# Class 11 RD Sharma Solutions- Chapter 23 The Straight Lines- Exercise 23.11

**Question 1: Prove that the following sets of three lines are concurrent:**

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

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

**Solution:**

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

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 = 0Given:

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.

**Question 2: For what value of **λ** are the three lines 2x – 5y + 3 = 0, 5x – 9y + **λ** = 0 and x – 2y + 1 = 0 concurrent?**

**Solution:**

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.

**Question 3: Find the conditions that the straight lines y = m**_{1}x + c_{1}, y = m_{2}x + c_{2} and y = m_{3}x + c_{3} may meet in a point.

_{1}x + c

_{1}, y = m

_{2}x + c

_{2}and y = m

_{3}x + c

_{3}may meet in a point.

**Solution:**

Given:

m

_{1}x – y + c_{1}= 0 …… (1)m

_{2}x – y + c_{2}= 0 …… (2)m

_{3}x – y + c_{3}= 0 …… (3)Solving equation (i) and (ii), we get,

m

_{1}x – y + c_{1}= m_{2}x – y + c_{2}m

_{1}x + c_{1}= m_{2}x + c_{2}m

_{1}x – m_{2}x = c_{2}– c_{1}x(m

_{1}– m_{2}) = c_{2}– c_{1}x = (c

_{2}– c_{1})/(m_{1}– m_{2})Now substituting the value of x in equation (i)

y = m

_{1}[(c_{2}– c_{1})/(m_{1}– m_{2})] + c_{1}y = m

_{1}c_{2}– m_{1}c_{1}+ m_{1}c_{1}– m_{2}c_{1}y = m

_{1}c_{2}– m_{2}c_{1}Now substituting the value of x and y in equation (iii), we get,

m

_{3}x – y + c_{3}= 0y = m

_{3}x + c_{3}m

_{1}c_{2}– m_{2}c_{1}= m_{3}[(c_{2}– c_{1})/(m_{1}– m_{2})] + c_{3}m

_{1}^{2}c_{2}– m_{1}m_{2}c_{1}+ m_{1}m_{2}c_{2}– m_{2}^{2}c_{1}= m_{3}c_{2}– m_{3}c_{1}+ m_{1}c_{3}– m_{2}c_{3}m

_{1}^{2}c_{2}– m_{1}c_{3}– m_{2}^{2}c_{1}+ m_{2}c_{3}– m_{3}c_{2}+ m_{3}c_{1}= 0m

_{1}(c_{2}– c_{3}) + m_{2}(c_{3}– c_{1}) + m_{3}(c_{1}– c_{2}) = 0Therefore, the required condition is m

_{1}(c_{2}– c_{3}) + m_{2}(c_{3}– c_{1}) + m_{3}(c_{1}– c_{2}) = 0

**Question 4: If the lines p**_{1}x + q_{1}y = 1, p_{2}x + q_{2}y = 1 and p_{3}x + q_{3}y = 1 be concurrent, show that the points (p_{1}, q_{1}), (p_{2}, q_{2}) and (p_{3}, q_{3}) are collinear.

_{1}x + q

_{1}y = 1, p

_{2}x + q

_{2}y = 1 and p

_{3}x + q

_{3}y = 1 be concurrent, show that the points (p

_{1}, q

_{1}), (p

_{2}, q

_{2}) and (p

_{3}, q

_{3}) are collinear.

**Solution:**

Given:

p

_{1}x + q_{1}y = 1 …… (i)p

_{2}x + q_{2}y = 1 …… (ii)p

_{3}x + q_{3}y = 1 …… (iii)Solving equation (i) and (ii), we get,

From equation (i) we get,

x = (1 – q

_{1}y)/p_{1}Now substituting the value of x in equation (ii)

p

_{2}[(1 – q_{1}y)/p_{1}] + q_{2}y = 1p

_{2}– p_{2}q_{1}y + p_{1}q_{2}y = p_{1}y(p

_{1}q_{2}– p_{2}q_{1}) = p_{1}– p_{2}y = (p

_{1}– p_{2})/(p_{1}q_{2}– p_{2}q_{1})Now substituting the value of y in x i.e.

