It is known that the bisector equation of the two lines l1 4x 3y 1 0l2 12x 5Y 13 0 is known

The upstairs method is correct, but the number is wrong, resulting in an incorrect result.

Let the point p(x,y) be any point on the bisector of the angle of the intersection of two straight lines, then p is equal to the distance of the two straight lines, so |4x-3y+1|/√(16+9)=|12x+5y+13|(144+25), which is |4x-3y+1|/5=|12x+5y+13|13, so (4x-3y+1) 5= (12x+5y+13) 13 , simplified to 2x+16y+13=0 , or 56x-7y+39=0

4x－3y＋1｜/√（16＋9）＝｜x＋5y＋13｜/√（1＋25）。

4x 3y 1) 5 (x 5y 13) 26, or (4x 3y 1) 5 (x 5y 13) 26.

From (4x 3y 1) 5 (x 5y 13) 26, get:

4√26x－3√26y＋√26＝5x＋25y＋65，∴（4√26－5）x－（25＋3√26）y＋√26－65＝0。

From (4x 3y 1) 5 (x 5y 13) 26, get:

4√26x－3√26y＋√26＝－5x－25y－65，∴（5＋4√26）x＋（25－3√26）y＋65＋√26＝0。

There are two equations for the angle bisector of two straight lines that meet the conditions, namely:

4√26－5）x－（25＋3√26）y＋√26－65＝0、（5＋4√26）x＋（25－3√26）y＋65＋√26＝0。

Find the slope of the connecting line k1 = 3 4 k2 = -5 12, calculate the sin cos of the angle between the two lines and the x-axis respectively, and then find the tan of the angle between one of the angle bisectors and the x-axis, that is, its slope, and find the slope of the other angle bisector, find the focus of the two lines, and find the two angle bisectors.

Find the intersection point first. 4x-3y+1=0,12x+5y+13=0 to get x=-11 14,y=-5 7

Then find the slope of the bisector and set it to k

Then (using the formula of the sandwich and hidden angle of the two cover lines, tan =|k2-k1)/(1+k1*k2)|）

4/3-k)/(1+4k/3)|=12/5-k)/(1-12k/5)|

The solution yields k = 8 or k = -1 8

So the angle bisector equation is y+5 7=8(x+11 14) or the goodwill is y+5 7=(-1 8)*(x+11 14).

That is, 56x-7y+39=0 or 14x+112y+91=0

Let the points p(x,y) on the bisector nanodie

then p to two straight lines are equal.

So the annihilation is slow|2x-y|/√5=|x-2y+3|/√52x-y|=|x-2y+3|

2x-y=x-2y+3

or 2x-y=-(x-2y+3).

The eggplant is sold to. x+y-3=0

x-y+1=0

Summary. 2x-4y+10=02x+y-5=0, i.e., 2x-4y=-102x+y=5, i.e., 5y=15y=3, so x=1, so the intersection is (1,3)<>

What is the intersection point of two straight lines 2x-4y+10=0 and 2x+y-5=0.

Kiss Good evening The intersection of two straight lines 2x-4y+10=0 and 2x+y-5=0 is (1,3)<>

2x-4y+10=02x+y-5=0, i.e., 2x-4y=-102x+y=5, i.e., 5y=15y=3, so x=1, so the intersection is (1,3)<>

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Dear, this question is mainly to investigate the solution of binary linear equations<>

Kiss, the solution of the binary system of equations, that is, the intersection of these two straight lines, <>

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x-2y+4=0 1

x+y-2=0 2

2-1.

3y-6=0

y=2 33 formulas into 2 formulas.

x=0 because the straight line l is perpendicular to the straight line l3: 3x-4y+5=0.

Then the equation for the straight line l is 4x+3y+b=0

Substitute x=0 y=2.

4*0+3*2+b=0

b=-6, so the linear l equation is 4x+3y-6=0

The intersection of l1:x-2y+4=0 and l2:x+y-2=0 is (0,2).

Since the straight line is perpendicular to l3:3x-4y+5=0 and the slope of l3 is 3 4, the slope of the straight line is -4 3

So l is y=-(4 3)x+2 and the general form is 4x+3y-6=0

y=-3 4*x+2 first find the intersection point (x=0, y=2), and then find the slope of the required straight line k=-3 4 Got it?

I'm answering this by first solving the system of equations.

3x+2y-1=0

5x+2y+1=0 The solution gives x=-1,y=2, so the slope of the intersection point is (-1,2)l3 3 5, and the pin resistance is because l is perpendicular to l3, so k*k3=-1, so k=-5 3

Let y=-5 3x+b

When x=-1, y=2, 2=-5 3*(-1)+b, the solution is b=1 3, so Qingdoutan l:3y+5x-1=0

The intersection of the straight line l1:2x+2y-5=0 and l2:3x-2y-3=0 is the simultaneous solution equation.

2x+2y-5=0

3x-2y-3=0 intersection coordinates are (8 5

Parallel to the equation of a straight line 2x+y-3=0, then the straight line can be set to 2x+y-b=0 and pass the point (8 5

9 10) substitute b = 41 10

So the equation for a straight line is 20x

10y-41=0

The straight line parallel to 2x+y-3=0 is 2x+y+b=0l1, the focus of l2 is x, y(8 5, 9 5) is substituted into 2x+y+b=0, and b=-5 is obtained

So the equation for a straight line is 2x+y-5=0

Connecting 2x+2y-5=0 and 3x-2y-3=0 yields:

x=8/5y=9/10

That is, its intersection points are (8 5, 9 10).

Because the straight line sought is parallel to the straight line 2x+y-3=0

Let the straight line be 2x+y-b=0

then substitute (8 5, 9 10) into the desired straight line.

b=41/10

So the straight line 2x+y- is found

Standard: 20x+10y-41=0

Solving systems of equations:

2x+2y-5=0

3x-2y-3=0

Get: x=8 5, y=9 10

That is, the intersection points are (8 5, 9 10).

Let the straight line be 2x+y+a=0

Substituting the intersection coordinates in: a=

So the straight line is 20x+10y+39=0

1, After the intersection of these two straight lines, it can be seen that the required straight line has the same value as these two lines, so that 2x+3y-5=0 is obtained

3x-2y-1=0

Solve this equation.

x=1y=1

If the equation for this line is: y=ax+b; Replace x=1; y=1 is substituted into the equation to get 1=a+b; and parallel to the straight line 2x+y-6=0; So their slopes are equal; It is also known that the slope of the straight line 2x+y-6=0 is 2; So a=2; i.e. b=-1;

The equation for this straight line is: y=2x-1