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1 year ago

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

Mathematics

1 answer:

AveGali [126]1 year ago

4 0

Answer:

Question (a)

Midpoint of a line segment:

$M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

Given:

J = (-2, 2)
L = (4, 0)

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

Given:

M = (3, 7)
K = (-1, -5)

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

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.

$\textsf{slope }m=\dfrac{y_2-y_1}{x_2-x_1}$

Given:

J = (-2, 2)
L = (4, 0)

$\implies \textsf{slope of }JL=\dfrac{0-2}{4+2}=-\dfrac13$

Given:

M = (3, 7)
K = (-1, -5)

$\implies \textsf{slope of }MK=\dfrac{-5-7}{-1-3}=3$

$\textsf{slope of }JL \times \textsf{slope of }MK=-\dfrac13 \times3=-1$

Hence segments JL and MK are perpendicular

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