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The distance from the point (- 1,2) to the line 3x + 4y-7 = 0 is___
Formula of distance from point to line
[3*(-1)+4*(2)-7]/(3*3+4*4)^(1/2)
=(-3+8-7)/5
=-2/5
The distance between point (2,1) and line 3x-4y + 7 = 0 is the distance between point and line
Using the formula of distance from point to line
|3*2-4*1+7|/√(3^2+4^2)=9/5
Find a point P on the x-axis so that the distance from the point to the line 3x + 4y-5 = 0 is 2, and find the coordinates of point P
Suppose P (x, 0)
2 = ｜ 3x-5 ｜ / radical (9 + 16) = ｜ 3x-5 ｜ / 5,
The solution is: x = 5 or - 5 / 3,
P (5,0) or (- 5 / 3,0)
Find a point P on the x-axis so that the distance from it to the line L1: 3x-4y + 6 = 0 is 6
Let P coordinate (x, 0)
The distance from P to the straight line is | 3x + 6 | / √ (3 & # 178; + 4 & # 178;) = 6
|3X+6|=30
3X+6=±30
X1=8,X2=-12
P1（8,0）、P2（-12,0）
Find a point P (x, y) in the rectangular coordinate system so that the distance from it to the x-axis, Y-axis and the straight line 3x-4y-6 = 0 is equal. How many points are there?
If the distances from P (x, y) to X axis and Y axis are equal, then x = y
The distance from P (x, y) to x-axis, Y-axis and line 3x-4y-6 = 0 are equal,
On the formula of distance from point to line
|3x-4x-6 | / 5 = | x |, the solution is reduced to 2x ^ 2-x-3 = 0, and the solution is x = 3 / 2, or - 1
The point P is (- 1, - 1), (3 / 2,3 / 2)
So there are two such points
Find the distance between the following two parallel lines 3x + 4Y + 10, 3x + 4Y = 0
2X+3Y-8=0,2X+3Y+18=0
The distance is two
Take any point on 3x + 4Y = 0, such as (1, - 3 / 4)
Then we can get C = 2 with the formula of distance from point to line
If you have any questions, please send me a message
The distance between two parallel lines 3x-4y + 13 = 0 and 3x-4y-7 = 0 is——
The distance is | 13 - (- 7) | / √ (3 & # 178; + 4 & # 178; = 4
Given two parallel lines l1:3x + 4y-10 = 0 and l2:3x + 4y-25 = 0, and the ratio of the distance between L and L1 to the distance between L and L2 is 2:3, the equation of line L is obtained
Let l be 3x + 4Y + B = 0, then the distance between L and L1 is D1 = | B + 10 | / 5, and the distance between L and L2 is D2 = | B + 25 | / 5. Because the ratio of the distance between L and L1 and the distance between L and L2 is 2:3, so | B + 10 | / 5: | B + 25 | / 5 = 2:3, so 2 | B + 25 | = 3 | B + 10 |, that is | 2B + 50 | = | 3B + 30 | so 2B + 50 = 3
We know that the distance between two parallel lines L1: 3x-4y = 0 and L2: 3x-4y + C = 0 is 1. Please answer immediately for the value of C. thank you!
Take any point at 3x-4y = 0
Like the origin
Then the distance from him to the other straight line is 1
So | 0-0 + C | / √ (9 + 16) = 1
c=±5
Given the straight line L1: 2x + Y-2 = 0 and L2: 3x + 4Y + 7 = 0, find the coordinates of the point on the line L1 whose distance from L2 is less than 3
Let the coordinates (x, y) of the point be y = 2-2x from L1
From the distance formula, we get | 3x + 4Y + 7 | / 5
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