What is the equation of the line passing through the point (4, 6) and parallel to the line y= 1/4x + 4?

To draw a line you either need twoof its points, or one of its points and its slope. Let's use this second approach.

We already have the point #(4,6)#. We derive the slope from the parallel line.

First of all, two lines are parallel if and only if they have the same slope. So, our line will have the same slope as the given line.

Secondly, to derive the slope from a line, we write its equation in the #y=mx+q# form. The slope will be the number #m#.

In this case, the line is already in this form, so we immediately see that the slope is #1/4#.

Recapping: we need a line passing through #(4,6)# and having slope #1/4#. The formula that gives the line equation is the following:

#y-y_0 = m(x-x_0)#

where #(x_0,y_0)# is the known point, and #m# is the slope. Let's plug our values:

#y-6= 1/4(x-4)#

Expanding the right hand side:

#y-6 = 1/4x-1#

Add #6# to both sides:

#y= 1/4x-1+6#

So the answer is

#y= 1/4x+5#

Parallel lines have the same slope, so the missing equation must have #1/4# as its slope.

Following the given, substituting #4# as #x# yields #y=6#, so as a shortcut, one can form the equation: #6=1/4(4)+b# to find #b#.

This becomes: #6=1+b#, where #b=5#.

Substituting into slope-intercept form, the final answer becomes:

#y=1/4x+5#

Source: http://www.mesacc.edu/~scotz47781/mat150/notes/eqn_line/Equation_Line_Parallel_Perpendicular_Notes.pdf