The equation of line `l_(1)` is `y=2x+3`, and the equation of line `l_(2)` is y=2x-5.

The equation of line l_(1) is y=(5)/(2)x-4 , and the equation of line l_(2) is y=-(2)/(5)x+9 .

Given equation of line L_(1) is y = 4. (iii) Find the equation of L_(2) .

In the given figure, Given equation of line L_(1) is y = 4. Write the slope of line L_(2)" if "L_(2) is the bisector of angle O.

An incident ray (L_(1)) is reflected by the mirror (L_(2))3x+4y=5 and the equation of reflected ray(L_(3)) is x+y=1 the equation of (L_(1)) is :

The bisector of two lines L and L are given by 3x^(2)-8xy-3y^(2)+10x+20y-25=0. If the line L_(1) passes through origin,find the equation of line L_(2).

The circle x^(2)+y^(2)-4x-2y=11 makes equal intercepts on the lines L_(1):x+2y+1=0 and L_(2). Then equation of L_(2) cannot be 2x+y+1=02x-y+3=02x+y=0(d)3x-2y+2=0

Lying in the plane x+y+z=6 is a line L passing through (1, 2, 3) and perpendicular to the line of intersection of planes x+y+z=6 and 2x-y+z=4 , then the equation of L is

Lying in the plane x+y+z=6 is a line L passing through (1, 2, 3) and perpendicular to the line of intersection of planes x+y+z=6 and 2x-y+z=4 , then the equation of L is

Let L be the line x-2y+3=0 .The equation of line L_1 is parallel to L and passing through (1,-2).Find the distance between L and L_1 .