Find perpendicular bisector + distance from intercepts to 3 points...

The triangle with the perimeter.

Hi Tony,

This question is long and tedious but reasonably straight forward.

We have points:  A(5, 3), B(– 2, 2) and C(2, – 6) and we trace the perpendicular bisectors  of segments AB and AC.

a) Find the coordinates of the intersection point of these perpendicular bisectors.

Find  the gradient of AB, the gradient of the perpendicular will be the negative reciprocal.

Find the midpoint of AB

Use the point and gradient to find the formual of the perpendicular bisector.

Do the same things for AC

Solve the two equations simultaneoulsy to find the point of intersection

b) Show that this point is at equal distance from all points (A, B and C)

Now use the distance formula 3 times to show this. 

I am sure that you can either do all of it, or do a chunk of it yourself :)

Midpoint of AB   =   ( (5 − 2) / 2  , ( +2) /2  )  =     ( 3/2  , 5/2)  = ( 1.5, 2.5)

Slope of AB  =   [ 3 − 2] / [ 5 − − 2]   =   [1 / 7]  .... − reciprocal slope.....  = −7

Equation of perp. bisec.   =    y   = − 7 (x  − 1.5) + 2.5    →  y = − 7x + 10.5 + 2.5 →

  y = − 7x + 13     (1)

Midpoint of BC   =   ( (2 − 2) / 2  , (−6 +2) /2  )  =     ( 0/2  , −4/2)  = ( 0, −2)

Slope of BC  =   [ 2 − − 6] / [  −2 − 2]   =   [8 / −4]  = −2  .... − reciprocal slope..... =  1/2

Equation of perp. bisec.   =    y   = (1/2) (x  − 0) − 2    →  y = (1/2)x  − 2   (2)

Midpoint of AC   =   ( (5 + 2) / 2  , (−6 +3) /2  )  =     ( 7/2  , −3/2)  = ( 3.5, −1.5)

Slope of AC  =   [ 3 − − 6] / [  5 − 2]   =   [9 / 3]  = 3  .... − reciprocal slope..... =  −1/3

Equation of perp. bisec.   =    y   = (−1/3) (x  − 3.5) − 1.5    →  y = (−1/3)x + 7/6  − 1.5 →

y = (−1/3)x   −  1/3  (3)

Intersection  of (1) and (2) 

− 7x + 13   =  (1/2)x  − 2    add 2, 7x to both sides.....

15  = (1/2)x + 7x

15  = 7.5x     divide both sides by  7.5

2  = x    and  y = − 7(2) + 13   =  −1         =    ( 2, −1)

And y = (−1/3)x   −  1/3   also passes  through the point   ( 2, −1)   → y = (−1/3)(2)   −  1/3  =

  −2/3  − 1/3  =  −1

Distance from ( 2, −1)  to  A  = √[ ( 2−5)^2  + ( −1 − 3)^2  ]  = 5

Distance from ( 2, −1)  to  B  = √[ ( 2− − 2)^2  + ( −1 − 2)^2  ]   = 5

Distance from ( 2, −1)  to  C  = √[ ( 2− 2)^2  + ( −1 − − 6)^2  ]   = 5

I'll leave it to you, Tony,  to check the distance computations above

Huge thanks to everyone!

I mostly try to do it myself first but then validate with what you guys provide.