What is the equation of the line parallel to y=x-1 that contains the point (1,2)

Question

I'm unsure of how to find the answer. i'd like a step-by-step demonstration

Answer 1

A line that is parallel to the line you were given must have the same slope, right?

The line y = x - 1 has a slope of 1, because the slope is equal to the coefficient of x, which is 1. [Remember y = mx + b, where m is the slope, and b is the y-intercept] 

 

So the new line is going to be y = x + b. We don't know b yet, but we can use the fact that the line goes through (1,2),

 

so   2 = 1 + b

so   b = 2-1 = 1

 

The new line is therefore: y = x + 1

Answer 2

Hi Austin,

Recall that the generic line formula is y=mx+b.

Since they are looking for a line that is parallel to y=x-1, we know the slope has to be the same (ie. m=1).

Given the slope of 1, we can plug the (1,2) point values into the line equation so that we have 2=1(1)+b.

Solving for b gives us a value of 1.

 

So the equation of the line is y=x+1.

Answer 3

For lines to be “parallel” means that they never intersect; that is, they have the same slope.  In slope-intercept form, the equation of a line is written as:

y = mx + b      where m is the slope and b is the y-intercept (value of y when x=0)

All lines parallel to that line y = x – 1 have the formula y = x + C   [note: same slope, different y-intercept).  Let’s find the line that contains the point (1,2) by finding C:
    y = x + C
    2 = 1 + C      (putting in values of x=1 and y=2)
    1  = C

The parallel line has the formula:    y = x + 1