write equation of a line parallel to the given line but passing the given point.

write equation of a line parallel to the given line but passing the given point.

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The line provided is in the form y=mx+b, where m is the slope. Therefore the slope of the line y=x+4 is equal to 1.

All parallel lines have the same slope, so we know that the slope of line we are looking for is 1.

Also provided is a point (-7,1) on the line we are looking for. The equation for a line where a point (x_{0,}y_{0}) is specified is: (y-y_{0})=m (x-x_{0}).

Since we know that the slope (m) is equal to 1, the value of x_{0 }is -7, and the value of y_{0} is 1, we can plug these into the above equation to get: (y-1)= 1(x-(-7)) or y-1=x+7

Solving this for y (by adding 1 to both sides), we get **y =x+8.**

Michelle,

Recall that when you are given a line l in the slope intercept form (y = mx + b) that every other line, with a different b, that has the slope m is parallel to l.

With that in mind we may assume that the line of interest, that is the one including (-7, 1), has the slope m = 1. Thus,

y = 1*x + b, or likewise

y = x + b

Since we have a point on that line, namely (-7, 1), the value of b may be found analytically by substituting the solution pair and solving for b:

1 = -7 + b

1 + 7 = -7 + b + 7

8 = b

Finally, the line parallel to y = x + 4 which contains (-7, 1) may be written in slope intercept form as:

y = x + 8

Do you happen to recall what the slope of a perpendicular line would be, given a line l in the form y = mx + b?

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## Comments

no but here is the question for perpendicular y=-1/2x+1; (4,2)

If m is the slope of the first line, then a perpendicular line will have the slope -(1/m). What do you get for y = -1/2x+1; (4,2) so far?

I should elaborate a bit on the comment. If m is not zero and m is the slope of the first line, then a perpendicular line will have the slope -(1/m). If m is zero, then perpendicular lines are of the form x = k, where k is some arbitrary real number like one.