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.


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