explain this formula: d=|Ax1+By1+C|/$$ \sqrt{\left(A^2+B^2\right)} $$

The formula you provided calculates the **distance** $ d $ from a point $ (x_1, y_1) $ to a line represented by the equation $ Ax + By + C = 0 $ in a two-dimensional plane. Here's a breakdown of how it works:

### Formula:
\[
d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}
\]

### Components:
- $ A $, $ B $, and $ C $ are the coefficients from the line equation $ Ax + By + C = 0 $.
- $ (x_1, y_1) $ is the point from which you want to calculate the distance to the line.
- The numerator $ |Ax_1 + By_1 + C| $ gives the perpendicular distance from the point to the line without considering direction.
- The denominator $ \sqrt{A^2 + B^2} $ normalizes the distance, making the formula independent of the slope or the specific form of the line equation.

### Step-by-Step Explanation:
1. **Line Equation**: A line in 2D can be expressed as $ Ax + By + C = 0 $. The coefficients $ A $, $ B $, and $ C $ determine the orientation and position of the line in the plane.
   
2. **Point**: The point $ (x_1, y_1) $ is the location in the plane from which you want to measure the distance to the line.

3. **Numerator**: The expression $ Ax_1 + By_1 + C $ calculates the value of the point $ (x_1, y_1) $ when substituted into the line equation. This gives a value proportional to the perpendicular distance between the point and the line, and taking the absolute value $ |Ax_1 + By_1 + C| $ ensures the distance is positive.

4. **Denominator**: $ \sqrt{A^2 + B^2} $ is the normalization factor. This represents the magnitude of the vector $ (A, B) $, which corresponds to the direction of the normal to the line. This factor ensures that the formula provides the correct perpendicular distance, regardless of the scaling of the line equation.

### Key Idea:
The formula measures the **perpendicular distance** between a point and a line in a two-dimensional space. The distance is always non-negative, and the normalization by $ \sqrt{A^2 + B^2} $ ensures that the distance is independent of the specific representation of the line equation.

### Example:
For the line equation $ 3x + 4y - 10 = 0 $ and the point $ (2, 1) $:

1. The numerator is $ |3(2) + 4(1) - 10| = |6 + 4 - 10| = |0| = 0 $.
2. The denominator is $ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $.

Thus, the distance is:
\[
d = \frac{0}{5} = 0
\]
This means the point $ (2, 1) $ lies exactly on the line.

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### Do you have any questions or want further details?

### 5 Related Questions:
1. How do we derive the distance formula from a point to a line?
2. How does this formula change in 3D space?
3. What are the geometric interpretations of the numerator and denominator?
4. Can we use this formula for vertical or horizontal lines?
5. How does this formula apply to lines in different coordinate systems?

### Tip: 
When working with distances to lines, it's important to check that the line equation is in standard form $ Ax + By + C = 0 $ before applying the formula.

Learn how to calculate the perpendicular distance from a point to a line in 2D using the formula d = |Ax1 + By1 + C| / sqrt(A^2 + B^2). This step-by-step guide explains each component of the formula and includes an example problem. Suitable for students in Grades 10-12.

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