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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.
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.
(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.
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.
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
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.
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.
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.
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:
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.
