Let #P(a,b) and Q(c,d)# be two points in the plane. Find the equation of the line #l# that is the perpendicular bisector of the line segment #bar(PQ)#?

Question

Caroline Aucoin · Accepted Answer

Equation of line #l# is #2(c-a)x+2(d-b)y+(a^2+b^2-c^2-d^2)=0# A line which is perpendicular bisector of the line joining #P(a,b)# and #Q(c,d)# and passes through their midpoint is locus of a point which is equidistant from these two points. Hence, equation is #(x-a)^2+(y-b)^2=(x-c)^2+(y-d)^2# or #x^2-2ax+a^2+y^2-2by+b^2=x^2-2cx+c^2+y^2-2dy+d^2# or #-2ax+2cx-2by+2dy+(a^2+b^2-c^2-d^2)=0# or #2(c-a)x+2(d-b)y+(a^2+b^2-c^2-d^2)=0#

Allison Allen · Answer

#2x(a-c)+2y(b-d)=(a^2+b^2)-(c^2+d^2), if, a!=c,b!=d#.

If, #a=c, b!=d", then eqn. is : "y=(b+d)/2#.

If, #a!=c, b=d", then eqn. is : "x=(a+c)/2# Let #M# be the mid-pt. of the line segment #bar(PQ)#, where, #P(a,b) and Q(c,d)#. Hence, #M((a+c)/2,(b+d)/2)#. The slope of #bar(PQ)=(d-b)/(c-a), c!=a# #rArr "the slope of the "bot-"bisector line l of "bar(PQ)# is given by, #(-1)-:(d-b)/(c-a)=(a-c)/(d-b), dneb#. #"Thus, the slope of line l is "(a-c)/(d-b), and, M in l#. Using, Slope-Pt. Form for #l#, its eqn. is, #y-(b+d)/2={a-c)/(d-b)(x-(a+c)/2)#, i.e., #2y(d-b)+(b+d)(b-d)=2x(a-c)-(a-c)(a+c)#, or, #2x(a-c)+2y(b-d)=(a^2+b^2)-(c^2+d^2), a!=c,b!=d#. Case : 1 : #a=c, b!=d# :- If, #a=c#, then, #bar(PQ)# is vertical , i.e., parallel to the Y-axis, and, so, the reqd. line #l# will be horizontal , i.e., parallel to the X- axis passing through #M((a+c)/2,(b+d)/2)#, and, as such, its eqn. will be # l : y=(b+d)/2#. Case : 2 : #b=d, a!=c# :- In this case, the eqn. of #l# is # l : x=(a+c)/2.# In both of the Cases, eqn. of #l# can be derived from #2x(a-c)+2y(b-d)=(a^2+b^2)-(c^2+d^2)# by taking either #a=c, or, b=d#. Finally, the Case #a=c, and, b=d# need not be considered, because, in that case, pts. #P and Q# coincide and hence segment #bar(PQ)# does not exist, so is the case with its perp. bsctr. Have fun with math!

Evelyn Arboleda · Answer

undefined The equation of the perpendicular bisector of the line segment $ \overline{PQ} $ can be found using the midpoint formula and the negative reciprocal of the slope of $ \overline{PQ} $.

First, find the midpoint of $ \overline{PQ} $ using the midpoint formula:
$$ M\left(\frac{a+c}{2}, \frac{b+d}{2}\right) $$

Then, find the slope of $ \overline{PQ} $ using the formula:
$$ m_{PQ} = \frac{d - b}{c - a} $$

The negative reciprocal of $ m_{PQ} $ is $ -\frac{1}{m_{PQ}} $. This represents the slope of the perpendicular bisector.

Using the point-slope form of a line, where $ (x_1, y_1) $ is a point on the line and $ m $ is the slope:
$$ y - y_1 = m(x - x_1) $$

Substitute $ M\left(\frac{a+c}{2}, \frac{b+d}{2}\right) $ for $ (x_1, y_1) $ and $ -\frac{1}{m_{PQ}} $ for $ m $. Then, simplify to find the equation of the perpendicular bisector.