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 write the equation of a line that is parallel to a given line, we need to find the slope of the given line first.
The given equation is 16x + 13y = -210. We can rewrite it in the slope-intercept form (y = mx + b) by isolating y:
13y = -16x - 210
y = (-16/13)x - (210/13)
Now we can identify the slope of the given line by looking at the coefficient of x, which is -16/13.
Since the line we want to find is parallel to the given line, it will have the same slope.
Next, we can use the slope-intercept form of a line (y = mx + b) to write the equation.
Let's use the point-slope form of the equation, where (x1, y1) is the given point (-31, 22):
y - y1 = m(x - x1)
Substituting the values into the equation, we get:
y - 22 = (-16/13)(x - (-31))
Simplifying further:
y - 22 = (-16/13)(x + 31)
Expanding the equation:
y - 22 = (-16/13)x - (16/13)(31)
y - 22 = (-16/13)x - (16/13)(31)
y - 22 = (-16/13)x - 496/13
Finally, rearrange the equation to the standard form:
(-16/13)x + y = 22 - 24/13
Multiply every term by 13 to get rid of the fraction:
-16x + 13y = (22*13) - 24
Simplifying further:
-16x + 13y = 286 - 24
-16x + 13y = 262
So, the equation of the line that passes through the point (-31, 22) and is parallel to 16x + 13y = -210 is -16x + 13y = 262.