It is known that the distance from A 1,0 ,B 3,2 3 to the straight line l is 2, and the equation for

There are 3 such straight lines:

1) connect ab, calculate the length of the straight line ab is 4, then one of the straight lines is perpendicular to ab and pass the midpoint of ab, so that the distance from the two points to the straight line is 2, the slope of ab is calculated (2 3-0) (3-1) = 3, the slope of the straight line is set to k, the two straight lines are perpendicular, the product of the slope is -1, then 3*k=-1, k=- 3 3, the straight line passes the midpoint of ab, the midpoint is (1, 3), then y- 3=- 3 3 (x-1).

2) The other two are parallel to ab, the slope is 3, but the distance between the straight line and ab is 2, exactly one is above ab, the other is below ab, the ab straight line is y= 3(x-1) = 3x- 3, let the straight line be y= 3x+m, using the formula of the distance between the two parallel straight lines, then |m-（-3）|[(1)2 +(3)2]=2, the solution is m=4-3 or -4-3, and the substitution will result in two other straight lines.

1.Straight line l ab and the distance is 2

kab=√3

Let the equation for the line l be: y = 3x+b

The distance from a(1,0) to the straight line l is 2,2=|√3+b|/23+b=±4 b=-√3±4 y=√3x-√3±42. |ab|=√(4+12)=4

So the distance from the perpendicular bisector of the straight line ab to a, b is 2, and the slope of the perpendicular bisector is k=- 3 3

ab midpoint coordinates (2, 3).

Point oblique. y-√3=-√3/3(x-2)

x+ 3y-5=0

There are three straight lines that meet the requirements.

The perpendicular bisector y=- 3 3x+5 3 32Two straight lines parallel to AB.

y= 3x- 3-4 and y= 3x- 3+4.

<> straight line l is the common tangent of two circles (two outside and one inside).

Concentric line: 3x-y- 3=0, tangent (2, 3) inner tangent: 3x+ 3y-5=0

Grandfather tangent: 3x-y- 3+4=0, 3x-y-( 3+4)=0

What is sought is the vertical wide space of AB and the bisecting loss guess line.

ab midpoint: (3, 3 5).

ab slope: 4 prudence only 3

l：y-3/5=-3/4(x-3)

y=-3/4x+57/20

It is known that the distance from the point (a,2) to the line l:x-y+3=0 is l, and the distance from the point to the line is d=|a-2+3|2=1

a+1|/√2=1

a> know that the answer 0) envy infiltration, then |a+1|=a+1

A+1 = 2, a= 2-1

It shows that the straight line sought by the early relatives is parallel to AB.

Since the slope of the straight line ab is (-4-0) (4-1)=-4 rolling3, the equation of the straight line can be set as.

3x+4y+m=0

Substitute (1,0) into the point-to-straight-line distance formula.

m=12 or -18

So the equation for the hall of the straight line is:

3x+4y+12=0

or 3x+4y-18=0

The farthest is the straight line perpendicular to the straight line ab.

Straight line ab slope 1 3

So the slope of the straight line perpendicular to ab is -3

Let y=-3x+b, and pass the point a(3,4).

The solution is b=13

So the equation is y= -3x+13

Find the line of ab first: y=1 3x + 3, and then find the line perpendicular to it. I just remember what a relation was and what multiplied to equal 1 or minus 1 (probably a coefficient relation to x), but I forgot the specifics.

Because after 3x-y-1=, the solution obtains: the straight line l passes through p(4 5, 7 5), and the equation l of the straight line l can be found as y-7 5=k(x-4 5), and then k is obtained by using the formula of the distance from the point to the straight line, and it can be substituted.

The intersection point can be obtained from the equation of the line L1 and the line L2 as (1,2), then the equation of the straight line can be set as y=k(x-1)+2, because the distance from the point a and the point b to the straight line is equal, so da=db can be solved from the distance from the point to the straight line k=1 6, k=-1 2 can be substituted into the set equation, x-6y+11=0 or x+2y-5=0

Solution 1: Let L1

The intersection point with l2 is p(x,y), and the slope of l is k, which is solved by p(1,2).

