中4數40分長答thx

a) L1 : Ax - y + 3 = 0

y = Ax + 3

L1 斜率 = A

L2 : 2x + 4y + C = 0

4y = -2x - C
y = (-1/2)x - C/4

L2 斜率 = -1/2

bi) L1┴L2 :
A * (-1/2) = - 1
A = - 1 / (-1/2)
A = 2

bii) L1重疊L2 :
Ax-y+3 = 0 = 2x+4y+C

A/2 = -1/4 = 3/C

A = -1/2 ,

C = - 12

a) Note that the slope of a straight line can be defined as (-a/b)

so,m(L1)=(-A/-1)=A and
m(L2)=(-2/4)=-1/2

b) Note that the slope of a ┴ lines = m(L1)*m(L2)=-1

so, m(L1)*m(L2)=-1
A(-1/2)=-1
A=2

c) L1 overlapped L2
i.e equation of L1=equation of L2
Ax-y+3=0
2x+4y+C=0
i.e Ax-y+3=2x+4y+C ...(*)

By consider the coefficient term of the L.H.S and the R.H.S in (*)
A=2 and C=3

a) L1: A
L2: -1/2


bi) A(-1/2) = -1
A= 2