Write an equation of the line that passes through (1, 5) and is parallel to the line y=−4x+2.

Answer 1

y = -4x + 2 is in slope intercept form y = mx + b, where m is the slope and b is the y-intercept. So slope of this line is -4.

We need to write the equation of the line parallel to that line and passing through (1,5). Parallel lines have the same slope, so for the line m = -4.

writing the equation of the line in point-slope form: y - y1 = m(x-x1) ---> y-5 = -4(x-1)

y - 5 = -4x + 4

y = -4x + 9

Answer 2

Dear Autumn,

Commit this to your memory: Parallel lines never intersect, and every group of parallel lines has the same slope.

Also, please do me the courtesy of reading everything that I have written so that you can work problems like these on your own in the future. These types of problems are on the SAT and ACT. Getting good at them will serve you well. Soon you will be asked to find the equation of a line that is perpendicular to a given line. This type of problem is similar, though not identical, to this problem. You will be ready for those perpendicular problems, too, if you understand the method below.

You were given the line y = −4x + 2.

You are asked to find the equation of a line that is parallel to the given line and on which the point (1,5) lies—among an infinite number of other points. Your given line does NOT have this point on it, since groups of parallel lines have NO points in common; if they did, these would be points of intersection and would mean that the lines were not parallel at all.

So, your parallel line will have the same slope as the given line. What is the slope of the given line? Well, since the equation of the line is written in slope-intercept form, you can read off the slope and need not do any calculations. The slope is the coefficient of the x term. The coefficient of x in this case is –4; that's the slope. That will be the slope of the parallel line too.

Thus, we have the beginnings of the equation for the parallel line. It is: y = –4x + c

We do not know what c is. Once we figure that out, we will have our equation. To get there, we fill in the values of x and y from the point that we've been given.

Since the (x,y) point (1,5) is on the parallel line, we know that when 1 is substituted for x and when 5 is substituted for y in the equation, it makes a true statement.

x = 1

y = 5

y = –4x + c

Now do the substitutions to get: 5 = –4(1) + c

You can simplify this to: 5 = –4 + c

Now solve for c by adding 4 to each side of the equation to get: 9 = c

Now go back to the equation that has x and y in it as well as c. We will substitute this specific value of c but leave x and y as they are.

y = –4x + 9 This is your parallel line that contains the point (1,5).