Question

Quadrilateral BECK is known to be a rhombus. Two of the vertices are B(3,5) and C(7,-3).

Answer 1

Answer:

a. The slope of EK is $\frac{1}{2}$

b. The equation of line EK is y = $\frac{1}{2}$ x -  $\frac{3}{2}$

Step-by-step explanation:

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

• m is the slope of the line

• b is the y-intercept

The rule of the slope is m = $\frac{y2-y1}{x2-x1}$ , where

• (x1, y1) and (x2, y2) are two points on the line

• The rule of the mid-point is ($\frac{x1+x2}{2},\frac{y1+y2}{2}$)

∵ 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 = $\frac{-3-5}{7-3}$ = $\frac{-8}{4}$ = -2

∴ m = -2

∴ The slope of BC = -2

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

∴ m⊥ = $\frac{1}{2}$

∴ The slope of EK = $\frac{1}{2}$

a. The slope of EK is $\frac{1}{2}$

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

∵ y = $\frac{1}{2}$ 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 = ($\frac{3+7}{2},\frac{5+-3}{2}$) = ($\frac{10}{2},\frac{2}{2}$) = (5, 1)

∴ The mid-point of EK = (5, 1)

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

∵ 1 =  $\frac{1}{2}$(5) + b

∴ 1 =  $\frac{5}{2}$ + b

→ Subtract  $\frac{5}{2}$ from both sides

∴  $-\frac{3}{2}$ = b

→ Substitute the value of b in the equation

∵ y = $\frac{1}{2}$ x + $-\frac{3}{2}$

∴ y = $\frac{1}{2}$ x -  $\frac{3}{2}$

b. The equation of line EK is y = $\frac{1}{2}$ x -  $\frac{3}{2}$