To find the equation of the line parallel to $y = \frac{1}{4}x + 4$ that passes through the point $(-1, 3)$, follow these steps:

Step 1: Recognize the slope of the parallel line

The slope of the given line is $\frac{1}{4}$ because the equation is in slope-intercept form ($y = mx + b$) where $m$ is the slope. Since parallel lines have the same slope, the slope of the desired line is also $\frac{1}{4}$.

Step 2: Use the point-slope form of a line

The point-slope form of a line is: $y - y_1 = m(x - x_1)$ Here:

• $m = \frac{1}{4}$
• $(x_1, y_1) = (-1, 3)$

Substitute these values: $y - 3 = \frac{1}{4}(x - (-1))$ $y - 3 = \frac{1}{4}(x + 1)$

Step 3: Simplify the equation

Distribute $\frac{1}{4}$: $y - 3 = \frac{1}{4}x + \frac{1}{4}$ Add 3 to both sides: $y = \frac{1}{4}x + \frac{1}{4} + 3$ $y = \frac{1}{4}x + \frac{13}{4}$

Final Answer:

The equation of the line is: $y = \frac{1}{4}x + \frac{13}{4}$

Would you like a graph of this line for visualization or further clarification?

Related Questions:

1. What is the slope of a line perpendicular to $y = \frac{1}{4}x + 4$?
2. How do you find the y-intercept of a line given a slope and a point?
3. What is the equation of a line passing through two given points?
4. Can you generalize how to find parallel and perpendicular lines in any form?
5. How do you rewrite $y = \frac{1}{4}x + \frac{13}{4}$ in standard form?

Tip:

Remember, parallel lines always share the same slope, while perpendicular lines have slopes that are negative reciprocals.