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

Let the equation of the line parallel to 3 x - y = 7 be 3 x - y = λ . Since, it passes through 1 , 2 , this point will satisfy the equation 3 x - y = λ . ∴   3 - 2 = λ ⇒ λ = 1 So, the line is 3 x - y = 1 . Now, x + y + 5 + 3 x - y = 0 + 1 ⇒ 4 x = - 4 ⇒ x = - 1 . Putting in x + y + 5 = 0 , we get y = - x - 5 = 1 - 5 = - 4 So, the point of intersection of x + y + 5 = 0 and 3 x - y = 1 is - 1 , - 4 . The required distance is the distance between 1 , 2 and - 1 , - 4 . By