The number of lines that can be drawn passing through point (2, 3) so that its perpendicular distance from (-1, 6) is equal to 6 is :

Answer

To solve the problem, we need to find the number of lines that can be drawn through the point (2, 3) such that the perpendicular distance from the point (-1, 6) to the line is equal to 6.

### Step-by-Step Solution:

1. **Equation of the Line**:
We can express the equation of the line passing through the point (2, 3) in slope-intercept form. Let the slope of the line be \( m \). The equation can be written as:
\[ y - 3 = m(x - 2) \]
Rearranging gives:
\[ y = mx - 2m + 3 \]
This can be rewritten in the standard form \( Ax + By + C = 0 \):
\[ mx - y + (3 - 2m) = 0 \]
Here, \( A = m \), \( B = -1 \), and \( C = 3 - 2m \).

2. **Distance Formula**:
The distance \( d \) from a point \( (x_1, y_1) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
In our case, the point is \( (-1, 6) \) and we want the distance to be 6. Plugging in the values:
\[ 6 = \frac{|m(-1) + (-1)(6) + (3 - 2m)|}{\sqrt{m^2 + 1}} \]
Simplifying the numerator:
\[ = \frac{| -m - 6 + 3 - 2m |}{\sqrt{m^2 + 1}} = \frac{| -3m - 3 |}{\sqrt{m^2 + 1}} \]
Thus, we have:
\[ 6 = \frac{3|m + 1|}{\sqrt{m^2 + 1}} \]

3. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 6\sqrt{m^2 + 1} = 3|m + 1| \]
Dividing both sides by 3:
\[ 2\sqrt{m^2 + 1} = |m + 1| \]

4. **Squaring Both Sides**:
Squaring both sides results in:
\[ 4(m^2 + 1) = (m + 1)^2 \]
Expanding both sides:
\[ 4m^2 + 4 = m^2 + 2m + 1 \]
Rearranging gives:
\[ 4m^2 - m^2 - 2m + 4 - 1 = 0 \]
Simplifying:
\[ 3m^2 - 2m + 3 = 0 \]

5. **Finding the Discriminant**:
We will now calculate the discriminant \( D \) of the quadratic equation \( 3m^2 - 2m + 3 = 0 \):
\[ D = b^2 - 4ac = (-2)^2 - 4 \cdot 3 \cdot 3 = 4 - 36 = -32 \]
Since the discriminant is negative, this indicates that there are no real roots for \( m \).

6. **Conclusion**:
Since there are no real values for \( m \), it means there are no lines that can be drawn through the point (2, 3) such that the perpendicular distance from (-1, 6) is equal to 6. Therefore, the number of such lines is:
\[ \text{Number of lines} = 0 \]