坐標問題 ++請詳細不要跌步回答++15分

己知p( -3,9)及Q(4,4)及R三點共線,且PQ=QR

1)求R坐標
設R=R(x,y)
由於Q為PR中點
則(-3+x)/2=4
-3+x=8
x=11
(9+y)/2=4
9+y=8
y=-1
所以R=R(11,-1)

2)求穿過R及S(9,5)的直線方程
PS的斜率=[5-(-1)]/(9-11)=-3
(y-5)/(x-9)=-3
y-5=-3(x-9)
y-5=-3x+27
3x+y-32=0

1) 4=(x-3)/2, 4=(Y+9)/2 兩個point既equation
4X2=[(X-3)/2]x2
8=X-3
8+3=X-3+3
X=11
4X2=[(Y+9)/2]x2
8=Y+9
8-9=Y+9-9
Y=-1
(11,-1)

2)首先要搵個slope
equation: (Y1-Y2) / (X1-X2)
=(-1-5)/ (11-9)
-6/2
=-3
跟住放落linear equation
(Y+1)=-3(X-11)

1) Let R(x,y)

As Q is the mid-pt of PR

4 = (-3+x)/2
8 = -3 + x
x = 11

4 = (9+y)/2
8 = 9+y
y = -1

So R(11,-1)

2) R(11, -1) and S(9,5)
Slope of RS = (5-(-1))/(9-11) = -3
Express the equation in two-point form,
Equation of RS:
y - 5 = -3(x-9)
y - 5 = -3x+27
3x + y - 32 = 0

2006-10-23 01:11:20 補充：
mid-pt. 係中間點Slope係斜度Express the equation in two-point form = 用兩點式表達直線方程