A(1,7),B(7,5) and C(0,-2),find the point of intersection of BC with the perpendicular bisector of AB ?

1 Answer
Sep 21, 2015

#(2,0)#

Explanation:

This will be pretty long. I'm not sure if there's a shorter solution, but this is how I would solve it.

We will start by finding the equation of the perpendicular bisector of #bar(AB)#.

Step 1 - Slope of the perpendicular bisector.

We will first find the slope of #bar(AB)# using the formula for slope.
#(y_1-y_2)/(x_1-x_2)#
#=[(7)-(5)]/[(1)-(7)]#
#=(2)/(-6)#
#=-1/3#

We will now look for the slope of the perpendicular bisector. To do this, we will get the negative reciprocal of the slope of #bar(AB)#
Slope of #bar(AB)#: #(-1/3)#
Negative reciprocal: #3#

The slope of the perpendicular bisector is 3.

Step 2 - Point of intersection of #bar(AB)# and the perpendicular bisector.

Since this is a bisector we're talking about, it should go through the middle of #bar(AB)#. To solve for the point of intersection, we will use the midpoint formula.
#((x_1+x_2)/2,(y_1+y_2)/2)#
#((1+7)/2,(7+5)/2)#
#(8/2,12/2)#
#(4,6)#

The point of intersection between #bar(AB)# and the perpendicular bisector is (4,6).

Step 3 - The equation of the perpendicular bisector.

Since we know the slope of the bisector and one point it intersects, we can write the equation of the line in point-slope form.
#y-y_0=m(x-x_0)#
#y-(6)=(3)[x-(4)]#
#y-6=3x-12#
#y=3x-12+6#
#y=3x-6#

The equation of the perpendicular bisector is #y=3x-6#.

We are done with the perpendicular bisector, so we can now proceed to finding the equation of the line that includes #bar(BC)#.

Step 1 - Slope of #bar(BC)#.
#(y_1-y_2)/(x_1-x_2)#
#=[(5)-(-2)]/[(7)-(0)]#
#=(7)/(7)#
#=1#

The slope of #bar(BC)# is 1.

Step 2 - Equation of the line including #bar(BC)#.

We know that the line passes through (7,5) and (0,-2), so we can write the equation in point-slope form using either of the two points. I will use (0,-2) since it is simpler.
#y-y_0=m(x-x_0)#
#y-(-2)=(1)[x-(0)]#
#y+2=x#
#y=x-2#

The equation of the line including #bar(BC)# is #y=x-2#.

FINAL STEP
Now that we have both the equations of the perpendicular bisector and the line including #bar(BC)#, we can now look for the point of intersection using substitution.
#y=x-2#
#3x-6=x-2#
#3x-x=-2+6#
#2x=4#
#x=2#

To solve for the value of #y#, just substitute the value #2# to either of the two equations.
#y=x-2#
#y=(2)-2#
#y=0#

The point of intersection of #bar(BC)# and the perpendicular bisector is #(2,0)#.