Given: HN←→ and PN←→ are perpendicular to one another.

Question

JN and NQ¯ are horizontal segments.
HJ and PQ are vertical segments.
△HNJ and △PNQ are right triangles.
Prove: The slopes of HN←→ and PN←→ are opposite reciprocals.
Lines H N and P N intersect at point N which is in the first quadrant.
If you can show two similar triangles, you can use the proportional relationship between the side lengths of the triangles with the definitions of slope and opposite reciprocals to show that the slopes of the two perpendicular lines are opposite reciprocals.
Which pair of triangles can be proved similar to complete the proof?

A.) △HJN∼△PQN
B.) △NHJ∼△NPQ
C.) △NJH∼△NQP
D.) △HNJ∼△NPQ

Answer 1

The correct answer is A.) △HJN∼△PQN.

To prove that the slopes of HN←→ and PN←→ are opposite reciprocals, we need to show two similar triangles. In this case, we can show that △HJN is similar to △PQN.

Since HJ and PQ are vertical segments, they are parallel, and thus △HJN and △PQN are right triangles. We know that HN←→ and PN←→ are perpendicular to one another, so ∠JNH = 90° and ∠NQP = 90°.

Since HN←→ and PN←→ are perpendicular, the slopes of these lines are the negative reciprocals of each other.

By proving that △HJN is similar to △PQN, we can show that the sides of these triangles are proportional, which implies that the slopes of HN←→ and PN←→ are opposite reciprocals.