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Anonymous asked in 科學及數學數學 · 1 decade ago

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

2 Answers

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  • 1 decade ago
    Favorite Answer

    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

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  • 1 decade ago

    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

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