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

Q.1 我試過,但ANS: x+2y+1=0

TO WANszeto

To Mr Kwok

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

### 3 個解答

• 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
• 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

• 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？