Find the equation of the line which is parallel to the line #3x+4y=1# and passes through the midpoint of the line segment joining #(1,2)# and #(-2,1)#. Find the distance of this line from the given line.?

2 Answers
Jun 28, 2017

Equation of the line: #6x+8y=9#
Distance = #7/10#

Explanation:

The midpoint of the line segment joining #(1, 2)# and #(-2, 1)# is:
#M = ((x_1+x_2)/2, (y_1+y_2)/2)#
#= (-1/2, 3/2)#

Now, the line passes through this point and parallel to the line #3x+4y=1# will have an equation #3x+4y=k# and as it passes through #(-1/2,3/2)#

#:. # The Equation of the line will be:
#3.(-1/2)+4.(3/2)=k#

or #-3+12=2k# or #k=9/2#

Hence equation of line is #3x+4y=9/2# or #6x+8y=9#

The distance between the lines will be the same as distance between #(-1/2,3/2)# and #3x+4y-1=0# i.e.

#|(3xx(-1/2)+4xx3/2-1)/sqrt(3^2+4^2)|#

#=|(-3/2+6-1)/5|#

#=7/10#

graph{(3x+4y-1)(6x+8y-9)((x-1)^2+(y-2)^2-0.003)((x+2)^2+(y-1)^2-0.003)((x+0.5)^2+(y-1.5)^2-0.003)=0 [-2.583, 2.417, 0.05, 2.55]}

Jun 28, 2017

Raj has written one set of answers, but I thought I'd approach the question from a different direction and see whether I find the same answers.

Rearranging the given line into standard form:

#3x+4y=1#

#4y=-3x+1#

#y = -3/4+1/4#

That is, the line has a gradient of #-3/4#. The new line we find should have the same gradient but a different intercept, if it is parallel to the original line.

Finding the midpoint between the points #(1,2)# and #(-2,1)#:

#(x,y)=((x_2+x_1)/2,(y_2+y_1)/2)=((-2+1)/2,(1+2)/2)=(-1/2,3/2)#

To find the new parallel line, we can substitute these two points and the known gradient into the general equation for a line, and find the new intercept:

#y=mx+c#

#3/2=-3/4(-1/2)+c#

#3/2=3/8+c#

#c=3/2-3/8=9/8#

Kind of ugly, but there it is. So the equation of the line we're looking for us:

#y = -3/4x+9/8#

(I can multiply through by 4 and rearrange to get an equation in the same terms as Raj wrote his:

#3x+4y=9/2#)

When we talk about the distance between two parallel lines, we mean the shortest possible distance, which means measuring perpendicular to the two lines. We have one point which is on one line: #(-1/2,3/2)#. We can use these to find a line perpendicular to both lines. If our two parallel lines have a gradient #m=-3/4#, then the slope of a perpendicular line will be #-1/m = 4/3#.

(I'm going to have to stop here for the moment because I've run out of time. I will return to complete my answer soon.)