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The problem asks to find the distance from the point \( P(2, -3) \) to three lines:

1. \( a) \ x + 2 = 0 \)
2. \( b) \ y = 49 \)
3. \( c) \ 2x + 5 = 0 \)

### Let's solve this step-by-step:

#### 1. Line \( a) \ x + 2 = 0 \)
This is a vertical line at \( x = -2 \). The distance \( d \) from a point \( (x_1, y_1) \) to a vertical line \( x = c \) is given by:
\[
d = |x_1 - c|
\]
For the point \( P(2, -3) \), we have:
\[
d = |2 - (-2)| = |2 + 2| = 4
\]
Thus, the distance is 4 units.

#### 2. Line \( b) \ y = 49 \)
This is a horizontal line at \( y = 49 \). The distance from a point \( (x_1, y_1) \) to a horizontal line \( y = c \) is:
\[
d = |y_1 - c|
\]
For the point \( P(2, -3) \), the distance is:
\[
d = |-3 - 49| = |-52| = 52
\]
Thus, the distance is 52 units.

#### 3. Line \( c) \ 2x + 5 = 0 \)
This is a slanted line, and we need to use the formula for the distance between a point \( (x_1, y_1) \) and a general line \( Ax + By + C = 0 \):
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]
Here, the line \( 2x + 5 = 0 \) can be rewritten as \( 2x + 0 \cdot y + 5 = 0 \), so \( A = 2 \), \( B = 0 \), and \( C = 5 \). For the point \( P(2, -3) \), we have:
\[
d = \frac{|2(2) + 0(-3) + 5|}{\sqrt{2^2 + 0^2}} = \frac{|4 + 5|}{\sqrt{4}} = \frac{9}{2} = 4.5
\]
Thus, the distance is 4.5 units.

### Final Answer:
- Distance to line \( x + 2 = 0 \): 4 units
- Distance to line \( y = 49 \): 52 units
- Distance to line \( 2x + 5 = 0 \): 4.5 units

Would you like more details or have any questions about these calculations?

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Here are 5 related questions:
1. How do you derive the distance formula for a point to a line?
2. What is the geometric meaning of vertical and horizontal lines?
3. How do you calculate the distance between two points?
4. What is the general form of a line equation, and how does it affect the distance formula?
5. How would the distance formula change if the line had a slope?

**Tip**: Always double-check the equation format when using the distance formula to ensure you're applying the correct coefficients.

Hallar la distancia del punto P(2, -3) a las rectas: a) x + 2 = 0 b) y = 49 c) 2x + 5 = 0

Learn how to calculate the distance from point P(2, -3) to three lines: x + 2 = 0, y = 49, and 2x + 5 = 0. This detailed explanation covers key geometry concepts, including distance formula applications for vertical, horizontal, and general lines. Suitable for students in Grades 10-12.

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