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HKers 發問於 科學及數學數學 · 7 年前

family straight lines

1) It is given that the straight lines L(1): 2x+y-1=0 , L(2) : 3x-2y-5=0 , L(3) : x-3=0

and L(4) : x+y-1=0

a) Write down the equation of the family of straight lines passing through the

point of intersection L(1) and L(2).

b) Hence, find the equation of the line L which passes through the point of

intersection of L(1) and L(2) and the point of intersection of L(3) and L(4)

2) A straight line L passes through the point of intersection of the

lines L(1) :3x-4y+7=0 and L(2) : x-3y+2=0 and makes an acute angle of

45 degrees with the line L(3): y=2x+1. Find the equations of L

THz

更新:

第一題是否要把(3,-2) 代入2x+y-1+k(3x-2y-5) = 0 嗎?

更新 2:

For Q.2 http://upload.lsforum.net/users/public/f309582qw22...

Q.1 我試過,但ANS: x+2y+1=0

更新 3:

TO WANszeto

為什麼我計算K時只計到K=-3/2 or -k-1=-k-2

而你間接用Let m1 再代番入k -->(1 + 3k) / (3 + 4k) = 1/3

反而計倒???

To Mr Kwok

I'm so sorry , for Q1, i calculated it wrong, you are right! thz for your time

3 個解答

評分
  • 7 年前
    最愛解答

    1)

    a)

    The required equation of the family of st. lines :

    2x + y - 1 + k(3x - 2y - 5) = 0

    (2 + 3k)x +(1 - 2k)y - (1 + 5k) = 0 ...... [1]

    b)

    Find the point of intersection of L(3) and L(4):

    L(3): x - 3 = 0 ...... [2]

    L(4): x + y - 1 = 0 ...... [3]

    From [2]:

    x = 3

    Put x = 3 into [3]:

    (3) + y - 1 = 0

    y = -2

    To find the required equation, put x = 3 and y = -2 into [1]:

    (2 + 3k)(3) + (1 - 2k)(-2) - (1 + 5k) = 0

    6 + 9k - 2 + 4k - 1 - 5k = 0

    8k + 3 = 0

    k = -3/8

    The required equation :

    [2 + 3(-3/8)]x + [1 - 2(-3/8)]y - [1 + 5(-3/8)] = 0

    (7/8)x + (14/8)y + (7/8) = 0

    x + 2y + 1 =0

    2)

    Let m1 be the slope of the required line,

    and m2 be the slope of L(3).

    The slope of L(3): m2 = 2

    |(m1 - m2) / (1 + m1m2)| = tan45°

    |(m1 - 2) / (1 + 2m1)| = 1

    (m1 - 2) / (1 + 2m1) = ±1

    m1 - 2 = 1 + 2m1 ..or.. m1 - 2 = -(1 + 2m1)

    m1 = -3 ..or.. m1 = 1/3

    The equation for the family of st. line passing through the point ofintersection of L(1) and L(2):

    (x - 3y + 2) + k(3x - 4y + 7) = 0

    (1 + 3k)x - (3 + 4k)y + (2 + 7k) = 0 ...... [1]

    Slope of the required line:

    (1 + 3k) / (3 + 4k) = -3 ..or.. (1 + 3k) / (3 + 4k) = 1/3

    1 + 3k = -9 - 12k ..or.. 3 + 9k = 3 + 4k

    15k = -10 ..or.. 5k = 0

    k = -2/3 ..or.. k = 0

    Put k = -2/3 into [1]:

    [1 + 3(-2/3)]x - [3 + 4(-2/3)]y + [2 + 7(-2/3)] = 0

    (-3/3)x - (1/3)y - (8/3) = 0

    3x + y + 8 = 0

    Put k = 0 into 1:

    (1 + 0)x - (3 + 0)y + (2 + 0) = 0

    x - 3y + 2 = 0

    The required equations:

    3x + y+ 8 = 0 ..and.. x - 3y + 2 = 0

    2013-10-07 22:21:11 補充:

    Check for Q.1 :

    The point of intersection of L(1) and L(2): (1, -1)

    The point of intersection of L(3) and L(4): (3 - 2)