x = (1 – q

_{1}y)/p_{1}x = (1 – q

_{1}[(p_{1}– p_{2})/(p_{1}q_{2}– p_{2}q_{1})])/p_{1}Now substituting the value of x and y in equation (iii), we get,

p

_{3}[(p_{1}q_{2}– p_{2}q_{1}– q_{1}(p_{1}– p_{2})(p_{1}q_{2}– p_{2}q_{1}))] + q_{3}p_{1}(p_{1 }– p_{2}) = 1(p

_{1}p_{3}q_{2 }– p_{2}p_{3}q_{1}– p_{1}p_{3}q_{1}+ p_{2}p_{3}q_{1})(p_{1}q_{1}– p_{2}q_{1}) + q_{3}p_{1}(p_{1}– p_{2}) = 1(p

_{1}p_{3}q_{2}– p_{1}p_{3}q_{1})(p_{1}q_{2}– p_{2}q_{1}) + q_{3}p_{1}^{2}– q_{3}p_{1}p_{2}= 1p

_{1}^{2}p_{3}q_{2}^{2}– p_{1}p_{2}p_{3}q_{1}q_{2}– p_{1}^{2}p_{3}q_{1}q_{2}+ p_{1}p_{2}p_{3}q_{1}^{2}+ q_{3}p_{1}p_{2}= 1 …… (iv)Also, if we assume points (p

_{1}, q_{1})(p_{2}, q_{2})(p_{3}, q_{3}) are collinearTherefore,

p

_{1}(q_{2}– q_{3}) + p_{2}(q_{3}– q_{1}) + p_{3}(q_{1}– q_{3}) = 0Now from equation (iv) we get,

p

_{1}[p_{1}p_{3}q_{2}^{2}– p_{2}p_{3}q_{1}q_{2 }– p_{1}p_{3}q_{1}q_{2}+ p_{2}p_{3}q_{1}^{2}+ q_{3}p_{2}] = 1p

_{1}[p_{3}q_{2}(p_{1}q_{2}– p_{2}q_{1}) – p_{3}q_{1}(p_{1}q_{2}– p_{2}q_{1}) + q_{3}(p_{1}– p_{2})] = 1Therefore, the given points, (p

_{1}, q_{1}), (p_{2}, q_{2}) and (p_{3}, q_{3}) are collinear.

**Question 5: Show that the straight lines L**_{1} = (b + c)x + ay + 1 = 0, L_{2} = (c + a)x + by + 1 = 0 and L_{3} = (a + b)x + cy + 1 = 0 are concurrent.

_{1}= (b + c)x + ay + 1 = 0, L

_{2}= (c + a)x + by + 1 = 0 and L

_{3}= (a + b)x + cy + 1 = 0 are concurrent.

**Solution:**

Given:

L

_{1}= (b + c)x + ay + 1 = 0 …… (i)L

_{2}= (c + a)x + by + 1 = 0 …… (ii)L

_{3}= (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 + a

^{2}x – b – b^{2}x + bcx + a = 0x(ac + a

^{2}– b^{2}+ bc) = b – ax(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.

**Question 6: If the three lines ax + a**^{2}y + 1 = 0, bx + b^{2}y + 1 = 0, and cx + c^{2}y + 1= 0 are concurrent, show that at least two of three constants a, b, c are equal.

^{2}y + 1 = 0, bx + b

^{2}y + 1 = 0, and cx + c

^{2}y + 1= 0 are concurrent, show that at least two of three constants a, b, c are equal.

**Solution:**

Given:

ax + a

^{2}y + 1 = 0 …… (i)bx + b

^{2}y + 1 = 0 …… (ii)cx + c

^{2}y + 1= 0 …… (iii)Solving equation (i) and (ii), we get,

From equation (i) we get,

x = (-1 – a

^{2}y)/aNow substituting the value of x in equation (ii)

b[(-1 – a

^{2}y)/a] + b^{2}y + 1 = 0-b – a

^{2}by + ab^{2}y + a = 0aby(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 – a

^{2}y)/ax = (-1 -a

^{2}(1/ab))/ax = (-b – a)/ba

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

c[(-b – a)/ba] + c

^{2}(1/ab) + 1 = 0-bc – ac + c

^{2}+ ab = 0c(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.

**Question 7: If a, b, c are in A.P. , prove that the straight lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent.**

**Solution:**

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.

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