What is the equation of the line that passes through #(0,-1)# and is perpendicular to the line that passes through the following points: #(8,-3),(1,0) #?

1 Answer
Feb 10, 2016

#7x-3y+1=0#

Explanation:

Slope of the line joining two points #(x_1, y_1)# and #(x_2, y_2)# is given by

#(y_2-y_1)/(x_2-x_1)# or #(y_1-y_2)/(x_1-x_2)#

As the points are #(8, -3)# and #(1, 0)#, slope of line joining them will be given by #(0-(-3))/(1-8)# or #(3)/(-7)#

i.e. #-3/7#.

Product of slope of two perpendicular lines is always #-1#. Hence slope of line perpendicular to it will be #7/3# and hence equation in slope form can be written as

#y=7/3x+c#

As this passes through point #(0, -1)#, putting these values in above equation, we get

#-1=7/3*0+c# or #c=1#

Hence, desired equation will be

#y=7/3x+1#, simplifying which gives the answer

#7x-3y+1=0#