A line segment is bisected by a line with the equation # 4 y - 6 x = 8 #. If one end of the line segment is at #( 1 , 8 )#, where is the other end?

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Answer 1 · 2018-06-13T13:51:11

#color(blue)((67/13,68/13)#

First rearrange #4y-6x=8# to the form #y=mx+b#

#y=3/2x+2 \ \ \ \[1]#

This will be perpendicular to the line through the point #(1,8)#

We need to find the equation of this line. We know that if two lines are perpendicular then the product of their gradients is #bb(-1)#

Gradient of #[1]# is: #3/2#

Let #bbm# be the gradient of the line through #(1,8)#

Then:

#m*3/2=-1=>m=-2/3#

Using point slope form of a line and point #(1,8)#:

#(y_2-y_1)=m(x_2-x_1)#

#y-8=-2/3(x-1)#

#y=-2/3x+26/3 \ \ \ [2]#

The intersection of #[1]# and #[2]# will be the midpoint of the line segment. Solving these simultaneously:

#3/2x+2=-2/3x+26/3=>x=40/13#

Substitute in #[1]#

#y=3/2(40/13)+2=86/13#

Co-ordinates of midpoint #(40/13,86/13)#

The co-ordinates of the midpoint are given by:

#((x_1+x_2)/2,(y_1+y_2)/2)#

So:

#((x_1+x_2)/2,(y_1+y_2)/2)=(40/13,86/13)#

If the end points are #(x_1,y_1)# and #(x_2,y_2)#

We have #(1,8)# and #(x_2,y_2)#

#((1+x_2)/2,(8+y_2)/2)=(40/13,86/13)#

Hence:

#(1+x_2)/2=40/13=>x_2=67/13#

and

#(8+y_2)/2=86/13=>y_2=68/13#

So other end point is:

#color(blue)((67/13,68/13)#