Number of points having distance `sqrt5` from the straight line `x-2y+1=0` and a distance `sqrt13` from the lines `2x+3y-1=0` is

Options

A. 1
B. 2
C. 4
D. 5

Answer

To solve the problem of finding the number of points that are at a distance of \(\sqrt{5}\) from the line \(x - 2y + 1 = 0\) and at a distance of \(\sqrt{13}\) from the line \(2x + 3y - 1 = 0\), we will follow these steps:

### Step 1: Distance from the first line
The distance \(d\) from a point \((h, k)\) to the line \(Ax + By + C = 0\) is given by the formula:
\[ d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}} \]
For the line \(x - 2y + 1 = 0\), we have \(A = 1\), \(B = -2\), and \(C = 1\). Thus, the distance from the point \((h, k)\) to this line is:
\[ d_1 = \frac{|h - 2k + 1|}{\sqrt{1^2 + (-2)^2}} = \frac{|h - 2k + 1|}{\sqrt{5}} \]
We want this distance to be \(\sqrt{5}\):
\[ \frac{|h - 2k + 1|}{\sqrt{5}} = \sqrt{5} \]
Multiplying both sides by \(\sqrt{5}\), we get:
\[ |h - 2k + 1| = 5 \]
This gives us two equations:
1. \(h - 2k + 1 = 5\)
2. \(h - 2k + 1 = -5\)

### Step 2: Solve the equations from Step 1
From the first equation:
\[ h - 2k + 1 = 5 \implies h - 2k = 4 \implies h = 2k + 4 \quad \text{(Equation 1)} \]
From the second equation:
\[ h - 2k + 1 = -5 \implies h - 2k = -6 \implies h = 2k - 6 \quad \text{(Equation 2)} \]

### Step 3: Distance from the second line
Now, we consider the second line \(2x + 3y - 1 = 0\). The distance from the point \((h, k)\) to this line is:
\[ d_2 = \frac{|2h + 3k - 1|}{\sqrt{2^2 + 3^2}} = \frac{|2h + 3k - 1|}{\sqrt{13}} \]
We want this distance to be \(\sqrt{13}\):
\[ \frac{|2h + 3k - 1|}{\sqrt{13}} = \sqrt{13} \]
Multiplying both sides by \(\sqrt{13}\), we get:
\[ |2h + 3k - 1| = 13 \]
This gives us two more equations:
1. \(2h + 3k - 1 = 13\)
2. \(2h + 3k - 1 = -13\)

### Step 4: Solve the equations from Step 3
From the first equation:
\[ 2h + 3k - 1 = 13 \implies 2h + 3k = 14 \quad \text{(Equation 3)} \]
From the second equation:
\[ 2h + 3k - 1 = -13 \implies 2h + 3k = -12 \quad \text{(Equation 4)} \]

### Step 5: Combine the equations
Now we have four equations:
1. \(h = 2k + 4\) (from Equation 1)
2. \(h = 2k - 6\) (from Equation 2)
3. \(2h + 3k = 14\) (from Equation 3)
4. \(2h + 3k = -12\) (from Equation 4)

### Step 6: Solve for points
1. Substitute \(h = 2k + 4\) into Equation 3:
\[ 2(2k + 4) + 3k = 14 \implies 4k + 8 + 3k = 14 \implies 7k = 6 \implies k = \frac{6}{7} \]
Substitute \(k\) back to find \(h\):
\[ h = 2\left(\frac{6}{7}\right) + 4 = \frac{12}{7} + \frac{28}{7} = \frac{40}{7} \]
So one point is \(\left(\frac{40}{7}, \frac{6}{7}\right)\).

2. Substitute \(h = 2k - 6\) into Equation 3:
\[ 2(2k - 6) + 3k = 14 \implies 4k - 12 + 3k = 14 \implies 7k = 26 \implies k = \frac{26}{7} \]
Substitute \(k\) back to find \(h\):
\[ h = 2\left(\frac{26}{7}\right) - 6 = \frac{52}{7} - \frac{42}{7} = \frac{10}{7} \]
So another point is \(\left(\frac{10}{7}, \frac{26}{7}\right)\).

3. Similarly, we can find points for Equation 4 by substituting \(h = 2k + 4\) and \(h = 2k - 6\).

### Conclusion
After solving all combinations, we find a total of **4 distinct points** that satisfy both conditions.

### Final Answer
The number of points is **4**.