一直線族通過 x + 3y + 1 = 0 及 3x - 7y + 4 = 0 的交點。
(a) 寫出該直線族的方程。
(b) 已知 L1為該直線族的其中一條直線 , 且平行於L2 : x - 2y + 3 = 0
求 L1 的方程。
A . Maths
2008-07-07 5:24 am
回答 (3)
2008-07-07 9:04 pm
2008-07-07 7:32 am
a) x+3y+1+k(3x-7y+4)=0
b)
Slope of L2=m2
x-2y+3=0
(1/2)x+3/2=y
m2=1/2
therefore slope of L1=m1=1/2
x+3y+1+k(3x-7y+4)=0
x+3y+1+3kx-7ky+4k=0
(1+3k)x+(3-7k)y+1+4k=0--------(1)
(3-7k)y=-(1+3k)x-(1+4k)
y= -(1+3k)/(3-7k) x - (1+4k)/ (3-7k)
-(1+3k)/(3-7k)=1/2
-2(1+3k)=3-7k
-2-6k=3-7k
k=5
SUB k=5 into (1)
(1+15)x +(3-35)y+1+20=0
16x-32y+21=0
L1: 16x-32y+21=0
b)
Slope of L2=m2
x-2y+3=0
(1/2)x+3/2=y
m2=1/2
therefore slope of L1=m1=1/2
x+3y+1+k(3x-7y+4)=0
x+3y+1+3kx-7ky+4k=0
(1+3k)x+(3-7k)y+1+4k=0--------(1)
(3-7k)y=-(1+3k)x-(1+4k)
y= -(1+3k)/(3-7k) x - (1+4k)/ (3-7k)
-(1+3k)/(3-7k)=1/2
-2(1+3k)=3-7k
-2-6k=3-7k
k=5
SUB k=5 into (1)
(1+15)x +(3-35)y+1+20=0
16x-32y+21=0
L1: 16x-32y+21=0
參考: me
2008-07-07 6:35 am
(a)x 3y 1 k(3x-7y 4)=0
(b)Let the equation of L1 be x 3y 1 k(3x-7y 4)=0
x(3k 1) y(3-7k) 1 4k=0
slope of L2=1/2
slope of L1=-(3k 1)/(3-7k)
-(3k 1)/(3-7k)=1/2
-6k-2=3-7k
k=5
sub k=5 into the equation
x[3(5) 1] y[3-7(5)] 1 4(5)=0
16x-32y 21=0
L1 is 16x-32y 21=0
(b)Let the equation of L1 be x 3y 1 k(3x-7y 4)=0
x(3k 1) y(3-7k) 1 4k=0
slope of L2=1/2
slope of L1=-(3k 1)/(3-7k)
-(3k 1)/(3-7k)=1/2
-6k-2=3-7k
k=5
sub k=5 into the equation
x[3(5) 1] y[3-7(5)] 1 4(5)=0
16x-32y 21=0
L1 is 16x-32y 21=0
收錄日期: 2021-04-13 15:47:32
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