2x−y+3=0……(i)
4x−2y+α=0⇒2x−y+2α=0……(ii)6x−3y+β=0⇒2x−y+3β=0……(iii)
d1=22+12∣2α−3∣=51⇒∣α−6∣=2
⇒α−6=2,−2⇒α=8,4
d2=22+12∣3β−3∣=52⇒∣β−9∣=6
⇒β−9=6,−6⇒β=15,3
Sum of all values of α and β=30.
If the line, 2x−y+3=0 is at a distance 51 and 52 from the lines 4x−2y+α=0 and 6x−3y+β=0 respectively, then the sum of all possible values of α and β is ____________.
Held on 5 Sept 2020 · Verified 6 Jul 2026.
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