Line L has equation #2x-3y= 5# and Line M passes through the point (2, 10) and is perpendicular to line L. How do you determine the equation for line M?

1 Answer
Dec 30, 2016

Answer:

In slope-point form, the equation of line M is #y-10=-3/2(x-2)#.
In slope-intercept form, it is #y=-3/2x+13#.

Explanation:

In order to find the slope of line M, we must first deduce the slope of line L.

The equation for line L is #2x-3y=5#. This is in standard form, which does not directly tell us the slope of L. We can rearrange this equation, however, into slope-intercept form by solving for #y#:

#2x-3y=5#

#color(white)(2x)-3y=5-2x"    "#(subtract #2x# from both sides)

#color(white)(2x-3)y=(5-2x)/(-3)"     "#(divide both sides by #-3#)

#color(white)(2x-3)y=2/3 x-5/3"     "#(rearrange into two terms)

This is now in slope-intercept form #y=mx+b#, where #m# is the slope and #b# is the #y#-intercept. So, the slope of line L is #2/3#.

(Incidentally, since the slope of #2x-3y=5# was found to be #2/3#, we can show that the slope of any line #Ax+By=C# will be #-A/B#. This may be useful to remember.)

Okay. Line M is said to be perpendicular to line L—that is, lines L and M create right angles where they cross.

The slopes of two perpendicular lines will be negative reciprocals of each other. What does this mean? It means that if the slope of a line is #a/b#, then the slope of a perpendicular line will be #-b/a#.

Since the slope of line L is #2/3#, the slope of line M will be #-3/2#.

Alright—now we know the slope of line M is #-3/2#, and we know a point that it passes through: #(2,10)#. We now simply pick an equation for a line that allows us to plug in this data. I will choose to insert the data into the slope-point equation for a line:

#y-y_1=m(x-x_1)#

#y-10=-3/2(x-2)#

Choosing slope-point form allows us to simply stop here. (You could choose to use #y=mx+b#, where #(x,y)=(2,10)# and #m=-3/2#, then solve for #b#, and finally use this #b# along with #m# in slope-intercept form again:

#y="        "mx"  "+b#

#10=-3/2 (2)+b#

#10="   "-3"    "+b#

#13=b#

#:.y=mx+b#
#=>y=-3/2 x + 13#

Same line, different form.)