The straight line is known a 2 y 3a 1 x 1. Verify that no matter what the value of a is, the straigh

a-2)y=(3a-1)x-1

i.e. y=[(3a-1) (a-2)]x-[1 (a-2)] when [(3a-1) (a-2)] 0, that is, the slope is greater than 0, must pass the first quadrant, and when [(3a-1) (a-2)]=0, a=1 3, y=3 5, must pass the first quadrant.

When [(3a-1) (a-2)] 0, that is, the slope is less than 0, as long as -[1 (a-2)] is judged to be positive, the first quadrant can be explained.

Then from [(3a-1) (a-2)] 0, the solution is 1 3 a 2 then a-2 0

So-[1 (a-2)] 0

So also over the first quadrant.

In summary, the straight line always passes the first quadrant.

In order to make this straight line no more than the second quadrant.

The slope [(3a-1) (a-2)] 0 and the y-intercept -[1 (a-2)] 0

Solving inequalities gives us a 2

Equation the straight line.

a-2)y=(3a-1)x-1

Into. a(3x-y)+(2y-x-1)=0, let.

3x-y=0

Moreover. 2y-x-1=0, solution.

x=1/5，y=3/5

This shows that no matter what value a takes, the straight line is always past the points in the first quadrant (1 5, 3 5), so the straight line always passes the first quadrant.

If a=2 then the linear equation is.

5x-1=0

x=1 5, which must have passed through the first quadrant.

If a is not = 2, the analysis is as follows:

a-2)y=(3a-1)x-1

y=(3a-1) (a-2)]x-1 (a-2)if a-2>0

i.e. a>2 then 3a-1

So (3a-1) (a-2) >0

So the straight line passes through the first.

1. Three quadrants.

If a-2<0 and 3a-1>0

i.e. 1 30, illustrating 00, in this case the straight line part is in the first quadrant.

If 3a-1<0, then a-2<0

So (3a-1) (a-2) >0

So the straight line passes through the first.

1. Three quadrants.

In summary, no matter what value a takes, the straight line always passes through the first quadrant.

1, that is, to prove that at least one of the intersection points of the straight line with the x and y axes is on the positive axis.

x=0y=1/(2-a)

y=0x=1/(3a-1)

If the intersection with the y-axis is on the negative axis, then 1 (2-a)0

That is, the intersection point with the x-axis is on the positive axis.

If the intersection point with the x-axis is on the negative axis, then 1 (3a-1).

There are two cases: 1 "When a = 2, the straight line is: x = 1 5 is satisfied without passing through the second quadrant.

2"When a is not = 2, the straight line is y=(3a-1) (a-2)x-1 (a-2), and when a is not = 1 3, then (3a-1) (a-2)> 0 -1 (a-2)2

In summary, the stool Zheng: a>=2

1, that is, to prove that at least one of the intersection points of the straight line with the x and y axes is on the positive axis.

x=0 y=1/(2-a)

y=0 x=1/(3a-1)

If the intersection point with the y-axis is on the negative axis, then 1 (2-a)<0 2-a<0 a>21 (3a-1)>0 is the intersection point with the x-axis on the positive axis.

If the intersection point with the x-axis is on the negative axis, then 1 (3a-1)<0 3a-1<0 a<1 3

1 (2-a)>1 (2-1 3)>0 is the intersection point with the y-axis on the positive axis.

At least one of the intersection points of the straight line with the x and y axes is on the positive axis, and the straight line always passes each quadrant.

2。It is sufficient to make the straight line intersect the x-axis, the positive axis, the y-axis intersect, and the negative axis.

1/(3a-1)＞0

1/(2-a)＜0

Regardless of the value of a, the line (a-2)y=(3a-1)x-1 always passes the intersection of the line x-2y+1=0 and 3x-y=0, and these two lines intersect at the point.

a(2 15, 2 5), so the straight line always passes the first quadrant.

2) The slope of a(2 15,2 5),o(0,0) is 1 3, when a=2, the slope of the straight line (a-2)y=(3a-1)x-1 does not exist, and when a is not 0, the slope of the straight line (a-2)y=(3a-1)x-1 is.

3a-1) (a-2) < 1 3, so when 1 8

The verification point (1 5, 3 5) is always in a straight line, so it always passes through the first quadrant.

If a=2 then the linear equation is.

