Quadrilateral BECK is known to be a rhombus. Two of the vertices are B(3,5) and C(7,-3). a. Find one slope of diagonal EK
b. Find an equation of line EK

a. The slope of EK is

b. The equation of line EK is y = x -  

Step-by-step explanation:

The form of the equation of a line is y = m x + b, where

The rule of the slope is m = , where

∵ BECK is a rhombus

∵ The diagonal is the line that joins two opposite vertices

∵ B and C are opposite vertices in the rhombus

∵ E and K are opposite vertices in the rhombus

∴ BC and EK are the diagonals of the rhombus BECK

∵ The diagonals of the rhombus are ⊥ and bisect each other

∴ EK is ⊥ bisector to BC

→ Let us find the slope and the mid-point of BC

∵ B = (3, 5) and C = (7, -3)

∴ x1 = 3 and y1 = 5

∴ x2 = 7 and y2 = -3

→ Substitute them in the rule of the slope above to find it

∵ m = = = -2

∴ m = -2

∴ The slope of BC = -2

→ To find the slope of EK reciprocal the slope of BC and change its sign

∴ m⊥ =

∴ The slope of EK =

a. The slope of EK is

→ Substitute the value of the slope in the form of the equation above

∵ y = x + b

→ To find b substitute x and y in the equation by the coordinates

   of a point on the line

∵ The mid-point of BC is the mid-point of EK

∵ The mid-point of BC = () = () = (5, 1)

∴ The mid-point of EK = (5, 1)

→ Substitute x by 5 and y by 2 in the equation

∵ 1 =  (5) + b

∴ 1 =   + b

→ Subtract   from both sides

∴   = b

→ Substitute the value of b in the equation

∵ y = x +

∴ y = x -  

b. The equation of line EK is y = x -  

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!