Which statement completes step 4 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 ?
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.

I think that it's one since in steep 2 you rotated it, and you have the two congruent triangles, I don't know what you did in step 3, but in step 4 I would tell the values of the points.

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JackelineCasarez JackelineCasarez · Answer 2 · 2022-01-18T18:50:10+01:00

The statement that most adequately accomplishes the proof of step 4 would be:

- JK lies on line f and has a slope of 3.

Given that,

A coordinate plane having line f located at y [tex]= 3x + 1[/tex]

and line g at y = [tex]-1 \frac{x}{3} +[/tex] [tex]1[/tex]

ΔJKL at J = [tex](-1, -2,)[/tex]

K [tex]= (0, 1)[/tex]

L [tex]= (0, -2)[/tex]

ΔJ'KL at J' = [tex]= (-3, 2)[/tex]

K [tex]= (0, 1)[/tex]

L' [tex]= (-3, 1)[/tex]

The one that would be for step 4 is

JK lying at line f with a slope of 3

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