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(-5,2) P2(6,-1) | 11x-3y-4=0 |
| 2 | P1(-1,7) P2(-9,-4) | 16x+22y+47=0 |
| 3 | P1(6,9) P2(8,-2) | 4x-22y+49=0 |
| 4 | P1(1,-3) P2(-4,-6) | 5x+3y+21=0 |
| 5 | P1(6,3) P2(-4,2) | 20x+2y-25=0 |
| 6 | P1(5,9) P2(-2,-1) | 14x+20y-101=0 |
| 7 | P1(5,1) P2(5,-2) | 0x-2y-1=0 |
| 8 | P1(-2,5) P2(3,-5) | 2x-4y-1=0 |
| 9 | P1(4,4) P2(-5,5) | 9x-1y+9=0 |
| 10 | P1(9,-3) P2(-3,-4) | 24x+2y-65=0 |
| 11 | P1(1,8) P2(-9,-2) | 1x+1y+1=0 |
| 12 | P1(-5,7) P2(-1,-1) | 1x-2y+9=0 |
| 13 | P1(-4,1) P2(7,-7) | 22x-16y-81=0 |
| 14 | P1(7,6) P2(-5,1) | 24x+10y-59=0 |
| 15 | P1(1,-1) P2(-3,-4) | 8x+6y+23=0 |
| 16 | P1(6,7) P2(-2,-2) | 16x+18y-77=0 |
| 17 | P1(9,7) P2(1,-8) | 16x+30y-65=0 |
| 18 | P1(-1,7) P2(-7,-4) | 12x+22y+15=0 |
| 19 | P1(-3,3) P2(-2,-5) | 2x-16y-11=0 |
| 20 | P1(6,9) P2(3,-6) | 1x+5y-12=0 |
| 21 | P1(9,-9) P2(-1,-3) | 5x-3y-38=0 |
| 22 | P1(1,7) P2(-7,-7) | 4x+7y+12=0 |
| 23 | P1(5,8) P2(-2,4) | 14x+8y-69=0 |
| 24 | P1(-9,4) P2(3,-4) | 3x-2y+9=0 |
| 25 | P1(4,9) P2(5,-1) | 2x-20y+71=0 |
| 26 | P1(5,-2) P2(-9,-7) | 28x+10y+101=0 |
| 27 | P1(9,9) P2(-3,-5) | 6x+7y-32=0 |
| 28 | P1(3,3) P2(-5,6) | 16x-6y+43=0 |
| 29 | P1(-1,1) P2(-2,-7) | 2x+16y+51=0 |
| 30 | P1(-5,8) P2(2,-5) | 7x-13y+30=0 |
