Aditya Gupta
Last Activity: 5 Years ago
easily solve both eqns together to find intersection point P(3, 1).
given L2= 3x= y+8 with slope m1= 3
after rotation it becomes L2’ which passes through P. its slope will become (m1+m2)/(1 – m1*m2), where m1 is slope of L2 and m2= tanpi/4= 1. so (m1+m2)/(1 – m1*m2)= 4/-2= – 2
so, eqn of line is L2’: y – 1= – 2(x – 3)
or y+2x= 7
any point on above line is of the form A or B(c, 7 – 2c).
so, PA= OP
PA^2= OP^2
(c – 3)^2+(7 – 2c – 1)^2= 3^2+1^2
(c – 3)^2+4(c – 3)^2= 10
or (c – 3)^2= 2
or c= 3 ± root2
so that 7 – 2c= 1 ∓ 2root2
kindly approve :)
