To write the equation of a line perpendicular to $y = 5x + 5$ and passing through the point $(10, 2)$, we can follow these steps:

Step 1: Identify the slope of the given line

The equation $y = 5x + 5$ is in slope-intercept form:
$y = mx + b$ Here, $m = 5$ is the slope of the given line.

Step 2: Find the slope of the perpendicular line

The slope of a line perpendicular to another line is the negative reciprocal of the original slope.
The negative reciprocal of $5$ is: $m_{\perp} = -\frac{1}{5}$

Step 3: Use the point-slope form of the line equation

The point-slope form of a line is: $y - y_1 = m(x - x_1)$ Where $m$ is the slope, and $(x_1, y_1)$ is a point on the line.
Here:

• $m = -\frac{1}{5}$
• $(x_1, y_1) = (10, 2)$

Substitute these values into the equation: $y - 2 = -\frac{1}{5}(x - 10)$

Step 4: Simplify to slope-intercept form $y = mx + b$

Distribute the slope $-\frac{1}{5}$: $y - 2 = -\frac{1}{5}x + 2$

Add $2$ to both sides: $y = -\frac{1}{5}x + 4$

Final Answer:

The equation of the line is: $y = -\frac{1}{5}x + 4$

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Related Questions:

1. How do you find the slope of a perpendicular line from a given slope?
2. How can you write an equation in point-slope form?
3. How do you convert a point-slope equation to slope-intercept form?
4. What is the slope-intercept form of a line?
5. How do you verify if two lines are perpendicular?

Tip: Remember, perpendicular slopes are always negative reciprocals of each other!