Find the direction in which a straight line must be drawn through the point (1,2) so that its point of intersection with the line x+y=4 may be at a distance of 2 3 from this point.

Step-by-step solution

Let the line pass through the point P(1, 2) at an angle with the positive x -axis. The equation of the line in parametric form is x-1 = y-2 = r , where r is the distance from P . Any point on this line is given by (1 + r , 2 + r ) . The intersection point lies on the line x + y = 4 . Substituting the coordinates of the point into this equation: (1 + r ) + (2 + r ) = 4 3 + r( + ) = 4 r( + ) = 1 Given r = 2 3 , we have 2 3 ( + ) = 1 . + = 3 2 = 6 2 . Multiplying by 1 2 on both sides: 1 2 + 1 2 = 6 2 2 = 3 2 . ( + 45^