A is the point (4,1). B is the point (10,15). Find the perpendicular bisector or of AB.

First, find the gradient, m, of the line AB. Gradient is given by: change in y / change in xChange in y = 15-1 = 14Change in x = 10-4 = 6Gradient = 14/6 = 7/3The perpendicular bisector is a straight line normal to the line AB.Perpendicular Gradient, m_p = - 1/m Therefore m_p = -3/7We know the perpendicular bisector is a straight line which has the general equation: y = mx + c.Where m is the gradient and c is the constant (y-intercept). We have found m, which in our case has been denoted as m_p = -3/7 Substitute into y = mx + cy = -(3/7)x + cTo find c, we need to find a point that the perpendicular bisector goes through. This point would be the mid point of the line.Midpoint in x = (4+10)/2 = 7Midpoint in y = (1+15)/2 = 8So point at which perpendicular bisector ‘bisects’ AB is: (7,8) <- Midpoint of ABUsing this midpoint, let x = 7 and y = 8 in our equation y = -(3/7)x + c...8 = -(3/7)*7 + cc = 11So we can conclude that the perpendicular bisector of the line AB is: y = -(3/7)x + 11