5x-1=0

x=1 5, which must have passed through the first quadrant.

If a is not = 2, the analysis is as follows:

a-2)y=(3a-1)x-1

y=(3a-1) (a-2)]x-1 (a-2)if a-2>0

i.e. a>2 then 3a-1

So (3a-1) (a-2) >0

So the straight line passes through the first.

1. Three quadrants.

If a-2<0 and 3a-1>0

i.e. 1 30, illustrating 00, in this case the straight line part is in the first quadrant.

If 3a-1<0, then a-2<0

So (3a-1) (a-2) >0

Therefore, the straight line passes through the first macro hole and the three quadrants.

As mentioned in the slag hand, no matter what value A takes, the straight line always passes through the first quadrant.

Solution: From the meaning of the question, it can be obtained: the intersection point of the line and the x-axis should be greater than 0; The intersection with the y-axis should be less than 0.

That is, when x=0, y<0;When y=0, x > 0. Substituting this into a straight line equation yields:

y=1/(a-2)<0；x=1 (3a-1)>0 solves: a<2 and a>1 3, i.e.,

1/3

Reducing the equation of the straight line (a-2)y=(3a-1)x-1 to a(3x-y)+(2y-x-1)=0, so that 3x-y=0 and 2y-x-1=0 are solved, and x=1 5, y=3 5 is solved, which shows that no matter what value a takes, the straight line is constant over the points in the first quadrant (1 5, 3 5), so the straight line always passes the first quadrant.

Solution: When a=2 the slope does not exist, so the straight line is x=1 5 This straight line crosses the first quadrant.

When a≠2 and the slope exists, then: y=(3a-1) (a-2)x - 1 (a-2).

At this time, it is divided into: when: (3a-1) (a-2) is greater than zero, less than zero is discussed in detailed classification, and can be obtained by combining a function image.

(a-2)y=(3a-1)x-1

y = (3a-1)/(a-2) x - 1/(a-2)

When a=1 3, i.e., 3a-1=0, the image is a straight line parallel to the x-axis, y=-1 (a-2)=-1 (1 3-2)=3 5 0, passing the first quadrant;

When a=2, i.e., a-2=0, the image is a straight line parallel to the y-axis, x=1 (3a-1)=1 (3*2-1)=1 5, passing the first quadrant;

When a 1 3 or a 2, the slope k=(3a-1) (a-2) 0, the image must pass the first quadrant;

When 1 3 a 2, the slope is less than zero, but the intersection coordinate with the y-axis =- 1 (a-2) 0, the image passes the first quadrant.

In summary, no matter what value K takes, the image passes the first quadrant.

a-2)y-(3a-1)x+1=0,ay-2y-3ax+x+1=0;

a(y-3x)+x-2y+1=0;

No matter what the value of a is; Over the fixed point y-3x=0;

x-2y+1=0;

x-6x+1=0;

5x=1;x=1/5;

y=3/5;

So over the fixed point Lie difference (1 is in the first quadrant, so it must pass through the first quadrant.

2）∴（3a-1）/(a-2)＜0;

x=0;y=1/(2-a)≥0;

1/3＜a＜2

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In the first question, the three intervals divided into a=2 and a=1 3 are classified and discussed, and the points where x>0 and y> liquid 0 exist exist in any interval.

In the second question, x<0 and y>0 is the range of a in the second quadrant, and then the sum of this range can be found by scattering.

1.The shift is disturbed, and it is turned into a straight line equation that is over a fixed point, and the fixed point is found so that Li Que must be in the first quadrant.

2.According to the fixed-point plot in 1, the limit value is found.

1: From (a-2)y=(3a-1)x-1: (y-3x)a=2y-x-1, because it is proved that no matter what value a is, the straight line always passes the first quadrant, so the coefficient in front of a is 0, and y-3x=0 and 2y-x-1=0 are bifurcated:

x= y= so the straight line is constant through the point (, the point is in the first quadrant, the proof is complete!) Slow down.

2: When the slope does not exist, a=2, is a straight line x = but the second quadrant, which is in line with the topic; When the slope exists, the straight line must pass the quadrant, then it is obtained by y=(3a-1) (a-2) -1 which (a-2), (3a-1) (a-2) >0 and -1 (a-2)<0 are solved: a>2

The value of a can be a>=2