    Equations for the st. line passing through (1,-1) and (3, -2):

    x + 2y + 1 = 0

    2013-10-07 22:28:43 補充:

    Check for Q.2:

    The pt. of intersection between L(1) and L(2): (-13/5, -1/5)

    The pt. lies on both 3x + y+ 8 = 0 and x - 3y + 2 = 0

    For L(3): Slope = 2, angle of inclination = 63.43°

    For 3x +y + 8 = 0: Slope = -3, angle of inclination = 108.43°

    Angle between the two lines = 108.43° - 63.43° = 45°

    2013-10-07 22:32:21 補充:

    (Cont'd).....Check for Q.2:

    For L(3): Slope = 2, angle of inclination = 63.43°

    For x - 3y + 2 = 0: Slope = 1/3, angle of inclination = 18.43°

    Angle between the two lines = 63.43° - 18.43° = 45°

    2013-10-07 23:45:09 補充:

    用 (3x - 4y + 7) + k(x - 3y + 2) = 0 會出現你說的問題,兩邊 k 會約去了。

    用 (x - 3y + 2) + k(3x - 4y + 7) = 0 便可以了。

    這樣的情況偶發會出現。

    資料來源: wanszeto, wanszeto
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  • 7 年前

    各位~

    對於 family of straight for passing through intersection of ax+by+c=0 and px+qy+r=0.

    Some would use

    (ax+by+c) + k(px+qy+r)=0 which would miss px+qy+r = 0

    Some would use

    k(ax+by+c) + (px+qy+r)=0 which would miss ax+by+c = 0

    My teacher in the past taught us there is one good method.

    2013-10-12 23:23:19 補充:

    k(ax+by+c) + (1-k)(px+qy+r)=0 which would include every member.

    ☆ヾ(◕‿◕)ノ

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  • CK
    Lv 7
    7 年前

    family of straight lines passing through the

    point of intersection L(1) and L(2).

    L(1): 2x+y-1=0 , L(2) : 3x-2y-5=0

    Required line : 2x+y-1+k(3x-2y-5) = 0

    用 L(3) : x-3=0 and L(4) : x+y-1=0 解交點

    x=3

    3+y-1=0; y = -2

    交點為 (3,-2)

    2013-10-07 17:18:23 補充:

    2x+y-1+k(3x-2y-5) = 0 會通過 (3,-2),即是點呢?

    2013-10-07 17:24:55 補充:

    L : 3x-4y+7+k( x-3y+2)=0

    L(3): y=2x+1.

    L(3) 和 x軸夾角為θ 則 tanθ = 2 ; θ = 63.43º

    θ+45º = ?????

    L : 3x-4y+7+k( x-3y+2)=0 slope 是什麼呢?

    2013-10-07 21:00:20 補充:

    是,因為 2x+y-1+k(3x-2y-5) = 0 會通過 L(1)和L(2)的交點,現在要求通過L(3)和L(4)的交點,即(3,-2)。

    因此,點(3,-2)需要在 2x+y-1+k(3x-2y-5) = 0 上,亦即把 x=3,y=-2代入 2x+y-1+k(3x-2y-5) = 0 之中,等於0的條件要成立。

    2013-10-07 21:11:10 補充:

    等於0的 [關係] 要成立。

    2013-10-07 21:33:38 補充:

    試試 63.43º -45º

    2013-10-07 21:38:39 補充:

    其實你做得好好,應該是其中一條線與x軸的夾角比L(3)大45º而另一條線細45º。

    2013-10-07 21:51:38 補充:

    把(3,-2) 代入2x+y-1+k(3x-2y-5) = 0

    2(3) + (-2) -1 + k[(3)(3) - 2(-2) - 5]=0

    6-2-1+k[9+4-5]=0

    3+8k=0

    8k=-3

    k=-3/8

    2x+y-1+(-3/8)(3x-2y-5) = 0

    全式 X 8 ==> 16x+8y-8+(3x-2y-5)(-3)=0

    16x+8y-8+(-9x+6y+15)=0

    7x+14y+7=0

    x+2y+1=0

    你是不是說 書的答案不是 x+2y+1=0 呢?

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