# 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

#### Answer:

In slope-point form, the equation of line M is

In slope-intercept form, it is

#### 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 **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

#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

*(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

Since the slope of line L is

Alright—now we know the slope of line M is **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.)*