# Q12. Two points have coordinates A(1,3) and C(7,7). Find the equation of perpendicular bisector of AC. B is the point on the y-axis equidistant from A and C and ABCD is a rhombus. Find the coordinates of B and D. Show that the area of the rhombus is 52 units^2 and hence calculate the perpendicular distance of A from BC. ``

Points A and C are opposite on the same diagonal. The slope of the line that joins AC is

`m = (Y_C-Y_A)/(X_C-X_A) =(7-3)/(7-1) =4/6 =2/3`

The point O at half distance between A and C has the coordinates

`O( X_A + (X_C-X_A)/2, Y_A+(Y_C-Y_A)/2) =O (1 +(7-1)/2 ,3+(7-3)/2) =O(4, 5)`

The slope of line that is perpendicular to AC is

`m_1 =-1/m =-3/2`

The equation of the line that passes through point `O(4,5)` and has the slope `-3/2` is

`(y-Y_O) =m_1*(x-X_0)`

`y-5 =(-3/2)(x-4)`

`y-5 = -3/2*x +6`

`y = -3/2*x +11`    (1)

Point B is on the y axis, it means `X_B =0` . This condition combined with equation (1) gives

`Y_B =+11`

Thus the coordinates of B are `B(0,11)`

Now to find the coordinates of point D.

The horizontal distance between points B and O is

`Delta(X) = X_O- X_B =4-0 =4`

The vertical distance between points B and O is

`Delta(Y) = Y_O -Y_B =5-11 =-6`

Point D is diametrally opposed to point B with respect to center O thus,

`X_D = X_B +2*Delta(X) =0 +2*4 =8`

`Y_D = Y_B +2*Delta(Y) = 11-2*6 =-1`

Coordinates of point D are `D(8,-1)`

Finally to find the area or figure we need to find the length of diagonals:

`|AC|=sqrt[(7-1)^2+(7-3)^2] =7.21`

`|BD| =sqrt[(8-0)^2 +(-1-11)^2] =14.42`

Area of figure is just

`S = (|AC|*|BD|)/2 =(7.21*14.42)/2 =52`

Answer: the points B and D have the coordinates `B(0.11)` and `D(8,-1)`