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# 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.

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.

thanks!

ans:

1)9x+2y-15=0 or 2x+y-5=0

2)26x+11y=0,2x+2y+5=0

Rating

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