To find the distance between a point and a line, we can use the following formula:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Where:

• $A$, $B$, and $C$ are the coefficients from the standard form of the line equation $Ax + By + C = 0$.
• $(x_1, y_1)$ are the coordinates of the given point.

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Step 1: Find the equation of the line $l$ through points $(1, 5)$ and $(4, -4)$.

The slope $m$ of the line is:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-4 - 5}{4 - 1} = \frac{-9}{3} = -3$

Using the point-slope form of a line equation:

$y - y_1 = m(x - x_1)$

Substitute $m = -3$ and the point $(1, 5)$:

$y - 5 = -3(x - 1)$

Simplify:

$y - 5 = -3x + 3 \quad \Rightarrow \quad y = -3x + 8$

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Step 2: Convert the line equation to standard form $Ax + By + C = 0$.

$y = -3x + 8 \quad \Rightarrow \quad 3x + y - 8 = 0$ So, the coefficients are:

• $A = 3$
• $B = 1$
• $C = -8$

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Step 3: Use the distance formula.

Now, use the point $(-1, 1)$ in the formula:

$d = \frac{|A(-1) + B(1) + C|}{\sqrt{A^2 + B^2}}$

Substitute $A = 3$, $B = 1$, $C = -8$, and the point $(-1, 1)$:

= \frac{|-3 + 1 - 8|}{\sqrt{9 + 1}} = \frac{|-10|}{\sqrt{10}} = \frac{10}{\sqrt{10}}$$ Simplify: $$d = \sqrt{10} \approx 3.16$$ --- ### Final Answer: The distance between the point $$P(-1, 1)$$ and the line $$l$$ is approximately: $$\boxed{3.16}$$ --- Do you need more details or further clarifications? Here are 5 related questions for further exploration: 1. What is the general equation for a line in point-slope form? 2. How can we derive the distance formula between two points in 2D? 3. What changes in the distance formula if we are working in 3D space? 4. How do we find the shortest distance between two parallel lines? 5. How do you determine if a point lies on a given line? **Tip:** Always check if your line equation is in standard form before applying the point-line distance formula!