Find the distance from P to l. Line l contains point (-2,1) and (4,1). Point P had coordinates (5,7).

To find the distance from a point \( P(x_1, y_1) \) to a line \( l \) passing through two points \( A(x_2, y_2) \) and \( B(x_3, y_3) \), we use the formula for the distance from a point to a line:

\[
d = \frac{|(y_3 - y_2)x_1 - (x_3 - x_2)y_1 + x_3y_2 - x_2y_3|}{\sqrt{(y_3 - y_2)^2 + (x_3 - x_2)^2}}
\]

We are given:

- Point \( P(5, 7) \)
- Points on the line: \( A(-2, 1) \) and \( B(4, 1) \)

### Step 1: Find the slope of the line
We calculate the difference in the y-coordinates and x-coordinates between \( A \) and \( B \):

\[
y_3 - y_2 = 1 - 1 = 0
\]
\[
x_3 - x_2 = 4 - (-2) = 6
\]

This indicates that the line is horizontal, as the slope is zero.

### Step 2: Apply the distance formula
For a horizontal line, the formula simplifies because we know the line has the equation \( y = 1 \), so the distance from a point \( P(x_1, y_1) \) to the line \( y = c \) is:

\[
d = |y_1 - c|
\]

Here, \( y_1 = 7 \) and \( c = 1 \):

\[
d = |7 - 1| = 6
\]

Thus, the distance from point \( P(5, 7) \) to the line \( l \) is **6 units**.

Would you like further clarification or additional details on this?

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Here are some follow-up questions:

1. What is the general formula to find the distance between a point and any non-horizontal line?
2. How would the solution change if the line was vertical?
3. Can you explain why the slope of a horizontal line is zero?
4. What is the equation of a vertical line passing through point \( A(-2, 1) \)?
5. How does the distance formula change when the line isn't axis-aligned?

**Tip:** When dealing with horizontal lines, the distance from any point is simply the difference in their y-values.

Learn how to find the distance between point P(5, 7) and a horizontal line passing through points (-2, 1) and (4, 1). This problem covers the distance formula and key geometric concepts, suitable for high school students in Grades 9-12.

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