + 4 root sign C = 16, (a is greater than or equal to B, C is greater than 0), and x = 4 times the root sign A-B minus 3 times the root sign C, and find the sum of all values of X
From 3 √ (a-b) + 4 √ C = 16,
Then, √ (a-b) = [16-4 √ C] / 3,
x=4√(a-b)-3√c
=4(16-4√c)/3-3√c
=(64-19√c)/3
So x < 64 / 3
Similarly √ C = [16-3 √ (a-b)] / 4,
Substituting,
x=4√(a-b)-3√c
=4√(a-b)-3[16-3√(a-b)]/4
=25√(a-b)/4-12
x>-12
So - 12
Homework help users on November 16, 2017
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A straight line passes through point a (√ 3, - 3), and its slope is equal to 2 times of the slope of the line y = 1 / root 3 times the slope of X A straight line passes through the point a (√ 3, - 3), and its inclination angle is equal to the line y = 1 / root sign 3 times the inclined angle of X. find the equation of this line. (2) the center of circle C is on the line L: x-2y-5 = 0, and through the origin and a (2,1), find the standard equation of circle C
(1) If the equation of the straight line is y = (√ 3) x + B, then - 3 = 3 + B, B = - 6. Therefore, the straight line is y = (√ 3) x - 6 (2) let the center of the circle be (2a + 5, a), then (2a + 5) 2 + a A2 = (2a + 3) 2 + (A-1) 2, a = - 1
The slope of the line x + y + root 3 equals 0 is 0
Slope calculation: ax + by + C = 0, k = - A / b
X + y + √ 3 = 0 slope k = - 1 / 1 = - 1
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