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  • If the line,  is at a distance <img alt="\frac{1}{\sqrt{5}}" src="http://entrancecorner.codecogs.com/gif.latex?%5Cfrac%7B1%7D%7B%5Csqrt%7B5%7D%7

If the line, 2x-y+3=0 is at a distance \frac{1}{\sqrt{5}} and \frac{2}{\sqrt{5}} from the lines 4x-2y+\alpha =0 and 6x-3y+\beta =0, respectively, then the sum of all possible values of \alpha and \beta is ______.
Option: 1 10
Option: 2 20
Option: 3 30
Option: 4 40

Answers (1)

best_answer

Apply distance between parallel line formula

Distance between the parallel line

\\2 x-y+3=0\;\;\;\;\;\;\;\;\;\ldots(1)\\4 x-2 y+\alpha=0\;\;\;\;\;\;\;\ldots(2)

Divide equation (2) by 2, to make coefficient of x and y same

\Rightarrow 2 x-y+\frac{\alpha}{2}=0

\\\Rightarrow\left|\frac{\frac{\alpha}{2}-3}{\sqrt{2^{2}+1^{2}}}\right|=\frac{1}{\sqrt{5}} \\ \Rightarrow\left|\frac{\frac{\alpha-6}{2}}{\sqrt{5}}\right|=\frac{1}{\sqrt{5}} \\ \Rightarrow|\alpha-6|=2 \\ \Rightarrow \alpha=8 \text { and } 4

Sum = 12

Again, Multiply equation (1) by 3

\\6 x-3 y+\beta=0 \\ 6 x-3 y+9=0 \\ \left|\frac{\beta-9}{3 \sqrt{5}}\right|=\frac{2}{\sqrt{5}} \\ |\beta-9|=6 \Rightarrow \beta=15,3 \\ \text { sum }=18

Sum = 12 + 18 = 30

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himanshu.meshram

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