Question 51380
Let's write the required equation of the line in the "slope-intercept" form: {{{y = mx+b}}} Where: m is the slope of the line and b is the y-intercept of the line.

Remember that parallel lines have the same slope.
So, your new line is parallel the the line whose equation is:
{{{y = (5/2)x+6}}} Comparing this with the slope-intercept form:
{{{y = mx+b}}} you can see that the slope of this line is {{{m = 5/2}}} and, because the new line is parallel to this, it too will have a slope of {{{m = 5/2}}}.
So, step 1, for the new line, you can write:
{{{y = (5/2)x + b}}} Now all you need to do is to find the value of b, the y-intercept of the new line.
You can do this by substituting the x- and y-coordinates of the given point (2, 1) into the equation {{{y = (5/2)x + b}}} and solve for b.

{{{1 = (5/2)(2) + b}}} Simplify this.
{{{1 = 5 + b}}} Subtract 5 from both sides of the equation.
{{{-4 = b}}} You can write the final equation now that you have the slope {{{m = 5/2}}} and the y-intercept {{{b = -4}}}:

{{{y = (5/2)x - 4}}} ...and that's it!