Triangle ABC has been rotated 90° to create triangle DEF. Using the image below, prove that lines BC and EF have the opposite and reciprocal slopes.

Proved: BC and EF have opposite and reciprocal slopes

Step-by-step explanation:

Two lines with slopes that are negative reciprocals of each other are said to perpendicular to each other.

Let m subscript 1 = m1

m subscript 2 = m2

So if m1 is the slope of the 1st (∆ABC)

m2 is the slope of the 2nd (∆DEF)

For both lines to be perpendicular,

m1×m2 = -1

So, the slope of the perpendicular line (DEF) is the negative reciprocal of the slope of the original line (ABC).

The diagram of the question can be found at .

Find attached the diagram.

From the diagram, the coordinates of the line segments BC and EF:

B(4,5) C(1,2)

E(5,-4) F(2,-1)

The slope of each of these lines:

m1 = (change in y)/(change in x)

= (2-5)/(1-4) = -3/-3

m1 = 1

m2 = (change in y)/(change in x)

= [-1-(-4)]/(2-5) = (-1+4)/(-3)

m2 = 3/-3

m2 = -1

m2 = negative reciprocal of m1

m2 = -1/1 = -1

From the solution, the slope of BC is the opposite reciprocal of EF.

Therefore, BC and EF have opposite and reciprocal slopes.

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

6mm^3 or .6cm^3

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6     6   6

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