amaths.equations of staright lines!

1) A straight line L passes through the point(1,3). If the area of the triangle enclosed by the line L and the positive axes is 25/4 sq. units,find the equations of L.
SOLUTION
Let the line L is x/a+y/b=1
from the question, L passes through the point(1,3)
So
1/a+3/b=1 (a,b>0)
b+3a=ab...(1)
Also the line pass through (a,0) and (0,b)
So
ab/2=25/4
ab=25/2...(2)
substitute into (1)
b+3a=25/2
b=(25/2)-3a
substitute into (2)
a[(25/2)-3a]=25/2
a(25-6a)=25
6a^2-25a+25=0
(2a-5)(3a-5)=0
a=5/2 or 5/3
b=5 or 15/2 [b=(25/2)-3a]
the equations of L are
(2/5)x+(1/5)y=1 or (3/5)x+(2/15)y=1
2x+y-5=0 or 9x+2y-15=0
2)Find the equations of 2 straight lines which are concurrent with the lines 2x-y-8=0 and 4x+y-3=0 and make equal intercepts with the axes.
If the intercepts of the equation is 0
Then let the equations of the line L is
ax+by=0
x=-(b/a)y...(1)
To find the intersection point of the lines ax+by=0 and 2x-y-8=0
We substitute (1) into 2x-y-8=0
2[-(b/a)y]-y-8=0
[(-2b/a)-1]y-8=0
(-a-2b)y-8a=0...(2)
To find the intersection point of the lines ax+by=0 and 4x+y-3=0
We substitute (1) into 4x+y-3=0
4[-(b/a)y]+y-3=0
[(-4b/a)+1]y-3=0
(a-4b)y-3a=0...(3)
Since 3 line are concurrent, the values of y should be equal.
From (2) and (3)
(8/3)(a-4b)y=(-a-2b)y
8a-32b=-3a-6b
11a=26b
a=(26/11)b
So the equations of L is
ax+by=0
(26/11)bx+by=0
26x+11y=0
If the intercepts of the line L is not equal to 0
Let the equations of the line L is
x/a+y/a=1
x+y=a
So we have x=a-y...(4)
To find the intersection point of the lines x+y=a and 2x-y-8=0
We substitute (4) into 2x-y-8=0
2(a-y)-y-8=0
2a-3y-8=0...(5)
To find the intersection point of the lines x+y=a and 4x+y-3=0
We substitute (4) into 4x+y-3=0
4(a-y)+y-3=0
4a-3y-3=0...(6)
Since 3 line are concurrent, the values of y should be equal.
(6)-(5):
2a+5=0
a=5/2
the equations of line L is
x/a+y/a=1
(2/5)x+(2/5)y=1
2x+2y+5=0

1.設L為y-3=m(x-1)
化作截距式得:
y-3=mx-m
mx-y=m-3
mx/(m-3)-y/(m-3)=1
x/[(m-3)/m]+y/(-m+3)=1
因the area of the triangle enclosed by the line L and the positive axes is 25/4
即1/2*[(m-3)/m]*(-m+3)=25/4
2(-m^2+6m-9)=25m
-2m^2+12m-18-25m=0
2m^2+13m+18=0
(2m+9)(m+2)=0
m=-9/2 或 m=-2
m=-9/2代入y-3=mx-m得:
y-3=(-9/2)x+9/2
2y-6=-9x+9
9x+2y-15=0
m=-2代入y-3=mx-m得:
y-3=-2x+2
2x+y-5=0
所以9x+2y-15=0或2x+y-5=0為所求的直線L

2.由
2x-y-8=0-----------(1)
4x+y-3=0----------(2)
(1)+(2)
6x-11=0
x=11/6代入(2)
22/3+y-3=0
y=-13/3
即交點為(11/6,-13/3)
又因所求直線與兩軸截距相等,即只有過原點或斜率等於-1的直線
(1)過原點(0,0)的直線,由兩點式得:
(y-0)/(-13/3-0)=(x-0)/(11/6-0)
11y=-26x
26x+11y=0
(2)斜率等於-1的直線,由點斜式得:
y+13/3=-1(x-11/6)
x+y+15/6=0
2x+2y+5=0
即26x+11y=0和2x+2y+5=0為所求