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) #?

Question

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)   #?

Kennedy Adams · Accepted Answer

#7x-3y+1=0#

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#

Ethan Adkins · Answer

undefined First, find the slope of the line passing through the points (8,-3) and (1,0):

m = (y2 - y1) / (x2 - x1)
m = (0 - (-3)) / (1 - 8)
m = 3 / (-7)
m = -3/7

The line perpendicular to this one will have a slope that is the negative reciprocal of -3/7, which is 7/3.

Now, use the point-slope form of the equation of a line with the slope of 7/3 and the point (0,-1):

y - y1 = m(x - x1)

Substitute the values into the formula:

y - (-1) = (7/3)(x - 0)

Simplify:

y + 1 = (7/3)x

Subtract 1 from both sides:

y = (7/3)x - 1

So, the equation of the line passing through (0,-1) and perpendicular to the line passing through the points (8,-3) and (1,0) is y = (7/3)x - 1.