what is y=x+4; (-7,1)

Question

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

 

Answer 1

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,y₀) is specified is:  (y-y₀)=m (x-x₀).

Since we know that the slope (m) is equal to 1, the value of x₀ is -7, and the value of y₀ 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.

Answer 2

Hey Michelle -- all parallel lines to y= x +4 will look like "y= x + (s)omething" ... simply plug in (-7,1) => 1= -7 +s ... s must be 8 ... y= x +8 works ... in general, for parallels to y= mx +b: "keep the m, find the new b"  :)

Answer 3

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?