A.math (Coordinates)

首先..因為P係係3x+y=6呢條線上面...

所以我地可以sub p的(a,b)入x同y度

所以3a+b=6

q都係一樣...

所以我地得出2b-a=5

然後兩條式相等...

所以將3a+6=6轉做b=6-3a

再sub落q o個條式度

所以2(6-3a)-a=5

12-6a-a=5

-7a=-7

a=1

so搵埋b=6-3*1=3

so p(1,3)..q(3,1)

搵slope of pq...

=3-1/1-3=-1

再搵y-intercept...3-c/1-0=-1

c=4

之係y-intercept係4...

所以y=-x+4

之係equation : x+y-4=0

***********************************************

計呢條數的方向係..

首先搵左o個兩條line同pq的關係先

再搵返a同b的數值

再搵pq的slope同y-intercept

從而得到答案

***************************************

或者會難明小小..but希望幫到你/\/\

3a + b = 6 ......(1)
2b - a = 5 ......(2)
from (2), a=2b-5 ......(3)
Sub. (3) into (1) : 3(2b-5)+b=6
7b=21
b=3
when b=3, a = 2(3)-5 = 1
So P(1, 3) & Q(3, 1)
Equation of PQ :
y-3 / x-1 = 1-3 / 3-1
y-3 = -1 (x-1)
x+y-4=0

put P into 3x+y=6
3a+b=6--------1
put Q into 2x-y=5
2b-a=5
a=2b-5-------2
put 2 into 1
6b-15+b=6
b=3 a=1
eqn of PQ : y-3/x-1 = 3-1/1-3
x+y-4=0