第 11 课 RD Sharma 解决方案-第 23 章直线-练习 23.11

问题1：证明以下三行是并发的：

(i) 15x – 18y + 1 = 0、12x + 10y – 3 = 0 和 6x + 66y – 11 = 0

(ii) 3x – 5y – 11 = 0, 5x + 3y – 7 = 0 和 x + 2y = 0

解决方案：

(i) 15x – 18y + 1 = 0, 12x + 10y – 3 = 0 and 6x + 66y – 11 = 0

Given: 

15x – 18y + 1 = 0   …… (i)

12x + 10y – 3 = 0   …… (ii)

6x + 66y – 11 = 0   …… (iii)

Solving equation (i) and (ii), we get,

From equation (i) we get,

x = (18y – 1)/15

Now substituting the value of x in equation (ii)

12 [(18y – 1)/15] + 10y – 3 = 0

216y – 12 + 150y – 45 = 0

366y = 57

y = 57/366 = 19/122

Now substituting the value of y in x i.e.

x = (18y – 1)/15

x = (18(19/122) – 1)/15

x = (342 – 122) / (122 × 15)

x = (342 – 122) / 1730

x = 220/1730

x= 22/173

Now substituting the value of x and y in equation (iii), we get,

6(22/173) + 66(19/122) – 11 = 0

(6 × 22 ×122) + (66 × 19 × 173) – (11 × 173 × 122) = 0

320 – 2052 + 732 = 0

0 = 0 

Therefore, the given lines are concurrent.

(ii) 3x – 5y – 11 = 0, 5x + 3y – 7 = 0 and x + 2y = 0

Given:

3x − 5y − 11 = 0   …… (i)

5x + 3y − 7 = 0   …… (ii)

x + 2y = 0   …… (iii)

Solving equation (ii) and (iii), we get,

From equation (iii) we get,

x = -2y

Now substituting the value of x in equation (ii)

5(-2y) + 3y – 7 = 0

-10y + 3y – 7 = 0

-7y = 7

y = -1

Now substituting the value of y in x i.e.

x = -2y

x = -2(-1)

x = 2

Now substituting the value of x and y in equation (i), we get,

3(2) − 5(-1) − 11 = 0

6 + 5 – 11 = 0

11 – 11 = 0 

0 = 0

Therefore, the given lines are concurrent.

问题 2：对于λ的值，2x – 5y + 3 = 0、5x – 9y + λ = 0 和 x – 2y + 1 = 0 三行并发？

解决方案：

Given:

2x − 5y + 3 = 0   …… (i)

5x − 9y + λ = 0   …… (ii)

x − 2y + 1 = 0   …… (iii)

Solving equation (i) and (iii), we get,

From equation (i) we get,

2x = 5y – 3

x = (5y – 3)/2

Now substituting the value of x in equation (iii)

[(5y – 3)/2] – 2y + 1 = 0

5y – 3 – 4y + 2 = 0

y = 1

Now substituting the value of y in x i.e.

x = (5y – 3)/2

x = (5 – 3)/2

x = 2/2

x = 1

Now substituting the value of x and y in equation (ii), we get,

5(1) – 9(1) + λ = 0

5 – 9 + λ = 0

λ = 4

Therefore, the value of λ is 4.

问题3：求直线y = m ₁ x + c ₁ ，y = m ₂ x + c ₂和y = m ₃ x + c ₃可能在一点相交的条件。

解决方案：

Given:

m₁x – y + c₁ = 0   …… (1)

m₂x – y + c₂ = 0   …… (2)

m₃x – y + c₃ = 0   …… (3)

Solving equation (i) and (ii), we get,

m₁x – y + c₁ = m₂x – y + c₂

m₁x + c₁ = m₂x + c₂

m₁x – m₂x = c₂ – c₁

x(m₁ – m₂) = c₂ – c₁

x = (c₂ – c₁)/(m₁ – m₂)

