Question

find the coordinates of those points on the line 3x+2y=5 which are equidistant from the lines 4x+3y-7=0 and 2y-5=0.

Answer 1

Let a point (x1 , y1)  be equidistant from the lines 4x + 3 y - 7 = 0 and 2 y - 5 = 0.
Hence,  | 4 x1 + 3 y1 -7 | / √(4²+3²)  =  | 2 y1 - 5 | / √2²

      =>    | 4 x1 + 3 y1 - 7 | = 5/2 * | 2 y1 - 5 |
The point (x1, y1) lies on the line 3 x + 2 y = 5  =>  y1 = 2.5 - 1.5 x1
      =>   | 4 x1 + 3 (2.5 - 1.5 x1) | = 2.5 * | 2 (2.5 - 1.5 x1) - 5 |
      =>   | 7.5 - 0.5 x1 | = 2.5 * | - 3 x1 |

There are two possibilities:
    7.5 - 0.5 x1 = 7.5 x1    =>  x1 = 15/16    => y1 = 2.5 - 1.5 * 15/16 = 35/32
    7.5 - 0.5 x1 = -7.5 x1  =>   x1 = -15/14   => y1 = 2.5 - 1.5 *-15/14 = 115/28

         (15/16,  35/32)    and  (-15/14, 115/28)

Answer 2

Let a point (x1 , y1)  be equidistant from the lines 4x + 3 y - 7 = 0 and 2 y - 5 = 0.

Hence,  | 4 x1 + 3 y1 -7 | / √(4²+3²)  =  | 2 y1 - 5 | / √2²

     =>    | 4 x1 + 3 y1 - 7 | = 5/2 * | 2 y1 - 5 |

The point (x1, y1) lies on the line 3 x + 2 y = 5  =>  y1 = 2.5 - 1.5 x1

     =>   | 4 x1 + 3 (2.5 - 1.5 x1) | = 2.5 * | 2 (2.5 - 1.5 x1) - 5 |

     =>   | 7.5 - 0.5 x1 | = 2.5 * | - 3 x1 |

There are two possibilities:

   7.5 - 0.5 x1 = 7.5 x1    =>  x1 = 15/16    => y1 = 2.5 - 1.5 * 15/16 = 35/32

   7.5 - 0.5 x1 = -7.5 x1  =>   x1 = -15/14   => y1 = 2.5 - 1.5 *-15/14 = 115/28

        (15/16,  35/32)    and  (-15/14, 115/28)