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

In Class 11 mathematics, Chapter 23 of RD Sharma's textbook focuses on "The Straight Lines". This chapter is fundamental for understanding various aspects of linear equations and their graphical representation. Exercise 23.11 is specifically designed to test students' proficiency in solving problems related to the concepts discussed in the chapter. This exercise includes problems that involve finding the equation of lines their slopes and other related properties.

The Straight Lines

Straight lines are a crucial component of coordinate geometry. They are defined by the linear equations and can be represented in various forms such as the slope-intercept form, point-slope form, and general form. The slope of a line indicates its steepness and direction while the y-intercept represents the point where the line crosses the y-axis. Understanding how to manipulate these equations and apply geometric properties is essential for solving problems involving straight lines.

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 = 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.

Read More:

• Standard Form of a Straight Line
• How to find the Equation of a Straight Line?

Conclusion

Exercise 23.11 in Chapter 23 of RD Sharma's textbook provides the comprehensive set of problems to the practice and reinforce the concepts of the straight lines. Mastery of this exercise will help students develop a deeper understanding of the linear equations and their applications in coordinate geometry. By solving these problems, students will enhance their problem-solving skills and prepare for the more advanced topics in mathematics.