If the line, 2 x - y +3=0 is at a distance (1/√5) and (2/√5) from the lines 4 x -2 y +α=0 and 6 x-3 y+β=0, respectively, then the sum of all possible values of α and β is .

Question

If the line, 2 x - y +3=0 is at a distance (1/√5) and (2/√5) from the lines 4 x -2 y +α=0 and 6 x-3 y+β=0, respectively, then the sum of all possible values of α and β is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Apply distance between parallel line formula 4 x-2 y+α=0 4 x-2 y+6=0 |(α-6/255)|=(1/55) |α-6|=2 ⇒ α=8,4 sum =12 again 6 x-3 y+β=0 6 x-3 y+9=0 |(β-9/3 √5)|=(2/√5) |β-9|=6 ⇒ β=15,3 sum=18 sum of all values of α and β is =30

Q. If the line, 2x−y+3=0 is at a distance 5​1​ and 5​2​ 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 ______.

Apply distance between parallel line formula 4x−2y+α=0 4x−2y+6=0 ∣∣​255α−6​∣∣​=551​ ∣α−6∣=2<br/><br/>⇒α=8,4 sum =12 again 6x−3y+β=0 6x−3y+9=0 ∣∣​35​β−9​∣∣​=5​2​ ∣β−9∣=6 ⇒β=15,3 sum=18 sum of all values of α and β is =30

Q. If the line, $2 x - y + 3 = 0$ is at a distance $\frac{1}{5}$ and $\frac{2}{5}$ from the lines $4 x - 2 y + α = 0$ and $6 x - 3 y + β = 0,$ respectively, then the sum of all possible values of $α$ and $β$ is ______.