For any point (x, y) on the line L, the point (4x + 2Y, x + 3Y) is still on the line

For any point (x, y) on the line L, the point (4x + 2Y, x + 3Y) is still on the line

Let the linear equation be y = KX + B
Substituting (x, y), (4x + 2Y, x + 3Y) into
y=kx+b
x+3y=k(4x+2y)+b
The results are as follows
y=(4k-1)x/(3-2k)+b/(3-2k)
k=(4k-1)/(3-2k)
b=b/(3-2k)
The solution is: k = - 1, k = 1 / 2, B = 0
The linear equation is y = - X or y = x / 2

For any point (x, y) of the line L, if the point (4x + 2Y, x + 3Y) is still on the line L, then the equation of L

For any point P (x, y) on line L, then point Q (4x + 2Y, x + 3Y) is also on line L. try to find the equation of line L

Let a and B be two variables
The linear equation is (B-Y) / (A-X) = (x + 2Y) / (3x + 2Y)
Simplification
b=(X+2Y)/(3X+2Y)a+(2Y²+XY-X²)/(3X+2Y)

The point P (x, y) is a point on the straight line L, and the point m (4x + 2Y, x + 3Y) is also on L. solve the L equation

The vector PM = (3x + 2Y, x + 2Y) is the direction vector of the line L, i.e
(3x+2y)/x=(x+2y)/y
(3x+2y)y=(x+2y)x
3xy+2y^2=x^2+2xy
x^2-xy-2y^2=0
(x+y)(x-2y)=0
X + y = 0 or x-2y = 0
The equation of L is y = - X or y = x / 2

The quadratic equation (1-I) x2 + (λ + I) x + (1 + I λ) = 0 (I is an imaginary unit, λ∈ R) has two imaginary roots if and only if the value range of λ is______ ．

The quadratic equation (1-I) x2 + (λ + I) x + (1 + I λ) = 0 has two imaginary roots, that is, the equation has no real roots. The original equation can be reduced to x2 + λ x + 1 - (x2-x - λ) I = 0. When λ∈ R, the equation has two complex roots. If it has real roots, then x2 + λ x + 1 = 0, and x2-x - λ = 0. Subtracting (λ + 1) (x + 1) = 0 When λ ≠ - 1, x = - 1, substituting into λ = 2. That is, when λ = 2, the original equation has real root x = - 1. So the range is λ ≠ 2. So the answer is: λ ≠ 2

No matter what the value of M is, the intersection point of two straight lines of function y = x + 2M and y = - x + 4 can not be in which quadrant? Why?

∵ y = - x + 4 passes through 1,2,4 quadrants
The intersection of two lines cannot be in the third quadrant

No matter what the value of M is, the intersection of y = x + 2M and y = - x + 4 cannot be in the quadrant

M in the line y = x + 2m is undetermined and may pass through four quadrants, while K0 in the line y = - x + 4 does not pass through three quadrants, so the intersection point cannot be in the third quadrant

No matter what the value of M is, which quadrant is the intersection of y = x + 2M and y = - x-4 impossible?
Please tell me how to solve this problem,

First quadrant
The intersection must be in y = - X - 4, and y = - X - 4 is only the first quadrant, so the intersection cannot be in the first quadrant

Given that m and N are two real roots of the equation x2-2x-1 = 0, the value of the algebraic formula 3m2-n2-8m + 1 is () A.9 B.7 C.1 D

If the root of the equation satisfies the equation m-2m-1 = 0, then M = 2m + 1 (1) is the same as n = 2n + 1 (2) according to the relationship between the root of the equation and the coefficient m + n = 2 (3), then 3 (2m + 1) - (2n + 1) - 8m + 1 = - 2m-2n + 3 = - 2 (M + n) + 3 = - 2 × 2 + 3 = - 1 (d)

It is known that m and N are two real numbers of the equation AX & # 178; + BX + C = 0. Let S1 = m + N, S2 = M & # 178; + n & # 178;, S3 = M & # 179; + n & # 179 Then the value of as2011 + bs2010 + cs2009 is
A,0 B,1 C,2010 D,2011

as2011+bs2010+cs2009=a(m^2011+n^2011)+b(m^2010+n^2010)+c(m^2009+n^2009)=(am^2+bm+c)m^2009+(an^2+bn+c)n^2009=0