find out the value of K,if the lines 3x+5y=7and 12x-Ky=5 are parallel.
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Step-by-step explanation:
The straight lines :
2x+ky=1
3x−5y=7
Comparing 2x+ky=1
with a
1
x+b
1
y+c
1
=0
a
1
=2,b
1
=k,c
1
=−1
Comparing 3x−5y=7
with a
2
x+b
2
y+c
2
=0
a
2
=3,b
2
=−5,c
2
=−7
∴ For a unique solution ,
a
2
a
1
=
b
2
b
1
⇒
3
2
=
−5
k
∴k
=
3
−10
For infinite solutions ,
a
2
a
1
=
b
2
b
1
=
c
2
c
1
⇒
3
2
=
−5
k
=
7
1
thereisnosuchvalueofkwhichwillsatisfytheequation
For no solutions :
a
2
a
1
=
b
2
b
1
=
c
2
c
1
⇒
3
2
=
−5
k
∴k=−
3
10
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