Basic Analytic Geometry here!!!!!

Show that (5,2) is on the perpendicular bisector of the segment AB where A=(1,3) and B=(4, -2)
please help me :(

Slope of AB = -2-3/4-1 = -5/3
Perpendicular Slope = 3/5
Mid point of AB ( 5/2, 1/2)
Equation of Perpendicular Bisector: y- 1/2 = 3/5(x - 5/2)
(2y-1)/2 = 3/5(2x-5)/2
2y-1 = 3/5(2x-5)
10y-5 = 6x-15 Sun\stitute (5,2)
20-5 = 30-15
15=15
Hence the point (5,2) is on the perpendicular bisector of AB

A (1 ; 3) B (4 ; - 2)


The typical equation of a line is: y = mx + b → where m: slope and where b: y-intercept

For the line (AB), the typical equation is: y = mx + b, where the slope is:

m = (yB - yA) / (xB - xA) = (- 2 - 3) / (4 - 1) = - 5/3

The bisector of the line (AB) is perpendicular to the line (AB)

Two lines are perpendicular if the product of their slope is - 1.

The slope of the line (AB) is (- 5/3), so the slope of the bisector is : 3/5

The equation of the bisector becomes: y = (3/5)x + b


The bisector passes through the middle of [AB]. The middle of [AB] is the point M:

xM = (xA + xB)/2 = (1 + 4)/2 = 5/2

yM = (yA + yB)/2 = (3 - 2)/2 = 1/2

→ M (5/2 ; 1/2)

…recall that the bisector passes through the point M → the coordinates must verify the equation

y = (3/5)x + b

b = y - (3/5)x → you substitute x and y by the coordinates of the point M

b = (1/2) - [(3/5) * (5/2)]

b = (1/2) - (3/2) = - 1


→ The equation of the bisector is: y = (3/5)x - 1


Let's look the point (5 ; 2). If this point is located on the perpendicular bisector of the segment AB, that means its coordinates verify the equation of the bisector.


y = (3/5)x - 1 → substitute x by 5

y = [(3/5) * 5] - 1

y = 3 - 1

y = 2 → this the ordinates of the point


The coordinates of the point (5 ; 2) verify the equation of the bisector, so you can say that this point belongs to the bisector.

Let the given point be called P(5, 2).

PA = √[(5 - 1)² + (2 - 3)²] = √(17)

PB = √[(5 - 4)² + (2 + 2)²] = √(17)

PA = PB
Therefore, P is on the perpendicular bisector of AB.