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Quadrilateral JKLM is graphed with vertices at J(-2,2), K(-1,-5), L(4,0), and M(3,7).

A) Prove that the diagonals JL and MK have the same midpoint.
B) Prove that segments JL and MK are perpendicular.​

Quadrilateral JKLM Is Graphed With Vertices At J22 K15 L40 And M37A Prove That The Diagonals JL And MK Have The Same MidpointB Prove That Segments JL And MK Are class=

Sagot :

semsee45

Answer:

Question (a)

Midpoint of a line segment:

[tex]M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)[/tex]

Given:

  • J = (-2, 2)
  • L = (4, 0)

[tex]\implies \textsf{midpoint of }JL=\left(\dfrac{-2+4}{2},\dfrac{2+0}{2}\right)=(1,1)[/tex]

Given:

  • M = (3, 7)
  • K = (-1, -5)

[tex]\implies \textsf{midpoint of }MK=\left(\dfrac{3-1}{2},\dfrac{7-5}{2}\right)=(1, 1)[/tex]

Question (b)

Find slopes (gradients) of JL and MK then compare.  If the product of the slopes of JL and MK equal -1, then JL and MK are perpendicular.

[tex]\textsf{slope }m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

Given:

  • J = (-2, 2)
  • L = (4, 0)

[tex]\implies \textsf{slope of }JL=\dfrac{0-2}{4+2}=-\dfrac13[/tex]

Given:

  • M = (3, 7)
  • K = (-1, -5)

[tex]\implies \textsf{slope of }MK=\dfrac{-5-7}{-1-3}=3[/tex]

[tex]\textsf{slope of }JL \times \textsf{slope of }MK=-\dfrac13 \times3=-1[/tex]

Hence segments JL and MK are perpendicular

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