### Question Description

line l passes through (4,-31) and line p is the graph of y+4=-5(x-4). If l is parallel to p what is the slope-intercept form equation of l?

## Final Answer

Let's look at the given equation first. y + 4 = -5(x - 4) is the equation of the graph of Line P. This equation is in Point-Slope form y - y1 = m(x - x1), where 'm' is the slope and (x1, y1) is the point. We can compare Line P's equation with the formula to get the slope

*y - y1 = m(x - x1) Point-Slope form*

* y + 4 = -5(x - 4) m is the slope*

* By comparison, m = -5*

Line L and Line P are parallel if they have the same slope. And since Line P has a slope of -5, that means that Line L also has a slope of -5. Therefore the slope of Line L is m = -5.

Line L also has the given point of (4, -31). This means we can let (x1, y1) = (4, -31), or rather x1 = 4 and y1 = -31. With the point and slope of Line L now known, we can plug back into the Point-Slope form. Keep in mind that we are instructed to put the equation into Slope-Intercept form y = mx + b, meaning that after we substitute, we would rearrange the equation to put in the form y = mx + b (or rather as an equation of y in terms of x)

*y - y1 = m(x - x1)*

* y - (-31) = -5(x - 4) Substitute the point and slope of Line L*

* y + 31 = -5(x - 4) Simplify if necessary, then convert to y = mx + b*

* y + 31 = -5x + 20 Distribute the parenthesis*

* y + 31 - 31 = -5x + 20 - 31 Get the y by itself*

* y = -5x - 11 Once y is isolated, it is in slope-intercept form y = mx + b*

**SOLUTION:**

** Slope-Intercept equation of Line L is y = -5x - 11**

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