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 :
Solutions · 1
Official solution
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
\[ \Rightarrow \] \[\left( {{{ - 1} \over {a - 1}}} \right)\left( { - {2 \over {{a^2}}}} \right) = - 1\]
\[ \Rightarrow \] a3−a2+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}} =2510=52