If the distance between two parallel lines 3x + 4Y + 5 = 0 and + 6x + ay + 30 = 0 is D, then a + D =? Find the specific process,

If the distance between two parallel lines 3x + 4Y + 5 = 0 and + 6x + ay + 30 = 0 is D, then a + D =? Find the specific process,

3 / 6 = 4 / a if a = 8, then 6x + 8y + 30 = 0, both sides multiply by 1 / 2 to get 3x + 4Y + 15 = 0
According to the distance formula between two parallel lines (3x + 4Y + 5 = 0, 3x + 4Y + 15 = 0)
d= 10/（3^2+4^2)^1/2 =2
So a + D = 10

Find the distance between the line 3x + 4y-5 = 0 and the line 3x + 4Y + 6 = 0

11/5

The distance between two parallel lines 3x + 4y-6 = 0 and 3x + 4Y + 12 = 0 is?
Detailed steps, detailed steps

The distance between two parallel lines ax + by + C1 = 0, ax + by + C2 = 0 is d = | C1-C2 | / √ a ^ 2 + B ^ 2
So the answer is distance d = | - 6-12 | / √ 3 ^ 2 + 4 ^ 2 = 18 / 5

What is the distance from point P (1,1) to the line 3x-4y + 6 = 0

Distance from point to line d = | ax0 + by0 + C | / √ (A & # 178; + B & # 178;)
=I3-4+6I/√(3²+4²)
=5/5
=1

Let x.y satisfy the condition 3x + 2Y
How do you get the coefficients together

C = 18.2

The intersection of line y-ax-1 = 0 and hyperbola 3x ^ 2-y ^ 2 = 1 and two points ab
1. When a is valued, AB is on two branches of hyperbola, and when a is valued, AB is on the same branch of hyperbola
2. When the value of a is, the circle with ab as the diameter passes through the origin of the coordinate

1) hyperbola 3x ^ 2-y ^ 2 = 1, asymptote slope is k = ± 1 / √ (1 / 3) = ± 3
From the solution of AB on the two branches of hyperbola - √ 30, we get √ 3

When the line y = ax + 1 and hyperbola 3x ^ 2-y ^ 2 = 1 intersect at two points AB, a and B are on two branches of hyperbola respectively

If y = ax + 1 generation: 3x ^ 2-y ^ 2 = 1
3x^2-(ax+1)^2=1
(3-a^2)x^2-2ax-2=0
Discriminant △ = 4A ^ 2 + 8 (3-A ^ 2) = 24-4a ^ 2 ≥ 0
-When √ 6 ≤ a ≤ √ 6, there is intersection point between straight line and hyperbola
When the intersection points are in the same branch, x1x2 = - 2 / (3-A ^ 2) > 0
a^2-3>0
a> √ 3, or, a

[high school mathematics ~ urgent ~ ~ reward points!] the straight line y = ax + 1 and the hyperbola 3x ^ 2-y ^ 2 = 1 intersect at two points a and B
Is there such a real number a that a and B are symmetric with respect to the line y = 1 / 2x
How can I make it equal to two?

If it exists, then the line of these two points should be perpendicular to the line y = 1 / 2x
That is, a = - 2
y=-2x+1
Substitute the above formula into 3x ^ 2-y ^ 2 = 1 to get
The univariate quadratic equation x ^ 2-4x + 2 = 1 of X
Δ=16-2×4＞0
So a = - 2 is the required value
I don't know Baidu Hi

It is known that the intersection of the line y = ax + 1 and the hyperbola 3x is a and B
(1) If a = 2, find | ab|

1) Simultaneous y = ax + 1; 3x ^ 2-y ^ 2 = 1 = = > (3-A ^ 2) x ^ 2-2ax-2 = 0
Because there are two intersections between straight line and hyperbola, so △ = 4A ^ 2 + 8 (3-A ^ 2) > 0
The solution is - √ 6y = 2x + 1
x1+x2=-4 ;x1x1=2 ;y1+y2=-6 ;y1y2=1
|AB|=√[(x1-x2)^2+(y1-y2)^2]=4√10

The distance from point P (- 1,3) to line 3x-4y + 12 = 0 is

Given the coordinates of a point and the equation of a straight line, the distance between a point and a straight line is usually calculated by the formula of the distance between a point and a straight line
d = | 3×(-1) - 4×3 + 12 |/√（3^2 + 4^2)
= 3/5