By comparing the ratios a1/a2, b1/b2 and c1/c2, find out for what value (s) of α, the lines representing the following equations have a unique solution, no solution or infinitely many solution: αx + 3y = α - 3 12x + αy = a
Answers
Answer:
Parallel
The given linear equation are
⇒6x−3y+10=0....eq1
⇒a
1
=6,b
1
=−3,c
1
=10
⇒2x−y+9=0...eq2
⇒a
2
=2,b
2
=−1,c
2
=9
⇒
a
2
a
1
=
3
6
⇒=
1
3
⇒
b
2
b
1
=
−1
−3
⇒
c
2
c
1
=
9
10
Answer:
SOLUTION :
Given :
a) 5x- 4y + 8 = 0 & 7x+ 6y - 9 = 0
b) 9x + 3y + 12 = 0 & 18x + 6y + 24 = 0
c) 6x - 3y + 10 = 0 & 2x - y + 9 = 0
(a)
On comparing with a1x + b1y +c1 = 0 & a2x + b2y + c2 = 0
a1= 5 , b1= - 4 , c1= 8
a2= 7, b2= 6 , c2 = -9
Now,
a1/a2 = 5/7 , b1/b2 = - 4/6, c1/c2= 8/-9
Since, a1/a2 ≠ b1/b2
Hence, the lines representing the pair of linear equations are INTERSECTING at a point and have exactly one solution.
(b) 9x + 3y + 12 = 0 & 18x + 6y + 24 = 0
On comparing with a1x + b1y +c1 = 0 & a2x + b2y + c2 = 0
a1= 9 , b1= 3, c1= 12
a2= 18, b2= 6 , c2 = 24
Now,
a1/a2 = 9/18= 1/2 , b1/b2 = 3/6= 1/2 , c1/c2= 12/24= 1/2
Since, a1/a2 = b1/b2=c1/c2
Hence, the lines representing the pair of linear equations are COINCIDENT LINES and have infinitely many solutions.
c) 6x - 3y + 10 = 0 & 2x - y + 9 = 0
On comparing with a1x + b1y +c1 = 0 & a2x + b2y + c2 = 0
a1= 6 , b1= -3, c1= 10
a2=2, b2= -1, c2 = 9
Now,
a1/a2 = 6/2 , b1/b2 = -3/-1= 3, c1/c2= 10/9
Since, a1/a2 = b1/b2 ≠ c1/c2
Hence, the lines representing the pair of linear equations are PARALLEL LINES and have no many solution.