What is the slope of any line perpendicular to the line passing through #(15,-12)# and #(24,27)#?

2 Answers
Jan 18, 2016

Answer:

#-3/13#

Explanation:

Let the slope of the line passing through the given points be #m#.

#m=(27-(-12))/(24-15)=(27+12)/9=39/9=13/3#

Let the slope of the line perpendicular to the line passing through the given points be #m'#.

Then #m*m'=-1 implies m'=-1/m=-1/(13/3)#

#implies m'=-3/13#

Hence, the slope of the required line is #-3/13#.

Jan 18, 2016

Answer:

The slope of any line perpendicular to the given one is: #-3/13#

Explanation:

The trick is to just remember that if the gradient of the first line is #m# the gradient of the one that is perpendicular to it (normal) has the gradient of #(-1)xx1/m#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Gradient (slope) of the first line")#

Let #m_1# be the gradient of the first line

Then
#m_1=(y_2-y_1)/(x_2-x_1)#

Given that
#(x_1,y_1)-> (15,-12)#
#(x_2,y_2)-> (24,27)#

We have:
#color(blue)(m_1=(27-(-12))/(24-15) color(white)(....)-> color(white)(....) 39/9)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Gradient (slope) of the second line")#

Let #m_2# be the gradient of the second line

Then
#m_2=(-1)xx1/m_1color(white)(....)-> color(white)(....)(-1)xx 9/39#

#color(blue)(m_2= -(9-:3)/(39-:3) =-3/13)#