The distance of the point (1,2) from the line x+y+5=0 measured along the line parallel to 3 x-y=7 is equal to

Step-by-step solution

The line passing through the point P(1, 2) and parallel to the line 3x - y = 7 has a slope m = 3 . The equation of this line is y - 2 = 3(x - 1) , which simplifies to y - 2 = 3x - 3 , or 3x - y - 1 = 0 . To find the intersection point Q of the lines x + y + 5 = 0 and 3x - y - 1 = 0 , we add the two equations: (x + y + 5) + (3x - y - 1) = 0 4x + 4 = 0 x = -1 . Substituting x = -1 into x + y + 5 = 0 , we get -1 + y + 5 = 0 , so y = -4 . Thus, the point of intersection is Q(-1, -4) . The distance between P(1, 2) and Q