Find the coordinates the those point on the line `3x+ 2y = 5` which are equisdistant from the lines ` 4x+ 3y - 7 = 0 and 2y - 5 = 0 `

Answer

To find the coordinates of the points on the line \(3x + 2y = 5\) that are equidistant from the lines \(4x + 3y - 7 = 0\) and \(2y - 5 = 0\), we can follow these steps:

### Step 1: Identify the equations of the lines
We have the following lines:
1. Line 1: \(3x + 2y = 5\)
2. Line 2: \(4x + 3y - 7 = 0\)
3. Line 3: \(2y - 5 = 0\)

### Step 2: Rewrite the equations in standard form
The equations of the lines can be rewritten as:
- Line 1: \(3x + 2y - 5 = 0\)
- Line 2: \(4x + 3y - 7 = 0\)
- Line 3: \(0x + 2y - 5 = 0\) (which simplifies to \(y = \frac{5}{2}\))

### Step 3: Use the formula for the distance from a point to a line
The distance \(d\) from a point \((h, k)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]

### Step 4: Set up the distance equations
We need to set the distances from the point \((h, k)\) on line 1 to lines 2 and 3 equal:
1. Distance from \((h, k)\) to line 2:
\[ d_1 = \frac{|4h + 3k - 7|}{\sqrt{4^2 + 3^2}} = \frac{|4h + 3k - 7|}{5} \]

2. Distance from \((h, k)\) to line 3:
\[ d_2 = \frac{|0h + 2k - 5|}{\sqrt{0^2 + 2^2}} = \frac{|2k - 5|}{2} \]

### Step 5: Set the distances equal
Set \(d_1 = d_2\):
\[ \frac{|4h + 3k - 7|}{5} = \frac{|2k - 5|}{2} \]

### Step 6: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 2|4h + 3k - 7| = 5|2k - 5| \]

### Step 7: Solve the absolute value equations
This can lead to two cases:
1. \(2(4h + 3k - 7) = 5(2k - 5)\)
2. \(2(4h + 3k - 7) = -5(2k - 5)\)

#### Case 1:
\[ 8h + 6k - 14 = 10k - 25 \]
Rearranging gives:
\[ 8h - 4k + 11 = 0 \quad \text{(Equation 1)} \]

#### Case 2:
\[ 8h + 6k - 14 = -10k + 25 \]
Rearranging gives:
\[ 8h + 16k - 39 = 0 \quad \text{(Equation 2)} \]

### Step 8: Solve the system of equations
We have two equations:
1. \(8h - 4k + 11 = 0\)
2. \(3h + 2k - 5 = 0\)

From Equation 2, we can express \(k\) in terms of \(h\):
\[ 2k = 5 - 3h \implies k = \frac{5 - 3h}{2} \]

Substituting \(k\) into Equation 1:
\[ 8h - 4\left(\frac{5 - 3h}{2}\right) + 11 = 0 \]
This simplifies to:
\[ 8h - 2(5 - 3h) + 11 = 0 \]
\[ 8h - 10 + 6h + 11 = 0 \]
\[ 14h + 1 = 0 \implies h = -\frac{1}{14} \]

Substituting \(h\) back to find \(k\):
\[ k = \frac{5 - 3\left(-\frac{1}{14}\right)}{2} = \frac{5 + \frac{3}{14}}{2} = \frac{\frac{70}{14} + \frac{3}{14}}{2} = \frac{\frac{73}{14}}{2} = \frac{73}{28} \]

Thus, one point is:
\[ \left(-\frac{1}{14}, \frac{73}{28}\right) \]

### Step 9: Repeat for the second case
Now we solve the second case:
From Equation 2:
\[ 8h + 16k - 39 = 0 \]
Using the same expression for \(k\):
\[ 8h + 16\left(\frac{5 - 3h}{2}\right) - 39 = 0 \]
This simplifies to:
\[ 8h + 8(5 - 3h) - 39 = 0 \]
\[ 8h + 40 - 24h - 39 = 0 \]
\[ -16h + 1 = 0 \implies h = \frac{1}{16} \]

Substituting \(h\) back to find \(k\):
\[ k = \frac{5 - 3\left(\frac{1}{16}\right)}{2} = \frac{5 - \frac{3}{16}}{2} = \frac{\frac{80}{16} - \frac{3}{16}}{2} = \frac{\frac{77}{16}}{2} = \frac{77}{32} \]

Thus, the second point is:
\[ \left(\frac{1}{16}, \frac{77}{32}\right) \]

### Final Answer
The coordinates of the points on the line \(3x + 2y = 5\) that are equidistant from the lines \(4x + 3y - 7 = 0\) and \(2y - 5 = 0\) are:
1. \(\left(-\frac{1}{14}, \frac{73}{28}\right)\)
2. \(\left(\frac{1}{16}, \frac{77}{32}\right)\)