Analytic Geometry: Equation of a Line
Given two points, the equation of a line whose points are equidistant from the two given points is determined.
| Problem | Given Points | Equation of the Line |
| 1 | P1(-3,5) P2(-2,-5) | 2x-20y+5=0 |
| 2 | P1(9,7) P2(1,-1) | 1x+1y-8=0 |
| 3 | P1(3,7) P2(-9,-4) | 24x+22y+39=0 |
| 4 | P1(4,8) P2(-1,5) | 5x+3y-27=0 |
| 5 | P1(9,-6) P2(-4,-9) | 13x+3y-10=0 |
| 6 | P1(-8,7) P2(7,-5) | 10x-8y+13=0 |
| 7 | P1(3,4) P2(-6,-8) | 6x+8y+25=0 |
| 8 | P1(4,-3) P2(-8,-4) | 24x+2y+55=0 |
| 9 | P1(1,8) P2(3,-9) | 4x-34y-25=0 |
| 10 | P1(-1,3) P2(8,-7) | 18x-20y-103=0 |
| 11 | P1(-3,4) P2(-8,-4) | 10x+16y+55=0 |
| 12 | P1(9,1) P2(-9,6) | 36x-10y+35=0 |
| 13 | P1(-5,9) P2(-8,-3) | 2x+8y-11=0 |
| 14 | P1(-2,2) P2(3,-1) | 5x-3y-1=0 |
| 15 | P1(2,1) P2(2,-4) | 0x-2y-3=0 |
| 16 | P1(8,7) P2(-3,-9) | 22x+32y-23=0 |
| 17 | P1(5,-3) P2(-8,-8) | 13x+5y+47=0 |
| 18 | P1(7,1) P2(-5,5) | 3x-1y-0=0 |
| 19 | P1(-9,1) P2(3,-5) | 2x-1y+4=0 |
| 20 | P1(1,3) P2(8,-1) | 14x-8y-55=0 |
| 21 | P1(3,-5) P2(-4,-9) | 14x+8y+63=0 |
| 22 | P1(-3,6) P2(-6,-6) | 2x+8y+9=0 |
| 23 | P1(5,7) P2(-5,-7) | 5x+7y-0=0 |
| 24 | P1(4,9) P2(-9,1) | 26x+16y-15=0 |
| 25 | P1(4,1) P2(-5,-4) | 9x+5y+12=0 |
| 26 | P1(5,6) P2(-3,9) | 16x-6y+29=0 |
| 27 | P1(-8,4) P2(-4,-7) | 8x-22y+15=0 |
| 28 | P1(-1,3) P2(4,-2) | 1x-1y-1=0 |
| 29 | P1(3,5) P2(6,-1) | 2x-4y-1=0 |
| 30 | P1(6,-2) P2(-7,-3) | 13x+1y+9=0 |
