Find the bisector of the angle between the lines `2x+y-6 = 0 and 2x- 4y + 7 = 0 ` which contains the point (1,2)

Options

A. `6x -2y -5=0`
B. `2x + 6y -19=0`
C. `6x+2y -5=0`
D. `2x + 6y + 19=0`

Answer

A

Explanation

Let us re-write the equations of the two lines in such a way that the values of the expression on the left hand side of the equality for x = 1, y = 2 become positve.
Re- writing the given equations, we obtion.
`-2x-y + 6=0` and `2x -4y + 7=0`
The bisector of the angle containing the point (1,2) is given by
`(-2x -y+6)/((-2)^(2)+(-1)^(2))=+(2x-4y + 7)/(sqrt(2^(2) + (-4)^(2)))`
`rArr 2(-2x-y + 4y + 7 rArr 6x -2y - 5=0`
ALITER we observe that the origin and the point (1,2) lie on the sam side of the line `2x + y - 6=0` and also on the same side of the line `2x -4y + 7=0` . Thus, the required bisector is the bisector of the angle containing the origin and is given by
`(-2x -y+6)/((-2)^(2)+(-1)^(2))=+(2x-4y + 7)/(sqrt(2^(2) + (-4)^(2)))rArr 6x -2y -5=0`