Question

If distance between lines 2x - y + 3 = 0 and 4x - 2y +a = 0 is 1/root5 while distance between 2x -y + 3 = 0 and 6x - 3y +B = 0 is 2/root5 then alpha+ beta be ?

Answer 1

this question can be answered in many ways I think this may be the best way

Answer 2

ANSWER.

=> a + b = 15 + 8 = 23

EXPLANATION.

$\sf \to \: distance \: between \: line \: 2x - y + 3 = 0 \: \: and \: \: \: 4x - 2y + \alpha \: is \: \dfrac{1}{ \sqrt{5} } \\ \\ \sf \to \: distance \: between \: line \: 2x - y + 3 = 0 \: and \: 6x - 3y + \beta \: is \: \frac{2}{ \sqrt{5} }$

$\sf\to \: by \: using \: formula \: of \: distance \: between \: lines \\ \\ \sf \to \: | \frac{ c_{2} - c_{1} }{ \sqrt{ {a}^{2} + {b}^{2} } } |$

$\sf \to \: from \: line \\ \\ \sf \to \: \: (1) \: equation \: 2x - y + 3 = 0 \: \: and \: \: 4x - 2y + \alpha = 0 = \frac{1}{ \sqrt{5} } \\ \\ \sf \to divide \: equation \: (2) \: by \: 2 \: for \: lines \: are \: equal \\ \\ \sf \to \: 2x - y + \dfrac{ \alpha }{2} = 0 \\ \\ \sf \to \: | \dfrac{ \dfrac{ \alpha }{2} - 3}{ \sqrt{ {2}^{2} + 1 {}^{2} } } | = \dfrac{1}{ \sqrt{5} } \\ \\ \sf \to \: | \dfrac{ \dfrac{ \alpha - 6}{2} }{ \cancel{ \sqrt{5} }} | = \dfrac{1}{ \cancel{ \sqrt{5} }} \\ \\ \sf \to \: | \alpha - 6| = 2 \\ \\ \sf \to \: \alpha = 8 \: \: and \: \: 4$

$\sf \to \: for \: line \\ \\ \sf \to \: (2) \: equation \: of \: line \: 2x - y + 3 = 0 \: \: \:is \: \: 6x - 3y + \beta \: \: is \: \frac{2}{ \sqrt{5} } \\ \\ \sf \to \: divide \: equation \: (2) \: \: by \: \: 3 \: \: we \: get \\ \\ \sf \to \: 2x - y + \frac{ \beta }{3} \\ \\ \sf \to \: | \dfrac{ \dfrac{ \beta }{3} - 3 }{ \sqrt{ {2}^{2} + {1}^{2} } } | = \dfrac{2}{ \sqrt{5} } \\ \\ \sf \to \: | \beta - 9| = 6 \\ \\ \sf \to \: \: \beta = 15 \: \: and \: \: \: 3$

$\sf \to \: value \: of \: \alpha + \beta \\ \\ \sf \to \: 8 + 15 = 23 = answer$