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PLEASE HELP! The vertices of square LMNO are L(0, 2), M(2, 0), N(0,−2), and O(−2, 0). Which of the following shows that its diagonals are congruent perpendicular bisectors of each other? I ATTACHED THE IMAGES.

PLEASE HELP The Vertices Of Square LMNO Are L0 2 M2 0 N02 And O2 0 Which Of The Following Shows That Its Diagonals Are Congruent Perpendicular Bisectors Of Each class=
PLEASE HELP The Vertices Of Square LMNO Are L0 2 M2 0 N02 And O2 0 Which Of The Following Shows That Its Diagonals Are Congruent Perpendicular Bisectors Of Each class=

Sagot :

Answer:

The fourth one is correct.

Step-by-step explanation:

To show the diagonals are congruent, calculate their lengths using the Distance Formula: d=(x2−x1)2+(y2−y1)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√.

Find the length of LN.

Substitute 0 for x1, 2 for y1, 0 for x2 and −2 for y2.

LN=(0−0)2+(−2−2)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

=02+(−4)2‾‾‾‾‾‾‾‾‾‾√

=8‾√

Find the length of MO.

Substitute 2 for x1, 0 for y1, −2 for x2 and 0 for y2.

MO=(−2−2)2+(0−0)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√

=(−4)2+02‾‾‾‾‾‾‾‾‾‾√

=8‾√

Since LN=MO, by the definition of congruent line segments, LN=MO.

To show the diagonals are perpendicular, calculate their slopes using the Slope Formula: m=y2−y1x2−x1.

Find the slope of LN.

Substitute 0 for x1, 2 for y1, 0 for x2 and −2 for y2.

m1=−2−20−0=−40⇒m1 cannot be defined.

Find the slope of MO.

Substitute 2 for x1, 0 for y1, −2 for x2 and 0 for y2.

m2=0−0−2−2

=0−4

=0

The slope of LN cannot be defined, this means that LN is parallel to y-axis. The slope of MO is equal to 0, this means that MO is parallel to x-axis. x and y axis are perpendicular, therefore LN⊥MO.

The figure shows the same square L M N O on a Cartesian plane as in the beginning of the task. Diagonal L N is perpendicular to diagonal M O.

To show the diagonals bisect each other, calculate their midpoints using the Midpoint Formula: P(x1+x22,y1+y22).

Find the midpoint of LN.

Substitute 0 for x1, 2 for y1, 0 for x2 and −2 for y2.

P1(0+02,2+(−2)2)=(0, 0)

Find the midpoint of MO.

Substitute 2 for x1, 0 for y1, −2 for x2 and 0 for y2.

P2(2+(−2)2,0+02)=(0, 0)

Both midpoints are the same point on the coordinate plane. This means the two diagonals intersect at their midpoints, or bisect each other.

The figure shows the same square L M N O on a Cartesian plane as in the beginning of the task. The diagonals intersect at point P. Segments L P, P N, M P, and P O are congruent.

Therefore, the diagonals are congruent perpendicular bisectors of each other.