The co-ordinates of those points on the line `3x+2y=5` which are equisdistant from the lines `4x+3y-7=0` and `2y-5=0` are ……and…………..

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 will follow these steps:

### Step 1: Identify the lines and their equations
We have:
1. The line \(L_1: 3x + 2y = 5\)
2. The line \(L_2: 4x + 3y - 7 = 0\)
3. The line \(L_3: 2y - 5 = 0\) (which simplifies to \(y = \frac{5}{2}\))

### Step 2: Find the distance from a point \((x_1, y_1)\) on line \(L_1\) to lines \(L_2\) and \(L_3\)
The distance \(D\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by the formula:
\[ D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

#### Distance to line \(L_2\):
For line \(L_2: 4x + 3y - 7 = 0\):
- Here, \(A = 4\), \(B = 3\), and \(C = -7\).
- The distance \(D_1\) is:
\[ D_1 = \frac{|4x_1 + 3y_1 - 7|}{\sqrt{4^2 + 3^2}} = \frac{|4x_1 + 3y_1 - 7|}{5} \]

#### Distance to line \(L_3\):
For line \(L_3: 2y - 5 = 0\):
- Here, \(A = 0\), \(B = 2\), and \(C = -5\).
- The distance \(D_2\) is:
\[ D_2 = \frac{|0 \cdot x_1 + 2y_1 - 5|}{\sqrt{0^2 + 2^2}} = \frac{|2y_1 - 5|}{2} \]

### Step 3: Set the distances equal
Since we want the points to be equidistant from both lines:
\[ D_1 = D_2 \]
This gives us:
\[ \frac{|4x_1 + 3y_1 - 7|}{5} = \frac{|2y_1 - 5|}{2} \]

### Step 4: Substitute \(y_1\) from line \(L_1\)
From the equation of line \(L_1\):
\[ 2y_1 = 5 - 3x_1 \implies y_1 = \frac{5 - 3x_1}{2} \]

### Step 5: Substitute \(y_1\) into the distance equation
Substituting \(y_1\) into the distance equation:
\[ \frac{|4x_1 + 3\left(\frac{5 - 3x_1}{2}\right) - 7|}{5} = \frac{|2\left(\frac{5 - 3x_1}{2}\right) - 5|}{2} \]
Simplifying both sides:
1. Left side:
\[ 4x_1 + \frac{15 - 9x_1}{2} - 7 = 4x_1 + \frac{15 - 14}{2} = 4x_1 + \frac{1}{2} \]
Thus:
\[ |4x_1 + \frac{1}{2}| = 5D_1 \]

2. Right side:
\[ |5 - 3x_1 - 5| = |-3x_1| = 3|x_1| \]

### Step 6: Solve the equation
Equating both sides:
\[ |4x_1 + \frac{1}{2}| = 3|x_1| \]
This gives us two cases to solve:
1. \(4x_1 + \frac{1}{2} = 3x_1\)
2. \(4x_1 + \frac{1}{2} = -3x_1\)

#### Case 1:
\[ 4x_1 - 3x_1 = -\frac{1}{2} \implies x_1 = -\frac{1}{2} \]
Substituting \(x_1\) back to find \(y_1\):
\[ y_1 = \frac{5 - 3(-\frac{1}{2})}{2} = \frac{5 + \frac{3}{2}}{2} = \frac{\frac{10 + 3}{2}}{2} = \frac{13}{4} \]

#### Case 2:
\[ 4x_1 + 3x_1 = -\frac{1}{2} \implies 7x_1 = -\frac{1}{2} \implies x_1 = -\frac{1}{14} \]
Substituting \(x_1\) back to find \(y_1\):
\[ y_1 = \frac{5 - 3(-\frac{1}{14})}{2} = \frac{5 + \frac{3}{14}}{2} = \frac{\frac{70 + 3}{14}}{2} = \frac{73}{28} \]

### Final Coordinates
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. \((- \frac{1}{2}, \frac{13}{4})\)
2. \((- \frac{1}{14}, \frac{73}{28})\)