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How to do the ninth question of group B on page 1 10 of compulsory two mathematics in People's Education Press?
*ac?BH，∴ kAC= from kBH= 1/2

-2 ,

So the AC equation is y- 1= -2(x-5), which is simplified to 2.

x+y- 1 1=0，

In the case of 2x-y-5=0, the c coordinate can be obtained by the following formula (

4,3),

Because b is on the high BH, let the coordinate of b be (

2y+5，y)，

Then the coordinates of point M in AB are (y+5, (y+ 1)/2.

), and m is on the straight line CM,

So 2(y+5)-(y+ 1)/2-5=0, and the solution is y= -3.

, so B(- 1, -3),

Therefore, the BC equation can be obtained as (y+3) from the two-point formula.

/(3+3)=(

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