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

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Straight Lines

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 ______.

Answer: 30

Solution

Put each line in the form

2 x - y + c = 0

4 x - 2 y + α = 0

gives

c = \frac{α}{2}

and

6 x - 3 y + β = 0

gives

c = \frac{β}{3}

. The distance of

2 x - y + 3 = 0

from the first is

\frac{∣ α /2 - 3∣}{5} = \frac{1}{5}

, so

∣ α - 6∣ = 2

and

α = 4

8

. The distance from the second is

\frac{∣ β /3 - 3∣}{5} = \frac{2}{5}

, so

∣ β - 9∣ = 6

and

β = 3

15

. Adding all the possible values,

4 + 8 + 3 + 15 = 30

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Integer answer

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