The equation of the bisector of that angle between the lines ` x+ y = 3 and 2x - y = 2 ` which contains the point (1,1) is

Options

A. `(sqrt5 - 2sqrt2) x + (sqrt5 + sqrt2) y - 3sqrt5 + 2 sqrt2 = 0`
B. `(sqrt5 + 2sqrt2) x + (sqrt5 - sqrt2)y - 3sqrt5 - 2 sqrt2 = 0`
C. 3x = 10
D. none of these

Answer

A

Explanation

First we re-write the equations of the two lines in such a way that the values of the expressions on the left hand sides of the equality for x = 1 , y = 1 become positive .
Re-writing the given equations , we obtain
-x-y + 3 = 0 and - 2x + y + 2 = 0.
Now , we obtain the bisector of the angle containing point (1,1) for positive sign. The required bisector is given by
`(-x-y + 3)/(sqrt((-1)^(2) + (-1)^(2))) = + (-2 x + y + 2)/(sqrt(-2)^(2) + 1^(2))`
`implies (sqrt5 - 2 sqrt2 ) x + (sqrt5 + sqrt2)y - 3sqrt5 + 2sqrt2 = 0` .