Assuming that point C is on the X axis, we might as well set point C (a, 0).

A( 1,1) is known; B(3，- 1)

Then, the slope of AC KAC =1-a; Slope of BC Kbc=a-3

As we all know, ACB = 90.

So, AC⊥BC

Then Kac*Kbc=- 1

= = = & gt( 1-a)*(a-3)=- 1

= = = & gt-a？ +4a-3+ 1=0

= = => Answer? -4a+2=0

= = = & gta=[4 √( 16-8)]/2=2 √2

Then, point C (2 √ 2,0)

2、

A(4， 1)； C( 1，4)

Then the midpoint of AC is (5/2, 5/2); The slope of AC connection line is k=- 1.

Then, the slope of the vertical line in AC is k= 1.

So the equation of the midline is: y-(5/2)= 1*[x-(5/2)].

= = = & gty=x

So point B (5, 5) is on the vertical line of AC.

3、

Connect the AD as shown.

Then we can get from Pythagorean theorem:

AE？ =AD？ De?

De? =BD？ -Really?

= = = & gtAE？ =AD？ -(BD？ -Really? )

= = = & gtAE？ =AD？ -BD？ +BE？

D is known as the midpoint of BC, so BD=CD.

= = = & gtAE？ =AD？ -CD？ +BE？

= = = & gtAE？ =AC？ +BE？

= = = & gtAE？ -Really? =AC？