The point A has coordinates (−1, a) and the point B has coordinates (3, b) The line AB has equation 5x + 4y = 17 Find the equation of the perpendicular bisector of the points A and B. - AQA - A-Level Maths Mechanics - Question 4 - 2019 - Paper 1

To find the midpoint (M) of points A(−1, a) and B(3, b), we use the midpoint formula:

M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{-1 + 3}{2}, \frac{a + b}{2} \right) = \left( 1, \frac{a + b}{2} \right).

The gradient of the perpendicular bisector is the negative reciprocal of the gradient of line AB. Therefore:

Gradient of the perpendicular bisector = -\frac{1}{(-\frac{5}{4})} = \frac{4}{5}.

Using the point-slope form of the equation of a line,

y - y_1 = m(x - x_1),

where (x_1, y_1) is the midpoint M(1, \frac{a + b}{2}) and m is the gradient we just found:

Substituting the values:

y - \frac{a + b}{2} = \frac{4}{5}(x - 1).

Multiplying through to rearrange:

y = \frac{4}{5}x - \frac{4}{5} + \frac{a + b}{2}.

This is the equation of the perpendicular bisector of points A and B.