Now substituting the value of x in equation (i)

y = m₁[(c₂ – c₁)/(m₁ – m₂)] + c₁

y = m₁c₂ – m₁c₁ + m₁c₁ – m₂c₁

y = m₁c₂ – m₂c₁

Now substituting the value of x and y in equation (iii), we get,

m₃x – y + c₃ = 0

y = m₃x + c₃

m₁c₂ – m₂c₁= m₃[(c₂ – c₁)/(m₁ – m₂)] + c₃

m₁²c₂ – m₁m₂c₁ + m₁m₂c₂ – m₂²c₁ = m₃c₂ – m₃c₁ + m₁c₃ – m₂c₃

m₁²c₂ – m₁c₃ – m₂²c₁ + m₂c₃ – m₃c₂ + m₃c₁ = 0

m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0

Therefore, the required condition is m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) = 0

问题 4：如果线 p ₁ x + q ₁ y = 1, p ₂ x + q ₂ y = 1 和 p ₃ x + q ₃ y = 1 是并发的，证明点 (p ₁ , q ₁ ), (p ₂ , q ₂ ) 和 (p ₃ , q ₃ ) 是共线的。

解决方案：

Given:

p₁x + q₁y = 1    …… (i)

p₂x + q₂y = 1    …… (ii)

p₃x + q₃y = 1    …… (iii)

Solving equation (i) and (ii), we get,

From equation (i) we get,

x = (1 – q₁y)/p₁

Now substituting the value of x in equation (ii)

p₂[(1 – q₁y)/p₁] + q₂y = 1  

p₂ – p₂q₁y + p₁q₂y = p₁

y(p₁q₂ – p₂q₁) = p₁ – p₂

y = (p₁ – p₂)/(p₁q₂ – p₂q₁)

Now substituting the value of y in x i.e.

x = (1 – q₁y)/p₁

x = (1 – q₁[(p₁ – p₂)/(p₁q₂ – p₂q₁)])/p₁

Now substituting the value of x and y in equation (iii), we get,

p₃[(p₁q₂ – p₂q₁ – q₁(p₁ – p₂)(p₁q₂ – p₂q₁))] + q₃p₁(p₁ – p₂) = 1

(p₁p₃q₂ – p₂p₃q₁ – p₁p₃q₁ + p₂p₃q₁)(p₁q₁ – p₂q₁) + q₃p₁(p₁ – p₂) = 1

(p₁p₃q₂ – p₁p₃q₁)(p₁q₂ – p₂q₁) + q₃p₁² – q₃p₁p₂ = 1

p₁²p₃q₂² – p₁p₂p₃q₁q₂ – p₁²p₃q₁q₂ + p₁p₂p₃q₁² + q₃p₁p₂ = 1    …… (iv)

Also, if we assume points (p₁, q₁)(p₂, q₂)(p₃, q₃) are collinear

Therefore, 

p₁(q₂ – q₃) + p₂(q₃ – q₁) + p₃(q₁ – q₃) = 0

Now from equation (iv) we get,

p₁[p₁p₃q₂² – p₂p₃q₁q₂ – p₁p₃q₁q₂ + p₂p₃q₁² + q₃p₂] = 1

p₁[p₃q₂(p₁q₂ – p₂q₁) – p₃q₁(p₁q₂ – p₂q₁) + q₃(p₁ – p₂)] = 1

Therefore, the given points, (p₁, q₁), (p₂, q₂) and (p₃, q₃) are collinear.

