Question

Find the sum of the sum of all possible values of \[\alpha \] and \[\beta \], If the line \[2x - y + 3 = 0\] is at a distance \[\dfrac{1}{{\sqrt 5 }}\] and \[\dfrac{2}{{\sqrt 5 }}\] from the lines \[4x - 2y + \alpha = 0\] and \[6x - 3y + \beta = 0\] respectively ?

Answer 1

Hint: We have to find the sum of all the possible values for which we will consider three equations of line and then we will calculate the distance of one line which is parallel to the other two.

Formula Used: We will use Distance formula \[d = \dfrac{{\left| {{c_2} - {c_1}} \right|}}{{\sqrt {{A^2} + {B^2}} }}\] to find the distances between the parallel lines .

Complete step by step solution
Let us consider three equation of line as-
\[2x - y + 3 = 0\] ---- \[{L_1}\]
\[4x - 2y + \alpha = 0\] ----\[{L_2}\]
\[6x - 3y + \beta = 0\] ---- \[{L_3}\]
Now simplify the Line equation \[{L_2}\] and \[{L_3}\] as such it looks parallel to \[{L_1}\].
\[4x - 2y + \alpha = 0\]----\[{L_2}\]
\[2x - y + \dfrac{\alpha }{2} = 0\] ---- (i)
Simplifying equation \[{L_3}\]:-
\[6x - 3y + \beta = 0\]---- \[{L_3}\]
\[2x - y + \dfrac{\beta }{3} = 0\] ---- (ii)
Now we have got two equations (i) and (ii) where both are parallel to \[{L_1}\].
Using the Distance Formula we have to find out the distance between the line equation (i) with \[{L_1}\] and equation (ii) with \[{L_1}\].
\[d = \dfrac{{\left| {{c_2} - {c_1}} \right|}}{{\sqrt {{A^2} + {B^2}} }}\]
For Equation (i) and \[{L_1}\]
 \[2x - y + 3 = 0\] ---- \[{L_1}\]
\[2x - y + \dfrac{\alpha }{2} = 0\] ---- (i)
Substitute the values in the formula \[d = \dfrac{{\left| {{c_2} - {c_1}} \right|}}{{\sqrt {{A^2} + {B^2}} }}\] where \[{C_2} = \dfrac{\alpha }{2},{C_1} = 3\] where distance will be equal to \[\dfrac{1}{{\sqrt 5 }}\] given in the question .
\[\dfrac{1}{{\sqrt 5 }} = \dfrac{{\left| {\dfrac{\alpha }{2} - 3} \right|}}{{\sqrt 5 }}\]
\[\left| {\dfrac{\alpha }{2} - 3} \right| = 1\]
Here comes the two cases where once we have to take mod as positive and in case 2 we have to take mod as negative.
CASE 1-
\[\dfrac{\alpha }{2} - 3 = 1\]
\[\dfrac{\alpha }{2} = 4\]
\[\alpha = 8\]
CASE 2-
\[ - \left( {\dfrac{\alpha }{2} - 3} \right) = 1\]
\[\dfrac{\alpha }{2} = 2\]
\[\alpha = 4\]
Now \[\alpha = 8,4\]
Now apply the same procedure for Equation (ii) with Line equation \[{L_1}\].
\[2x - y + 3 = 0\] ---- \[{L_1}\]
\[2x - y + \dfrac{\beta }{3} = 0\] ---- (ii)
Substitute the values in the formula \[d = \dfrac{{\left| {{c_2} - {c_1}} \right|}}{{\sqrt {{A^2} + {B^2}} }}\] where \[{C_2} = \dfrac{\beta }{3},{C_1} = 3\] where distance will be equal to \[\dfrac{2}{{\sqrt 5 }}\] given in the question .
\[\dfrac{2}{{\sqrt 5 }} = \dfrac{{\left| {\dfrac{\beta }{3} - 3} \right|}}{{\sqrt 5 }}\]
\[\left| {\dfrac{\beta }{3} - 3} \right| = 2\]
Here comes the two cases where once we have to take mod as positive and in case 2 we have to take mod as negative.
CASE 1-
\[\dfrac{\beta }{3} - 3 = 2\]
\[\dfrac{\beta }{3} = 5\]
\[\beta = 15\]
CASE 2-
\[ - \left( {\dfrac{\beta }{3} - 3} \right) = 2\]
\[\dfrac{\beta }{3} = 1\]
\[\beta = 3\]
Now \[\beta = 15,3\]
Now Sum of All the possible values of \[\alpha \] and \[\beta \] are = \[8 + 4 + 15 + 3 = 30\]

Hence the answer is 30

Note: The key point to remember is to equate the coefficients of x and y of the lines \[{L_2}\] and \[{L_3}\] as such it starts looking parallel to \[{L_1}\] which would help the student to find out the Variables for calculating the distance using Distance Formula.