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

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 ​

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Yến Oanh 3 months 2021-08-15T07:59:03+00:00 1 Answers 5 views 0

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    2021-08-15T08:01:01+00:00

    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}

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