On the X2 power + (5 + 5b) xy-x + 2x + 7 of the polynomial (A-2) of x y, excluding the quadratic term, find the value of a-2b? Answer before 20 o'clock! Fast

(5 + 5b) xy-x + 2x + 7, without quadratic term
Then: quadratic coefficient = 0
a-2=0
5+5b=0
The solution is: a = 2; b = - 1
a-2b=2+2=4

The coefficients and times of X-Y, - A's quadratic, B's cubic + A's cubic, B's quadratic, 2 / 2 x-3 y, 3A's quadratic + 2B's cubic

Coefficient: the constant in front of the unknown is the coefficient
Times: sum of times of all unknowns in the same term
X-Y, the coefficient of X term is 1, the coefficient of - y term is - 1, and the degree of both terms is 1
[(- a) ^ 2] (b ^ 3), coefficient 1, times 5
X / 2-y / 3, X / 2 coefficient is 1 / 2, and - Y / 3 coefficient is - 1 / 3, and the number of two terms is 1
3A ^ 2 + 2B ^ 3, 3A ^ 2 coefficient is 3, times is 2, 2b ^ 3 coefficient is 2, times is 3

(-1.3a+2b)

Square of 1.3a-5.2ab

Given points a (- 5,4) and B (3,2), what is the equation of the straight line passing through point C (- 1,2) and having the same distance from a and B

Solution: the distance between a and B is equal respectively, so there are two linear equations. A (- 5,4) and B (3,2) are located on the same side of the line, and the slope k = (4-2) / (- 5-3) = - 1 / 4, so the line is Y-2 = - 1 / 4 * (x + 1), x + 4y-7 = 0 (2) if a (- 5,4), B (3,2)

A mathematical problem, seek master solution, solve P (3,2), and in the two axis equal to the tangent equation

Let the equation be
x/a+y/a=1
Due to P (3,2) - substitution
3/a+2/a=1
5/a=1
a=5
So the equation is x / 5 + Y / 5 = 1

If the line L passes through the point P (1,2), and the distances between M (2,3), n (4, - 5) and the line L are equal, then the equation of L is (process,

K does not exist
Then it is perpendicular to the X axis
It's x = 1
Then the distance is not equal
The slope is K
y-2=k(x-1)
kx-y+2-k=0
Equal distance
|2k-3+2-k|/√(k²+1)=|4k+5+2-k|/√(k²+1)
|k-1|=|3k+7|
Then k-1 = 3K + 7 or k-1 = - (3K + 7)
So k = - 4, k = - 3 / 2
So 4x + y-6 = 0, 3x + 2y-7 = 0

Given that the line L passes through the point P (- 1,2), and the distance between the point m (- 4,1), n (2,5) and l is equal, the equation of the line L is obtained

I use the combination of number and shape
The distance between M (- 4,1), n (2,5) and l is equal. There are two possibilities
The first possibility is that l is parallel to the line Mn, so KL = Kmn = (YM yn) / (XM xn) = 2 / 3
So the oblique formula of L is Y-2 = 2 / 3 (x + 1), which is sorted out to be 2x-3y + 8-0
The second possibility is that the line L passes through the midpoint of Mn, and the coordinates of the midpoint of Mn are (- 1,3)
The linear equation is x = - 1

If a straight line is drawn through the point P (1,2) so that the distances between a (2,3), B (4, - 5) and it are equal, then the equation of the straight line is?

Let Y-2 = K (x-1)
kx-y+2-k=0
The distance between the point m (x0, Y0) and the line ax + by + C = 0 is
[Ax0+By0+C]/sqrt(A^2+B^2)
The above is the distance formula from the point to the line, where [] represents the absolute value and sqrt represents the root sign
therefore
The solution of [2k-3 + 2-k] = [4K + 5 + 2-k] is obtained
k1=-1.5,k2=-4

Draw a straight line through P (2.2) and make it equal to the distance between two points a (2.3) and B (4.5). Find the equation of this straight line

y=-x+4

If a straight line is drawn through P (1,2) so that the distance between a (2,3) B (4, - 5) and a (2,3) B (4, - 5) is equal, then the equation of the straight line is?

Let the linear equation be K (x-1) - y + 2 = 0, because the distances from a (2,3), B (4, - 5) to the straight line are equal, so | K (2-1) - 3 + 2 | / (k ^ 2 + 1) ^ 1 / 2 = | K (4-1) - (- 5) + 2 | / (k ^ 2 + 1) ^ 1 / 2 | K (2-1) - 3 + 2 | = | K (4-1) - (- 5) + 2 | k-1 | = | 3K + 7 or k-1 = - (3K + 7) so k = - 4 or K = - 1.5

On the X2 power + (5 + 5b) xy-x + 2x + 7 of the polynomial (A-2) of x y, excluding the quadratic term, find the value of a-2b? Answer before 20 o'clock! Fast