1.Find the equation of a straight line passing through
the point of intersection of the line L1:4x-2y+3=0 and
L2:3x+7y-3=0 and with y-intercept 4.
Answer is 23x-3y+12=0
2.Find the equation(s) of the line which is concurrent
with the lines 13x-11y=9 and 7x-5y=10 and makes
equal intercepts on both axes.
67x-65y=0,x+y-11=0
請列明每一步!!!!!!!!!!!
A.Math一問
2008-04-04 10:45 pm
回答 (2)
2008-04-04 11:50 pm
✔ 最佳答案
Please refer to the following solution:圖片參考：http://hk.geocities.com/stevieg_1023/CONCU1.gif
圖片參考：http://hk.geocities.com/stevieg_1023/CONCU2.gif
2008-04-04 11:31 pm
1)
the family line is:
4x-2y+3+k(3x+7y-3)=0
when x=0, y=4
4(0)-2(4)+3+k[3(0)+7(4)-3]=0
-8+3+k(28-3)=0
25k=5
k=1/5
the equation is:
4x-2y+3+1/5(3x+7y-3)=0
20x-10y+15+3x+7y-3=0
23x-3y+12=0
2)
13x-11y=9
13x-11y-9=0----(1)
7x-5y=10
7x-5y-10=0---(2)
13x-11y-9+k(7x-5y-10)=0
13x-11y-9+7kx-5ky-10k=0
(13+7k)x+(-11-5k)y-9-10k=0
when x=0, y=a
(-11-5k)a-9-10k=0
-11a-5ak-9-10k=0---(1)
when x=a, y=0
13a+7ak-9-10k=0----(2)
(1)-(2)
-24a-12ak=0
-12a(2+k)=0
2+k=0
k=-2
when k=-2
-11a-5a(-2)-9-10(-2)=0
-11a+10a-9+20=0
-a+11=0
a=11
y=11
the equation
(13+7(-2))x+(-11-5(-2))y-9-10(-2)=0
-x-y+11=0
x+y-11=0
the family line is:
4x-2y+3+k(3x+7y-3)=0
when x=0, y=4
4(0)-2(4)+3+k[3(0)+7(4)-3]=0
-8+3+k(28-3)=0
25k=5
k=1/5
the equation is:
4x-2y+3+1/5(3x+7y-3)=0
20x-10y+15+3x+7y-3=0
23x-3y+12=0
2)
13x-11y=9
13x-11y-9=0----(1)
7x-5y=10
7x-5y-10=0---(2)
13x-11y-9+k(7x-5y-10)=0
13x-11y-9+7kx-5ky-10k=0
(13+7k)x+(-11-5k)y-9-10k=0
when x=0, y=a
(-11-5k)a-9-10k=0
-11a-5ak-9-10k=0---(1)
when x=a, y=0
13a+7ak-9-10k=0----(2)
(1)-(2)
-24a-12ak=0
-12a(2+k)=0
2+k=0
k=-2
when k=-2
-11a-5a(-2)-9-10(-2)=0
-11a+10a-9+20=0
-a+11=0
a=11
y=11
the equation
(13+7(-2))x+(-11-5(-2))y-9-10(-2)=0
-x-y+11=0
x+y-11=0
收錄日期: 2021-04-25 17:21:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080404000051KK01399