For graphics (1)

∠a=∠AOB

OB=8, AO= 10, so AB=6.

therefore

The slope of the straight line AC k = tan ∠ A = tan ∠ AOB = AB/OB = 6/8 = 3/4.

There is also a straight line intersection point A(0, 10)

So according to the point oblique type, there are

y- 10=3/4(x-0)

That is, AC: y = 3x/4+ 10.

For figure (2)

∠a=∠OAB+90

OB=8, AO= 10, so AB=6.

therefore

The AC slope of a straight line k = tan ∠ a = tan (∠ AOB+90) =-ob/ab = 8/6 = 4/3.

There is also a straight line intersection point A(0, 10)

So according to the point oblique type, there are

y- 10=-4/3(x-0)

That is, AC: y =-4x/3+ 10.

That is, there are two answers.

Ac: y = 3x/4+ 10 or AC: y =-4x/3+ 10.

5, according to the formula

D=|Ax0+By0+C|/ radical a? +B？ =|3×0-4×(- 1)+6|/ radical number 3? +(-4)? = 10/5=2

So choose d

Is there anything wrong or unclear?

Can be questioned