How do you solve this pre calculus problem?

Question

Given A(2,0) and B(8,4), show that P(3,5) is on the perpendicular bisector of line AB by these two methods:

Method 1. Show line AB is perpendicular to line PM where M is the midpoint of line AB.

Method 2. Show PA = PB.

Please help!!

Thanks!!!!

Answer 1

hi there...

no1.

find the slope AB = (4-0)/(8-2) = 2/3

find the midpoint of AB = M =((2+8)/2 , (0+4)/2 )= (5, 2)

find slope PM = (2-5)/(5-3) = -3/2

to show to lines are perpendicular, kust multiply the slopes = -1

then m1 x m2 = -1

       2/3 x -3/2 = -1, so these lines are perpendicular..

no 2.

PA = PB ( use equal distance formulae)

|PA|= |PB|

(3-2)^2 + (5-0)^2 = (3-8)^2+(5-4)^2

      1 + 25    =  25 + 1

  26   =  26,done..

see ya...friend.

  

Answer 2

You want to find the slope of AB first:

(4-0)/(8-2)

That would equal:

2/3

To find PM, you have to use the midpoint formula to find the middle of AB:

(2+8)/2 for the x-coordinate

(0+4)/2 for the y-coordinate

So after you do the math, the midpoint of AB is:

(5,2)

A line that is perpendicular to another has a negative and reciprocal slope. For example, if the slope is 5, then the perpendicular slope is -1/5 or if the slope is 5/2, the perpendicular slope is -2/5. So find the slope of PM:

(5-2)/(3-5)

Do the math and you get:

-3/2

The second one, all you have to do is the distance formula between those two lines. Too much typing to show it to you. Sorry >.<  

Answer 3

1) gradient of AB is (4-0)/(8-2)=2/3

midpt of AB is ((2+8)/2,(0+4)/2)=(5,2)

gradient of PM is (5-2)/(3-5)=-3/2

since (2/3) * (-3/2) = -1, AB & PM are perpendicular

2) PA=sqrt[(5-0)^2+(3-2)^2]=sqrt(26)

PB=sqrt[(5-4)^2+(3-8)^2]=sqrt(26) (QED)