Answer:
a. The slope of EK is [tex]\frac{1}{2}[/tex]
b. The equation of line EK is y = [tex]\frac{1}{2}[/tex] x - [tex]\frac{3}{2}[/tex]
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 = [tex]\frac{y2-y1}{x2-x1}[/tex] , 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 = [tex]\frac{-3-5}{7-3}[/tex] = [tex]\frac{-8}{4}[/tex] = -2
∴ m = -2
∴ The slope of BC = -2
→ To find the slope of EK reciprocal the slope of BC and change its sign
∴ m⊥ = [tex]\frac{1}{2}[/tex]
∴ The slope of EK = [tex]\frac{1}{2}[/tex]
a. The slope of EK is [tex]\frac{1}{2}[/tex]
→ Substitute the value of the slope in the form of the equation above
∵ y = [tex]\frac{1}{2}[/tex] 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 = ([tex]\frac{3+7}{2},\frac{5+-3}{2}[/tex]) = ([tex]\frac{10}{2},\frac{2}{2}[/tex]) = (5, 1)
∴ The mid-point of EK = (5, 1)
→ Substitute x by 5 and y by 2 in the equation
∵ 1 = [tex]\frac{1}{2}[/tex](5) + b
∴ 1 = [tex]\frac{5}{2}[/tex] + b
→ Subtract [tex]\frac{5}{2}[/tex] from both sides
∴ [tex]-\frac{3}{2}[/tex] = b
→ Substitute the value of b in the equation
∵ y = [tex]\frac{1}{2}[/tex] x + [tex]-\frac{3}{2}[/tex]
∴ y = [tex]\frac{1}{2}[/tex] x - [tex]\frac{3}{2}[/tex]
b. The equation of line EK is y = [tex]\frac{1}{2}[/tex] x - [tex]\frac{3}{2}[/tex]
