Question

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 ​

Answer 1

a. The slope of EK is
`(1)/(2)`

b. The equation of line EK is y =
`(1)/(2)` x -
`(3)/(2)`

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 =
`(y2-y1)/(x2-x1)` , where

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

• The rule of the mid-point is (
  `(x1+x2)/(2),(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 =
`(-3-5)/(7-3)` =
`(-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⊥ =
`(1)/(2)`

∴ The slope of EK =
`(1)/(2)`

a. The slope of EK is
`(1)/(2)`

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

∵ y =
`(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 = (
`(3+7)/(2),(5+-3)/(2)`) = (
`(10)/(2),(2)/(2)`) = (5, 1)

∴ The mid-point of EK = (5, 1)

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

∵ 1 =
`(1)/(2)`(5) + b

∴ 1 =
`(5)/(2)` + b

→ Subtract
`(5)/(2)` from both sides

∴
`-(3)/(2)` = b

→ Substitute the value of b in the equation

∵ y =
`(1)/(2)` x +
`-(3)/(2)`

∴ y =
`(1)/(2)` x -
`(3)/(2)`

b. The equation of line EK is y =
`(1)/(2)` x -
`(3)/(2)`