Two points have coordinates A (2,3) and B (6,7). If C(7,t) lies on the perpendicular bisector

of AB, find the value of t. Find the coordinates of D such that the line AB is the perpendicular bisector of CD.

A (2 ; 3) B (6 ; 7) C (7 ; t) lies on the perpendicular bisector


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) = (7 - 3) / (6 - 2) = 4/4 = 1



You know that 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 (1), so the slope of the bisector is : - 1

The equation of the bisector becomes: y = - x + b


You know that the bisector passes through the middle of [AB], so the coordinates of the point M must verify the equation of the bisector.

The middle of [AB] is the point M:

xM = (xA + xB)/2 = (2 + 6)/2 = 4

yM = (yA + yB)/2 = (3 + 7)/2 = 5

→ M (4 ; 5)

…recall that the bisector passes through this point

y = - x + b

b = y + x → you substitute x and y by the coordinates of the point M

b = 5 + 4 = 9

→ The equation of the bisector is: y = - x + 9


But you know that the point C belongs to the bisector, so the coordinates of this point C must verify the equation of the bisector.

y = - x + 9 → you substitute x by the abscissa of the point C, i.e. 7

y = - 7 + 9

y = 2 → this is the ordinate of the point C

→ t = 2

→ C (7 ; 2)



If the line (AB) is perpendicular to the bisector of the line (CD).

The point D belongs to the line (MC), and if you make a drawing, you can deduce that the point M is the middle point of [CD]

xM = (xC + xD)/2 = (7 + xD)/2 = 4 → 7 + xD = 8 → xD = 8 - 7 → xD = 1

yM = (yC + yD)/2 = (2 + yD)/2 = 5 → 2 + yD = 10 → yD = 10 - 2 → yD = 8