Straight Lines and Pair of Straight Lines

‌‌

|6+12−3|

5

=‌

|12+4a−3|

5

=‌

|3α+4β−3|

5

‌‌ (i) ‌‌‌‌‌‌‌‌‌‌‌‌‌‌ (ii) ‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌ (iii) ‌
From (i) and (ii) we get
‌|4a+9|=15
‌⇒4a+9=15,−15
‌⇒‌‌4a=6,−24⇒a=‌

3

2

,−6
From (i) and (iii), we get
3α+4β−3=±15
Case (I) When 3α+4β−3=15‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
Also, 4α−3β+1=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) (given)
On solving Eqs. (i) and (ii) we get
α=2β=3
Case (II) When 3α+4β−3=−15
⇒‌‌3α+4β=−12‌‌‌⋅⋅⋅⋅⋅⋅⋅(iii)
‌ and ‌4α−3β+1=0‌‌‌⋅⋅⋅⋅⋅⋅⋅(iv)
On solving Eqs. (iii) and (iv) we get,
α=‌

−8

5

,β=‌

−9

5

∴ Sum of all possible values of a,α,β= ‌

3

2

+2+3=‌

3

2

+5=‌

13

2

‌‌‌⋅⋅⋅⋅⋅⋅⋅(v)
‌ or ‌−6−‌

8

5

−‌

9

5

=‌

−30−17

5

=‌

−47

5

‌‌‌⋅⋅⋅⋅⋅⋅⋅(vi)
On adding Eqs. (v) and (vi)
∴ Total sum =‌

−29

10