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To solve for the distance from the point \( (12, 5) \) to the line \( 5y + 12x = 0 \), we can use the distance formula from a point \((x_1, y_1)\) to a line \(ax + by + c = 0\):

\[
\text{Distance} = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}
\]

For the line \(5y + 12x = 0\):
- \(a = 12\)
- \(b = 5\)
- \(c = 0\)

For the point \((12, 5)\):
- \(x_1 = 12\)
- \(y_1 = 5\)

Plugging in these values:

\[
\text{Distance} = \frac{|12 \cdot 12 + 5 \cdot 5 + 0|}{\sqrt{12^2 + 5^2}}
\]

Calculate each part:

1. **Numerator**:
   \[
   |12 \cdot 12 + 5 \cdot 5| = |144 + 25| = |169| = 169
   \]

2. **Denominator**:
   \[
   \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13
   \]

So,

\[
\text{Distance} = \frac{169}{13} = 13
\]

The correct answer is **13 units**. 

If you'd like more details or have questions, please let me know!

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Here are 5 related questions for practice:

1. How would you find the distance from a point to a vertical line (e.g., \( x = 3 \))?
2. What changes in the formula if the line equation is in slope-intercept form \( y = mx + b \)?
3. How would you find the distance from a point to a line if the coordinates are negative?
4. How can this formula be used in three-dimensional space for a plane?
5. How does the formula adjust if the line equation has fractions or decimals?

**Tip:** When calculating distances, ensure all values are simplified to avoid common errors with radicals or large numbers.

Pada bidang Kartesius, jarak dari titik (12, 5) ke garis 5y + 12x = 0 adalah ____ unit.

Learn how to calculate the distance from the point (12, 5) to the line 5y + 12x = 0 using the distance formula. The solution involves applying the perpendicular distance theorem, ideal for students in Grades 10-12.

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