line l passes through (4,-31) and line p is the graph of y+4=-5(x-4).

Algebra
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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?

Jul 27th, 2014
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


Jul 27th, 2014

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