Since slope is calculated using the formula m = StartFraction v 2 minus v 1 Over x 2 minus x 1 EndFraction, the slope of both lines is equivalent to . It is given that the lines are parallel, and we calculated that the slopes are the same. Therefore, parallel lines have the same slopes.

The product of these slopes is [(d-b)/(c-a)] × [(c-a)/(-d+b)] = -1

Proved: This product shows that the slopes are negative reciprocals

Question:

The complete question as seen on ( question ID = 13270010):

Determine the missing information in the paragraph proof.

Given: Line PQ is rotated 90° counterclockwise to form line P’Q’. The lines are perpendicular. Line PQ contains the points (a, b) and (c, d). Line P’Q’ contains the points (–b, a) and (–d, c).

Prove: The slopes of perpendicular lines are negative reciprocals.

The slopes of lines PQ and P’Q’ can be determined using the formula m = StartFraction v 2 minus v 1 Over x 2 minus x 1 EndFraction

The product of these slopes is ________. This product shows that the slopes are negative reciprocals. It is given that the lines are perpendicular and we have shown that the slopes of the lines are negative reciprocals.

Step-by-step explanation:

Line PQ contains the points (a, b) and (c, d)

Line P’Q’ contains the points (–b, a) and (–d, c)

The two lines are said to be perpendicular.

For two lines are said to be perpendicular, the product of the slope of the two lines must be equal to -1.

Let's determine if it's true for this two lines.

Find attached the the diagram and explanations.

The product of these slopes is [(d-b)/(c-a)] × [(c-a)/(-d+b)] = -1

This product shows that the slopes are negative reciprocals

A

Step-by-step explanation:

just took the test

answer: x=60

answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

hope this ^_^