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