identify the equation of the line through (-2, 1) and parallel to y = 3x + 1
y-1= 3x-(-2)
y = 3x + 3
Since the new equation has the same slope it must look like
y = 3x+ b
sub in the point (-2,1)
1 = 3(-2) + b
b = 7
new equation: y = 3x + 7
To find the equation of a line parallel to a given line, we need to know that parallel lines have the same slope.
The given line has the equation y = 3x + 1. The slope of this line is 3, which is the coefficient of x.
To find the equation of the line parallel to y = 3x + 1, we start by determining the slope of the parallel line, which is also 3.
Now, we have the slope of the line and a point (-2, 1) that the line goes through. We can use the point-slope form of a linear equation to find the equation of the parallel line.
The point-slope form is given by: y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line.
Using the slope m = 3 and the point (-2, 1), we substitute these values into the point-slope form to get: y - 1 = 3(x - (-2)).
Simplifying, we have: y - 1 = 3(x + 2).
Expanding the equation, we get: y - 1 = 3x + 6.
Finally, rearranging the equation to the slope-intercept form (y = mx + b), we have: y = 3x + 7.
So, the equation of the line through (-2, 1) and parallel to y = 3x + 1 is y = 3x + 7.