Find the values of p in (i) to (iv) and p and q in (v) for the following pair of equations
(i) 3x - y - 5 = 0 and 6x - 2y - p = 0, if the lines represented by these equations are parallel.
(ii) -x + py = 1 and px - y - 1 = 0, if the pair of equations has no solution.
(ii) -3x + 5y = 7 and 2px - 3y = 1, if the lines represented by these equations are intersecting at a unique point.
(iv) 2x +3y - 5 = 0 and px - 6y - 8 = 0, if the pair of equations has a unique solution.
(v) 2+3y =7 and 2px + py = 28 - qy, if the pair of equations has infinitely many solutions.
(i) Given pair of linear equations is
3x - y - 5 = 0 …(i)
and 6x - 2y - p = 0 …(ii)
On comparing with ax + by + c = 0 we get
Here, a1 = , b1 = - 1, c1 = - 5;
And a2 = 6, b2 = - 2, c2 = - p;
a1 /a2 = /6 = 1/2
b1 /b2 = 1/2
c1 /c2 = 5/p
Since, the lines represented by these equations are parallel, then
a1/a2 = b1/b2 c1/c2
Taking last two parts, we get
So, p
Hence, the given pair of linear equations are parallel for all real values of p except 10 i.e.,
P
(ii) Given pair of linear equations is
- x + py = 1 …(i)
and px - y - 1 = 0 …(ii)
On comparing with ax + by + c = 0, we get
Here, a1 = , b1 = p, c1 =- 1;
And a2 = p, b2 = - 1, c2 =- 1;
a1 /a2 =
b1 /b2 = - p
c1 /c2 = 1
Since, the lines equations has no solution i.e., both lines are parallel to each other.
a1/a2 = b1/b2 c1/c2
= - p 1
Taking last two parts, we get
p
Taking first two parts, we get
p2 = 1
p = 1
Hence, the given pair of linear equations has no solution for p = 1.
(iii) Given, pair of linear equations is
- 3x + 5y = 7
and 2px - 3y = 1
On comparing with ax + by + c = 0, we get
Here, a1 = , b1 = 5, c1 = - 7;
And a2 = 2p, b2 = - 3, c2 = - 1;
a1 /a2 =
b1 /b2 = - 5/3
c1 /c2 = 7
Since, the lines are intersecting at a unique point i.e., it has a unique solution
a1/a2 b1/b2
so,
p 9/10
Hence, the lines represented by these equations are intersecting at a unique point for all real values of p except
(iv) Given, pair of linear equations is
2x + 3y - 5 = 0
and px - 6y - 8 = 0
On comparing with ax + by + c = 0 we get
Here, a1 = , b1 = 3, c1 = - 5;
And a2 = p, b2 = - 6, c2 = - 8;
a1 /a2 =
b1 /b2 = - 3/6 = - 1/2
c1 /c2 = 5/8
Since, the pair of linear equations has a unique solution.
a1/a2 b1/b2
so - 1/2
p - 4
Hence, the pair of linear equations has a unique solution for all values of p except - 4
i.e., p
(v) Given pair of linear equations is
2x + 3y = 7
and 2px + py = 28 - qy
or 2px + (p + q)y - 28 = 0
On comparing with ax + by + c = 0 we get
Here, a1 = , b1 = 3, c1 = - 7;
And a2 = 2p, b2 = (p + q), c2 = - 28;
Since, the pair of equations has infinitely many solutions i.e., both lines are coincident.
a1/a2 = b1/b2 = c1/c2
= =
Taking first and third parts, we get
p = 4
Again, taking last two parts, we get
=
p + q = 12
Since p = 4
So, q = 8
Here, we see that the values of p = 4 and q = 8 satisfies all three parts.
Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8.