JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 3)
If the line, 2x - y + 3 = 0 is at a distance
$${1 \over {\sqrt 5 }}$$ and $${2 \over {\sqrt 5 }}$$ 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 :
$${1 \over {\sqrt 5 }}$$ and $${2 \over {\sqrt 5 }}$$ 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 :
Answer
30
Explanation
Apply distance between parallel line formula
$$4x - 2y + \alpha = 0$$
$$4x - 2y + 6 = 0$$
$$\left| {{{\alpha - 6} \over {25}}} \right| = {1 \over {55}}$$
$$|\alpha - 6|\, = 2 \Rightarrow \alpha = 8,4$$
sum = 12
Again
$$6x - 3y + \beta = 0$$
$$6x - 3y + 9 = 0$$
$$\left| {{{\beta - 9} \over {3\sqrt 5 }}} \right| = {2 \over {\sqrt 5 }}$$
$$|\beta - 9|\, = 6 \Rightarrow \beta = 15,3$$
sum = 18
$$ \therefore $$ Sum of all values of $$\alpha$$ and $$\beta$$ is = 30
$$4x - 2y + \alpha = 0$$
$$4x - 2y + 6 = 0$$
$$\left| {{{\alpha - 6} \over {25}}} \right| = {1 \over {55}}$$
$$|\alpha - 6|\, = 2 \Rightarrow \alpha = 8,4$$
sum = 12
Again
$$6x - 3y + \beta = 0$$
$$6x - 3y + 9 = 0$$
$$\left| {{{\beta - 9} \over {3\sqrt 5 }}} \right| = {2 \over {\sqrt 5 }}$$
$$|\beta - 9|\, = 6 \Rightarrow \beta = 15,3$$
sum = 18
$$ \therefore $$ Sum of all values of $$\alpha$$ and $$\beta$$ is = 30
Comments (0)
