If the line- 2x-y - 3 - 0 is at a distance 1-5 and 2-5 from the lines 4x - 2y - -0 and 6x- 3y - - - 0- respectively- then the sum of all possible values of - and - is

L_1:2x y+3=0 L_2:4x 2y+α = 0 L_3:6x 3y+β = 0 | α2 3 |√(5) = 1√(5) ⇒ nbsp; α2 3 = 1, 1 ⇒α = 8, 4 sum =12 | β3 3 |√(5) = 2√(5) ⇒ nbsp; β3 3 = 2, 2 ⇒ ...

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Standard XII

Mathematics

Condition for Two Lines to Be Perpendicular

If the line, ...

Question

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

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Solution

$L_{1} : 2 x - y + 3 = 0$
$L_{2} : 4 x - 2 y + α = 0$
$L_{3} : 6 x - 3 y + β = 0$
$\frac{∣ ∣ ∣ \frac{α}{2} - 3 ∣ ∣ ∣}{\sqrt{5}} = \frac{1}{\sqrt{5}}$
$\Rightarrow \frac{α}{2} - 3 = 1, - 1$
$\Rightarrow α = 8, 4$
sum $= 12$

$\frac{∣ ∣ ∣ \frac{β}{3} - 3 ∣ ∣ ∣}{\sqrt{5}} = \frac{2}{\sqrt{5}}$
$\Rightarrow \frac{β}{3} - 3 = 2, - 2$
$\Rightarrow β = 15, 3$
sum $= 18$
Total sum $= 12 + 18 = 30$

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If the line, 2x−y+3=0 is at a distance 1√5 and 2√5 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

L1:2x−y+3=0 L2:4x−2y+α=0 L3:6x−3y+β=0 ∣∣∣α2−3∣∣∣√5=1√5 ⇒α2−3=1,−1 ⇒α=8,4 sum =12 ∣∣∣β3−3∣∣∣√5=2√5 ⇒β3−3=2,−2 ⇒β=15,3 sum =18 Total sum =12+18=30

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