Finishing: kx-y+2-k=0

The distance from the origin (0.0) to the straight line d= 1.

D= absolute value (2-k)/ radical number k 2+1=1.

The solution is k=3/4, so the straight line y=3/4(x- 1)+2.

Finishing: 3x-4y+5=0

When k does not exist, x= 1 also satisfies the problem.

There are two straight lines: 3x-4y+5=0 and x= 1.

2, the set point is P(x, y)

Root number (x 2+y)/root number (x-3) 2+y) = 1/2.

Simplification: 2x 2+2y 2 = x 2-6x+9+y 2

x^2+6x-9+y^2=0

(x+3)^2+y^2= 18

This curve represents a circle with a center of (-3,0) and a radius of 2 root numbers 3.

3. The straight line l 13x-4y-7=0. The slope of the straight line k=3/4.

The line l2 is perpendicular to the slope of the line 3x-4y-7 k 1=-4/3.

Let the equation of l2 line be: y =-4x/3+b.

Let x=0, then y = b.

Let y=0. Then x=4b/3.

Because the triangle formed by the two coordinate axes is a right triangle. So the third side of the triangle = root sign (b 2+ 16b 2/9) = 5b/3.

Because the perimeter is equal to 10.

Yes: absolute value (b+4b/3+5b/3)= absolute value (4b)= 10. The solution is b= plus or minus 5/2.

So the equation L2 of the straight line: y =-4x/3 plus or minus 5/2.