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 Pure - Question 4 - 2019 - Paper 1

The midpoint M of points A and B can be calculated using the midpoint formula:

M = igg( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \bigg)

Substituting the coordinates of points A (-1, a) and B (3, 0.5), we have:

$M = \bigg( \frac{-1 + 3}{2}, \frac{a + 0.5}{2} \bigg)$

This simplifies to:

$M = \bigg( 1, \frac{a + 0.5}{2} \bigg)$

To find the gradient of line segment AB, we can rearrange the equation of line AB:

$5x + 4y = 17$ $4y = -5x + 17$ $y = -\frac{5}{4}x + \frac{17}{4}$

Thus, the gradient (m) of AB is -5/4. The gradient of the perpendicular bisector is the negative reciprocal:

$m_{perpendicular} = \frac{4}{5}$

Using point-slope form for the equation of a line, the equation for the perpendicular bisector passing through point M(1, \frac{a + 0.5}{2}) is:

$y - y_1 = m(x - x_1)$ Substituting the values:

$y - \frac{a + 0.5}{2} = \frac{4}{5}(x - 1)$

Rearranging gives:

$y = \frac{4}{5}x - \frac{4}{5} + \frac{a + 0.5}{2}$

This is the required equation of the perpendicular bisector.