The equation of a line is y=-2x+1. what is the equation of the line that is parallel to the first line and passes through (2,2)?

The equation in slope-intercept form is y=-2x+6.

The equation in standard form is 2x+y=6.

The equation in point-slope form is y-2=-2(x-2).

Step-by-step explanation:

The slope-intercept form of a line is y=mx+b where m is the slope and b is the y-intercept.

Parallel lines will have the same slope and different y-intercept.

Anyways the slope of y=-2x+1 is -2.

So the equation of our line we are looking for is -2.

So we know our equation is in the form y=-2x+b.

We must inf b using y=-2x+b with (x,y)=(2,2).

y=-2x  +b  with (x,y)=(2,2)

2=-2(2)+b

2=-4+b

Add 4 on both sides:

2+4=b

Simplify:

6=b

The equation is y=-2x+6.

Now it didn't say what form it wanted.

There are some forms I can give you like standard and point-slope form.

There is also general form but it is not too much different from standard form.

Standard form is ax+by=c where a,b, and c are integers if possible.

Point-slope form is y-y1=m(x-x1) where (x1,y1) is a point on the line and m is the slope.

So let's go for standard form (ax+by=c) first:

y=-2x+6

add 2x on both sides:

2x+y=6

This is standard form because it is in the form

ax+by=c.

Ok we know point (2,2) is on our line and we also know we have slope,m, is -2.

Point-slope form is

y-y1=m(x-x1)

y-2=-2(x-2)

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