RD Chapter 23 The Straight Lines Ex 23.11 Solutions
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
Answer - 1 : -
(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)
Now, consider thefollowing determinant:
=> 1320 – 2052 +732 = 0
Hence proved, thegiven 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)
Now, consider thefollowing determinant:
Hence, the given linesare 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?
Answer - 2 : -
Given:
2x − 5y + 3= 0 … (1)
5x − 9y+ λ = 0 … (2)
x − 2y + 1 =0 … (3)
It is given that thethree lines are concurrent.
Now, consider thefollowing determinant:
2(-9 + 2λ) + 5(5– λ) + 3(-10 + 9) = 0
-18 + 4λ + 25 –5λ – 3 = 0
λ = 4
∴ The value ofλ is 4.
Question - 3 : - Find the conditions that the straight lines y = m1x + c1,y = m2x + c2 and y = m3x + c3 maymeet in a point.
Answer - 3 : -
Given:
m1x – y + c1 =0 … (1)
m2x – y + c2 =0 … (2)
m3x – y + c3 =0 … (3)
It is given that thethree lines are concurrent.
Now, consider thefollowing determinant:
m1(-c3 +c2) + 1(m2c3-m3c2) + c1(-m2 +m3) = 0
m1(c2-c3)+ m2(c3-c1) + m3(c1-c2)= 0
∴ The requiredcondition 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.
Answer - 4 : -
Given:
p1x + q1y= 1
p2x + q2y= 1
p3x + q3y= 1
The given lines can bewritten as follows:
p1 x +q1 y – 1 = 0 … (1)
p2 x +q2 y – 1 = 0 … (2)
p3 x +q3 y – 1 = 0 … (3)
It is given that thethree lines are concurrent.
Now, consider thefollowing determinant:
Hence proved, thegiven three 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.
Answer - 5 : -
Given:
L1 =(b + c)x + ay + 1 = 0
L2 =(c + a)x + by + 1 = 0
L3 =(a + b)x + cy + 1 = 0
The given lines can bewritten as follows:
(b + c) x + ay + 1 = 0… (1)
(c + a) x + by + 1 = 0… (2)
(a + b) x + cy + 1 = 0… (3)
Consider the followingdeterminant.
Hence proved, thegiven lines are concurrent.