To find the equation of line B, which passes through point P and is parallel to line A, we need to do the following:

1. Identify the slope of line A: Since parallel lines have the same slope, we first need to determine the slope of line A. The slope $m$ is given by the change in $y$ over the change in $x$ between two points on the line. We can read two points from the graph for line A, for example, the points where the line intersects the grid.

2. Determine the slope for line B: Since line B is parallel to line A, it will have the same slope as line A.

3. Find the equation of line B: Use the point-slope form of the equation for a line: $y - y_1 = m(x - x_1)$ where $m$ is the slope, and $(x_1, y_1)$ is the point $P$ that line B passes through. Plug in the values for $m$, $x_1$, and $y_1$, then simplify the equation into the slope-intercept form $y = mx + c$.

I will analyze the graph to extract the necessary information. PIL import Image

Load the image to analyze the graph

img_path = "/mnt/data/1737394524459f472xbn0.jpg" img = Image.open(img_path)

Display the image for analysis

img.show()I've displayed the graph for analysis. I'll now manually interpret the graph to find the slope of line A and the coordinates of point P, so we can calculate the equation of line B.

Let's identify the slope of line A by locating two points on it, and then use point P to construct the equation of line B. I’ll proceed with that analysis.

Please give me a moment.Upon inspecting the graph:

1. Slope of line A: From the graph, line A passes through points like (−4, −3) and (0, 1). The slope $m_A$ is calculated using the formula: $m_A = \frac{y_2 - y_1}{x_2 - x_1}$ where $(x_1, y_1) = (-4, -3)$ and $(x_2, y_2) = (0, 1)$. $m_A = \frac{1 - (-3)}{0 - (-4)} = \frac{4}{4} = 1$ So, the slope of line A is $m_A = 1$.

2. Coordinates of point P: Point P appears to be at (3, 2).

Since line B is parallel to line A, it will have the same slope of 1. The equation of line B is therefore: $y - y_1 = m(x - x_1)$ Substituting $m = 1$, $x_1 = 3$, and $y_1 = 2$: $y - 2 = 1(x - 3)$ Simplifying: $y - 2 = x - 3$ $y = x - 1$

Thus, the equation of line B is $y = x - 1$.

Let me know if you'd like any further clarifications or details!