It is known that the intersection of the line L1 y x m and the line L2 y 2x 6 is found in the fourth

The intersection of the line l1:y=-x+m and the line l2:y=2x-6 is (x, y)-x+m=2x-6

3x=m+6

The intersection is in the fourth quadrant.

x＞0 m+6＞0

m＞－6y=-x+m x=m-y

y=2x-6 x=（1/2）y+3

m-y=（1/2）y+3

2/3）y=m－3

The intersection is in the fourth quadrant.

y＜0 m－3＜0

m3 is 6 m3

The line L2: Y=2x-6 intersects with the x-axis (3,0) and the Y-axis intersects (0,-6) the line L1:Y=-x+m is parallel to the line Y=-x, Y=-x+m is obtained by Y=-x translation, since the intersection is in the fourth quadrant.

Therefore, the two limit positions of y=-x+m are the two straight lines that cross (3,0) and (0,-6) respectively.

Substituting the coordinates of the two points into the y=-x+m equation yields m=3 and m=-6 and therefore -6

Draw a diagram and shift y=-x along the y-axis until the intersection point with l2 is in the fourth quadrant, at this time m=-6, and then the downward transition point is in the fourth quadrant.

So m<-6;

Since the intersection is in the fourth quadrant, so x>0, y<0 so 2+1 3m>0 , 2 3m-2<0

Solve the equation to get 3

Simplification of simultaneous equations: x=(m+6) 3;y=2m/3-2;

If the intersection is in the fourth quadrant, then x>0 and y<0

m+6) 3>0 m+6>0 then m>-6

2m 3-2<0 then m<3

So -6

m<-6 This kind of problem is the most intuitive and easiest to draw.

Solve systems of equations. y＝-2x+m

y＝2x-1

Get the coordinates of the chaotic obstruction of the intersection of the auspicious node.

x (m+1) state positive 4 and y (m-1) 2 because, the intersection is in the fourth quadrant.

So, (m+1) 4>0, (m-1) 20, m-1

Solve the system of equations y -2x+my 2x-1 to obtain the coordinates of the intersection x (m+1) 4, y (m-1) 2

Because You Chang digs the core of the gods, the intersection is in the fourth swift quadrant, (m+1) 4>0, (m-1) 20, m-1

The first shirt before the macro finds the intersection position or book The intersection coordinates are (m+3, 2m+3).

Because in the fourth quadrant m+3 0 2m+3 regret and 0

Then the range of m is -3 m

Solve systems of equations. y＝-2x+m

y＝2x-1

Get the coordinates of the intersection point.

x＝(m+1)/4，y＝(m-1)/2

Because, the intersection is in the fourth quadrant.

So, (m+1) 4>0, (m-1) 2<0, i.e., m+1>0, m-1<0

So, the range of m is -1

2x-my+4=0 (1)

2mx+3y-6=0 (2)

2)-(1)*m, get.

3+m^2)y=6+4m

y=(6+4m)/(3+m^2)

1)*3+(2)*m, get.

6+2m^2)x=6m-12

x=(6m-12)/(6+2m^2)

The coordinates of the intersection are (x,y), and to make the intersection in the second quadrant, then x<0,y>0, note that the denominator of x,y is 》0, so 6m-12<0,m<2

6+4m>0,m>-3/2

Therefore, the value range of m is -3 2