How to find the slope k and intercept B of other straight lines when 2y-6x-3 = 0

How to find the slope k and intercept B of other straight lines when 2y-6x-3 = 0

2y-6x-3=0
2y=6x+3
y=3x+1.5
Slope k = 3, intercept B = 1.5

The equation of a line with intercept 2 on the x-axis and perpendicular to the line 3x-2y + 1 = 0 is?
Who knows? I want an equation

The intercept on the X axis is 2
Then the line can be set as y = K (X-2)
It is also perpendicular to the line 3x-2y + 1 = 0
So k * (3 / 2) = - 1
So k = - 2 / 3
So the linear equation is y = (- 2 / 3) * (X-2)
That is, 2x + 3y-4 = 0
If you don't understand, please hi me, I wish you a happy study!

The intercept of the line L on the y-axis is - 3 and is parallel to the line 3x + 2y-1 = 0

First, let the linear equation be y = ax + B. from the intercept of the line, we can get that the line passes (0, - 3), then B = - 3, and then the line is parallel to the known line, then the slope is a = - 3 / 2. So the linear equation is y = - 3x / 2-3, that is, 3x + 2Y + 3 = 0

The equation x ^ 2 / (| m | - 1) + y ^ 2 / (2 - M) = 1 represents an ellipse with focus on the y-axis, then the value range of M is

2 - m>|m| - 1>0
3-m>|m|
If M > 0, 2M1 or m

It is known that the vertical bisector of the chord ab of the left focus f passing through the ellipse with x square / 2 + y square = 1 intersects the X axis at P (m, 0), and the value range of M is obtained

Let the equation of string AB be y = K (x + 1), and the midpoint be C, then the equation of CP is y = - 1 / K (x-m). Substituting the line AB into the elliptic equation, we get the intersection coordinates: X1 = (- 2K & # 178; - √ (2k & # 178; + 2)) / (2k & # 178; + 1) x2 = (- 2K & # 178; + √ (2k & # 178; + 2)) / (2k & # 178; +...)

Let P (x, y) be the moving point (1) on the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1 to find the value range of X + 2Y
Let P (x, y) be a moving point on the ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1
(1) Find the value range of X + 2Y
(2) Crossing point Q (2 √ 3,0), make a straight line with inclination angle α intersect with ellipse x ^ 2 / 4 + y ^ 2 / 3 = 1 at different two points m, N, and find the value range of QM and QN

X ^ 2 / 4 + y ^ 2 / 3 = 11 parameter equation a = 2, B = √ 3x = 2cosuy = √ 3sinux + 2Y = 2cosu + 2 √ 3sinu = 4sin (U + π / 6) - 1 ≤ sin (U + π / 6) ≤ 1-4 ≤ x + 2Y ≤ 42mn linear equation: y = K (X-2 √ 3) k = Tana m (x1, Y1) n (X2, Y2) 3x ^ 2 + 4Y ^ 2-12 = 03x ^ 2 + 4K ^ 2 (X-2 √ 3) ^ 2-12 = 0 (3

Given that P (x, y) is a point on the ellipse x 24 + y 2 = 1, the range of M = x + 2Y is obtained

The parametric equation of ∵ x24 + y2 = 1 is x = 2cos θ y = sin θ (θ is a parameter) ∵ let P (2cos θ, sin θ) (4 points) ∵ M = x + 2Y = 2cos θ + 2Sin θ = 22sin (θ + π 4) & nbsp; (7 points) ∵ M = x + 2Y in the range of [− 22, 22]. (10 points)

The square of 3x + the square of 2Y = 2x find the value range of the square of X + the square of Y

3x^2+2y^2=2x
2x^2+2y^2+x^2-2x=0
2(x^2+y^2)+(x-1)^2=1
x^2+y^2=[1-(x-1)^2]/2
(x-1) ^ 2 is greater than or equal to 0
So [1 - (x-1) ^ 2] / 2 is less than or equal to 1 / 2
So the range of x ^ 2 + y ^ 2 is [0,1 / 2]

Let x > = 0, Y > = 0, and X + 2Y = 1, then the value range of 2x + 3Y square is
Let x > = 0, Y > = 0, and X + 2Y = 1, then the value range of 2x + (3Y Square) is

X = 1 - 2Y substituting:
(2x +3y)^2 = [ 2(1-2y) +3y]^2
= (2 -y)^2
x= 1- 2y >=0 ,0=

If the equation x2m2 + Y2 (m − 1) 2 = 1 denotes an ellipse whose quasilinear is parallel to the X axis, then the range of M is ()
A. M ＞ 12b. M ＜ 12C. M ＞ 12 and m ≠ 1D. M ＜ 12 and m ≠ 0

Because the equation x2m2 + Y2 (m − 1) 2 = 1 represents an ellipse whose directrix is parallel to the X axis, the intersection point of the ellipse is on the Y axis, so 0 ＜ M2 ＜ (m-1) 2, the solution is m ＜ 12 and m ≠ 0