How do I write an equation of a line that passes through the point (-31,22) and is parallel to this equation: 16x + 13y= -210

To find the equation of a line that is parallel to another line, we need to know that parallel lines have the same slope.

First, we need to determine the slope of the given equation, 16x + 13y = -210. To do this, we can rearrange the equation into slope-intercept form, which is y = mx + b, where m is the slope.

Rearranging the equation, we have:
13y = -16x - 210

Dividing both sides by 13, we get:
y = (-16/13)x - 210/13

The slope of the given equation is -16/13.

Since the line we want to find is parallel to the given equation, it will have the same slope of -16/13.

Now, we can use the point-slope form of a linear equation to write the equation of the line that passes through the point (-31, 22) with the same slope.

The point-slope form of a linear equation is: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope.

Plugging in the values, we get:
y - 22 = (-16/13)(x - (-31))

Simplifying further, we have:
y - 22 = (-16/13)(x + 31)

To convert this equation to slope-intercept form, we can distribute the (-16/13) to (x + 31) and simplify:

y - 22 = (-16/13)x - (16/13)(31)
y - 22 = (-16/13)x - (496/13)
y = (-16/13)x - (496/13) + 22
y = (-16/13)x - (496/13) + (286/13)
y = (-16/13)x - (210/13)

Therefore, the equation of the line parallel to 16x + 13y = -210 and passing through the point (-31, 22) is y = (-16/13)x - (210/13).