Find the equation of a straight line which is perpendicular to the line x + 3Y + 1 = 0 and whose intercept on the x-axis is 2

Find the equation of a straight line which is perpendicular to the line x + 3Y + 1 = 0 and whose intercept on the x-axis is 2

The slope of X + 3Y + 1 = 0 is - 1 / 3, so the slope of the equation is 3. Let y = 3x + B, because the intercept on the X axis is 2, then x = 2, y = 0, B = - 6,
3x-y-6 = 0 (the intercept is 2, so it is impossible to be in the negative half axis, and the intercept has positive and negative values)

If the intercept of the line L on the Y axis is 2 and it is perpendicular to the line L ′: x + 3y-2 = 0, then the equation of L is______ ．

The slope of the straight line L ′: x + 3y-2 = 0 is equal to - 13, so the slope of the straight line L is equal to 3. According to the intercept of the straight line L on the Y axis is 2, so the equation of L is & nbsp; y = 3x + 2, that is, 3x-y + 2 = 0, so the answer is 3x-y + 2 = 0

If the intercept of the line 2x + 3Y + C = 0 on the x-axis is 1 larger than that on the y-axis, then C is

Intercept on X-axis = | C | / 2
Intercept on Y-axis = | C | / 3
|C|/2-|C|/3=1
|C|=6
C=+-6

When m is the value, the image of the linear function y = (2m ^ 2 + m) x + m ^ 2 - 2 is parallel to the straight line y = 3x + 5, and the intercept of the image on the Y axis is - 1?
Finding the analytic expression of function satisfying the condition

The image of y = (2m ^ 2 + m) x + m ^ 2 - 2 is parallel to the line y = 3x + 5
∴2m^2+m=3
(2m+3)(m-1)=0
Ψ M = - 3 / 2, or 1
Because the intercept of the image on the Y axis is - 1
∴ m^2 -2=-1
M = - 1 or 1
To sum up, M = 1

On the image of the linear function y = 1 / 2-1 / 2 and on the Y axis, the point whose distance is equal to 1 is
... is y = 1 / 2x-1 / 2

The original problem is "a function y = (1 / 2) X-1 / 2"!
If it is, then the coordinates of the point with a distance of 1 on the image and Y axis are (1,0)

The intercept of the graph of the first-order function on the y-axis is - 4, and the relation of the first-order equation can be obtained through the points (1, - 2)

y=ax+b
b=-4
a+b=-2
a=2
So the formula is: y = 2x-2

1. If the image of a function of degree passes through point a (- 2,4), and the ordinate of the intersection point with y axis is - 3, then the relation of this function is_________ .
2. If points a (- 2, B1) and B (2, B2) are on the line y = - 4x + 5, then B1______ B2. (fill in ">"“

1. If the image of a function of degree passes through point a (- 2,4), and the ordinate of the intersection point with y axis is - 3, then the relation of this function is__ y=-7x/2-3_______ Let y = kx-3 ∵ the ordinate of the intersection point of y-axis be - 3, and let the image of y = kx-3 ∵ first-order function pass through a (- 2,4) ∵ 4 = - 2k-3 ∵ k = - 7 / 2 ∵ y = - 7x / 2-32, if a (- 2, B1) and B (...)

If the function y = KX + B passes through the point (1,5) and the intercept on the Y axis is 3, then k =, B =

The intercept on the y-axis is 3, so the intersection of the image and the y-axis is (0,3), then you will know

If the image of the function y = - x + B passes through the point (5,1), then its intercept on the Y axis is?

y(5)=-5+b=1
B = 6
So its intercept on the y-axis is 6

In the linear function y = 3x + 12, if the value range of Y is - 6 ≤ y ≤ 6, find the value range of X

-6≤3x+12≤6
-6≤x≤-2
Didn't your teacher teach you?