JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 19)
If the two lines x + (a – 1) y = 1 and
2x + a2y = 1 (a$$ \in $$R – {0, 1}) are perpendicular, then
the distance of their point of intersection from the
origin is :
$${2 \over \sqrt5}$$
$${\sqrt2 \over 5}$$
$${2 \over 5}$$
$$\sqrt{2 \over 5}$$
Explanation
Line 1 : x + (a – 1) y = 1
Slope of this line (m1) = $${{ - 1} \over {a - 1}}$$
Line 2 : 2x + a2y = 1
Slope of this line (m2) = $$ - {2 \over {{a^2}}}$$
Line 1 and Line 2 are perpendicular to each other.
$$ \therefore $$ m1 m2 = -1
$$ \Rightarrow $$ $$\left( {{{ - 1} \over {a - 1}}} \right)\left( { - {2 \over {{a^2}}}} \right) = - 1$$
$$ \Rightarrow $$ $${a^3} - {a^2} + 2 = 0$$
$$ \Rightarrow $$ $$\left( {a + 1} \right)$$$$\left( {{a^2} - 2a + 2} \right)$$ = 0
$$ \therefore $$ $${a = - 1}$$
So lines are
Line 1 : x - 2y + 1 = 0
Line 2 : 2x + y - 1 = 0
Solving these equation we get point of intersection
P$$\left( {{3 \over 5}, - {1 \over 5}} \right)$$
Now distance of P from origin
OP = $$\sqrt {{{\left( {{3 \over 5}} \right)}^2} + {{\left( { - {1 \over 5}} \right)}^2}} $$
= $$\sqrt {{{10} \over {25}}} $$
= $$\sqrt {{2 \over 5}} $$
Slope of this line (m1) = $${{ - 1} \over {a - 1}}$$
Line 2 : 2x + a2y = 1
Slope of this line (m2) = $$ - {2 \over {{a^2}}}$$
Line 1 and Line 2 are perpendicular to each other.
$$ \therefore $$ m1 m2 = -1
$$ \Rightarrow $$ $$\left( {{{ - 1} \over {a - 1}}} \right)\left( { - {2 \over {{a^2}}}} \right) = - 1$$
$$ \Rightarrow $$ $${a^3} - {a^2} + 2 = 0$$
$$ \Rightarrow $$ $$\left( {a + 1} \right)$$$$\left( {{a^2} - 2a + 2} \right)$$ = 0
$$ \therefore $$ $${a = - 1}$$
So lines are
Line 1 : x - 2y + 1 = 0
Line 2 : 2x + y - 1 = 0
Solving these equation we get point of intersection
P$$\left( {{3 \over 5}, - {1 \over 5}} \right)$$
Now distance of P from origin
OP = $$\sqrt {{{\left( {{3 \over 5}} \right)}^2} + {{\left( { - {1 \over 5}} \right)}^2}} $$
= $$\sqrt {{{10} \over {25}}} $$
= $$\sqrt {{2 \over 5}} $$
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