问题 5：证明直线 L ₁ = (b + c)x + ay + 1 = 0, L ₂ = (c + a)x + by + 1 = 0 和 L ₃ = (a + b)x + cy + 1 = 0 是并发的。

解决方案：

Given:

L₁ = (b + c)x + ay + 1 = 0    …… (i)

L₂ = (c + a)x + by + 1 = 0    …… (ii)

L₃ = (a + b)x + cy + 1 = 0    …… (iii)

Solving equation (i) and (ii), we get,

From equation (i) we get,

y = (-1 – (b + c)x)/a

Now substituting the value of y in equation (ii)

(c + a)x + b[(-1 – (b + c)x)/a] + 1 = 0

(c + a)x + b[(-1 – (bx + cx)/a] + 1 = 0

cx + ax + b[(-1 – (bx + cx)/a] + 1 = 0

acx + a²x – b – b²x + bcx + a = 0

x(ac + a² – b² + bc) = b – a

x(c(a – b) + (a – b)(a + b)) = b – a

x(a – b)(c + a + b) = -(a – b)    [Dividing both side by (a – b)]

x(c + a + b) = -1

x = -1/(a + b + c)

Now substituting the value of x in y i.e.

y = (-1 – (b + c)x)/a

y = (-1 – (b + c)[-1/(a + b + c)])/a

y = (-(a + b + c) + b + c)/a(a + b + c)

y = (-a – b – c + b + c)/a(a + b + c)

y = (-a)/a(a + b + c)

y = -1/(a + b + c)

Now substituting the value of x and y in equation (iii), we get,

(a + b)[-1/(a + b + c)] + c[-1/(a + b + c)]+ 1 = 0 

-a – b – c + a + b + c = 0

0 = 0

Therefore, the given lines are concurrent.

问题 6：如果三行 ax + a ² y + 1 = 0、bx + b ² y + 1 = 0 和 cx + c ² y + 1= 0 是并发的，证明三个常数 a、 b、c 相等。

解决方案：

Given: 

ax + a²y + 1 = 0    …… (i)

bx + b²y + 1 = 0    …… (ii)

cx + c²y + 1= 0    …… (iii)

Solving equation (i) and (ii), we get,

From equation (i) we get,

x = (-1 – a²y)/a

Now substituting the value of x in equation (ii)

b[(-1 – a²y)/a] + b²y + 1 = 0

-b – a²by + ab²y + a = 0

aby(b – a) = b – a  [Dividing both side by (b – a)]

aby = 1

y = 1/ab

Now substituting the value of y in x i.e.

x = (-1 – a²y)/a

x = (-1 -a²(1/ab))/a

x = (-b – a)/ba

Now substituting the value of x and y in equation (iii), we get,

c[(-b – a)/ba] + c²(1/ab) + 1 = 0

-bc – ac + c² + ab = 0

c(c – b) – a(c – b) = 0

(c – b)(c – a) = 0

c – b = 0

c = b

or

c – a = 0

c = a

Therefore, at least two of three constants a, b, c are equal.

问题 7：如果 a,b,c 在 AP 中，证明直线 ax + 2y + 1 = 0, bx + 3y + 1 = 0 和 cx + 4y + 1 = 0 是并发的。

解决方案：

Given if a, b, c are in A.P.

Thus, b – a = c – b

2b = a + c  [Common difference]     …… (i)

Also given:

ax + 2y + 1 = 0    …… (ii)

bx + 3y + 1 = 0    …… (iii)

cx + 4y + 1 = 0    …… (iv)

Solving equation (ii) and (iii), we get,

From equation (ii) we get,

x = (-1 – 2y)/a

Now substituting the value of x in equation (iii)

b[(-1 – 2y)/a] + 3y + 1 = 0

-b – 2by + 3ay + a = 0

y(3a – 2b) = b – a

y = (b – a)/(3a – 2b)

Now substituting the value of y in x i.e.

x = (-1 – 2y)/a

x = (-1 – 2[(b – a)/(3a – 2b)])/a

x = (-(3a – 2b) – 2b + 2a)/a(3a – 2b)

x = -1/(3a – 2b)

Now substituting the value of x and y in equation (iv), we get,

c[-1/(3a – 2b)] + 4[(b – a)/(3a – 2b)] + 1 = 0

-c + 4b – 4a + 3a – 2b = 0

-a + 2b – c = 0

From equation (i) we know, 2b = a + c,

Thus, -a + a + c – c = 0

0 = 0 

Therefore, the given lines are concurrent.