Question

Which statement completes step 6 of the proof? Coordinate plane with line f at y equals 3 times x plus 1 and line g at y equals negative one third times x plus 1. Triangle JKL is at J negative 1 comma negative 2, K 0 comma 1, and L 0 comma negative 2. Triangle J prime K L prime is at J prime negative 3 comma 2, K 0 comma 1, and L prime negative 3 comma 1. Step 1 segment KL is parallel to the y-axis, and segment JL is parallel to the x-axis. Step 2 ΔJKL was rotated 90° clockwise to create ΔJ'KL'. Point K did not change position, so it remains point K. Therefore, ΔJKL ≅ ΔJ'KL'. Step 3 segment K L prime is perpendicular to the y-axis, and segment J prime L prime is perpendicular to the x-axis. Step 4 segment JK lies on line f and has a slope of 3. Step 5 segment J prime K lies on line g and has a slope of negative one third. Step 6 ? The product of the slopes of segment JK and segment J prime K is −1; therefore, lines f and g are perpendicular. The slopes of segment JK and segment J prime K are congruent; therefore, lines f and g are parallel. The product of the slopes of segment KL and segment K L prime is −1; therefore, lines f and g are perpendicular. The slopes of segment KL and segment K L prime are congruent; therefore, lines f and g are parallel.

Answer 1

Answer:

"The product of the slopes of segment JK and segment J prime K is −1; therefore, lines f and g are perpendicular."

Step-by-step explanation:

it literally says it on my chart because i have to find step 5 not 6.

Step 1 segment KL is parallel to the y-axis, and segment JL is parallel to the x-axis.

Step 2 ΔJKL was rotated 90° clockwise to create ΔJ'KL'. Point K did not change position, so it remains point K. Therefore, ΔJKL ≅ ΔJ'KL'.

Step 3 segment K L prime is perpendicular to the y-axis, and segment J prime L prime is perpendicular to the x-axis.

Step 4 segment JK lies on line f and has a slope of 3.

Step 5 ?

Step 6 The product of the slopes of segment JK and segment J prime K is −1; therefore, lines f and g are perpendicular.