Forces of `1` and `2 units` act along the lines `x = 0` and ` y = 0`. The equation of the line of action of the resultant is

A ray of light is incident along a line which meets another line, 7x-y+1 = 0 , at the point (0, 1). The ray isthen reflected from this point along the line, y+ 2x=1 . Then the equation of the line of incidence of the ray of light is (A) 41x + 38 y - 38 =0 (B) 41 x - 38 y + 38 = 0 (C) 41x + 25 y - 25 = 0 (D) 41x - 25y + 25 =0

A ray of light is incident along a line which meets another line 7x-y+1=0 , at the point (0,1). The ray is then reflected from this point along the line y+2x=1 . Then the equation of the line of incidence of the ray of light is

Two forces F_(1)=1N and F_(2)=2N act along the lines x=0 and y=0, respectively. Then find the resultant of forces.

Two forces F_(1)=1N and F_(2)=2N act along the lines x=0 and y=0, respectively. Then find the resultant of forces.

A ray of light is incident along a line which meets another line,7x-y+1=0, at the point (0,1). The ray isthen reflected from this point along the line,y+2x=1. Then the equation of the line of incidence of the ray of light is (A) 41x+38y-38=0 (B) 41x-38y+38=0( C) 41x+25y-25=0 (D) 41x-25y+25=0

A ray of light travels along the line 2x-3y+5=0 and strikes a plane mirror lying along the line x+y=2 . The equation of the straight line containing the reflected ray is

A circle has two of its diameters along the lines 2x + 3y - 18 =0 and 3x - y - 5 =0 and touches the line x + 2y + 4=0. Equation of the circle is

A ray of light along the line x- 2y + 5 = 0 is reflected from the line 3x - 2y + 7= 0 . Find the equation of the line containing the reflected ray .