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

• Last Updated : 19 Jan, 2021

### 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

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Solution:

(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.

### 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 = m1x + c1, y = m2x + c2 and y = m3x + c3 may meet in a point.

Solution:

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

### Question 4: If the lines p1x + q1y = 1, p2x + q2y = 1 and p3x + q3y = 1 be concurrent, show that the points (p1, q1), (p2, q2) and (p3, q3) are collinear.

Solution:

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.

### Question 5: Show that the straight lines L1 = (b + c)x + ay + 1 = 0, L2 = (c + a)x + by + 1 = 0 and L3 = (a + b)x + cy + 1 = 0 are concurrent.

Solution:

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.

### Question 6: If the three lines ax + a2y + 1 = 0, bx + b2y + 1 = 0, and cx + c2y + 1= 0 are concurrent, show that at least two of three constants a, b, c are equal.

Solution:

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.

### 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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