1) When l ab, there is k = k

ab=- then the equation for l is y-2=-

x-1), i.e., x+2y-5=0

2) When L passes the midpoint m of AB, it is easy to obtain m(4,.The equation for l is.

i.e. x-6y+11=0

The equation for the straight line l is x+2y-5=0 or x-6y+11=0

Solution 2: Let the equation for l be (3x-y-1) + x+y-3) = 0, that is, (3+ ) x + ( 1) y-1-3 = 0∵da

db,∴.The solution is =-7 or =-

Substituting the l equation yields x+2y-5=0 or x-6y+11=0

Because l1 is over (0,3), let y=kx

3 bring (3,0) into k=-1

So l1:y=-x

3 Because let l2:y=ax

b Bring two points in.

ab=22ab=-3

The solution yields a=5 3 and b=1 3

So l2:3y=5x

1 Column System of Equations:

y=-x33y=5x

1 solution yields x=1, y=2

Find d(1,2).

Let l be y=ax+b

Bringing in two points of cd, we get a = 1 6 and b = 11 6 so the equation for l is x-6y + 11 = 0

k(ab)=2v3 2=v3, straight ab:y=v3(x-1)=v3x-v3 v3x-y-v3=0

The straight line is l, and when lllab, let the straight line l be v3x-y+c=0, |c-(-v3)|/v(3+1)=|c+v3|/2=1, |c+v3|=2

c+v3=+-2 c=-v3+-2, so the l equation is v3x-y-v3+-2=0

When l intersects ab, l must pass the midpoint m(2,v3) of ab, and let l be y-v3=k(x-2), i.e., kx-y-2k+v3=0

The distance from point a(1,0) to l = 1, so |k*1-0-2k+v3|v(k 2+1)=1, get k=v3 3

The distance from point b(3,2v3) to l = 1, so |3k-2v3-2k+v3|v(k 2+1)=1, get k=v3 3

So the equation for l is v3 3*x-y-v3 3=0, and both sides are multiplied by v3, i.e., x-v3y-1=0

In summary, there are three straight lines that meet the conditions: v3x-y-v3+-2=0 and x-v3y-1=0

Let the equation for the straight line be y=kx+b and use the formula from the point to the straight line distance |k+b|/√k^2+1)=1|3k-2√3+b|k 2 + 1) = 1 3 3 + b = (k + b) when taking + 2k = 2 3 k = 3 because the distance between the two points and l is equal, so the midpoint of the two points must be on l (with triangles congruent or anything can be proved) both (2, 3) on l 2k- 3+b=0b=- 3, when taking - 2k+b= 3, and then use |k+b|k 2 + 1) = 1 changes to b 2 + 2kb = 1 The solution of the two formulas gives b = wide trillion 3 3 k = 3 3 so the straight line l: y= 3x- 3 y= 3 3x + 3 3

When the slope of the straight line l does not exist without the withered reeds in the mu, x=2 satisfies the meaning of the question if the slope of the straight line exists y=kx+ba(1,0), and the distance from b(3,2 3) to the straight line l is equal to 1|k+b|Root number (k 2+1) = 1|3k+b-2 root number 3|(root number k 2 + 1) = 1k = root number 3b = 2 + - root number 3y = root number resistance 3x + (2 + - root number 3).

A straight line with a distance equal to 1 to the point a(1,0) is a tangent of a circle with a(1,0) as the center and 1 as the radius.

Similarly. A straight line with a distance equal to 1 to the point b(3,2) is a tangent of a circle with b(3,2) as the center and 1 as the radius.

The distance between the two points a(1,0) and b(3,2) and the straight line l is equal to 1, then it is the common tangent of the two circles.

ab|=2 2>1 The two circles are separated.

So there are 4 more tangent lines.

y=1 and x=2

and y=x+(-1 2).

The distance between the two points a(1,0) and b(3,2) to the line l is equal to 1, so the line ab is parallel to the line l.

The equation for the straight line ab is.

y=x-1 Let the equation for the line l be y=x+b

The distance from point A to the line L is.

1+b|/√2=1

1+b)²=2

b=-1-√2

or b=-1+ 2

So. y=x-1+√2

Or. y=x-1-√2

And when A and B are on either side of a straight line, L.

Ask for ...... againThere are 4 in total.