Jefferson is plotting the vertices of an isosceles right triangle on a coordinate graph. He has plotted the two points shown. Where sh

Question

Jefferson is plotting the vertices of an isosceles right triangle on a coordinate graph. He has
plotted the two points shown. Where should he plot the third point?
I (-2,0)
(-2,-9)
(3,1)
(4,2)

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Josephine 1 week 2021-09-09T08:33:06+00:00 1 Answer 0

Answers ( )

    0
    2021-09-09T08:34:11+00:00

    Answer:

    Option (4).

    Step-by-step explanation:

    Given question is incomplete; here is the complete question.

    Jefferson is plotting the vertices of an isosceles right triangle on a coordinate graph. He has plotted the two points shown. Where should he plot the third point?

    The point is on the graph is (-2,-4) (4,-4)


    The options are


    (-2,-9)

    (3,1)

    (-2,0)

    (4,2)

    Distance between (-2, -4) and (4, -4) = 6 units

    Now we will check every option,

    1). For point (-2, -9),

    Distance between (-2, -4) and (-2, -9) = \sqrt{(-2+2)^2+(-4+9)^2}=5

    Similarly, distance between (4, -4) and (-2, -9) = \sqrt{(4+2)^2+(-4+9)^2}=\sqrt{61}

    None side is equal. So it’s not an isosceles triangle.

    2). For a point (3, 1),

    Distance between (3, 1) and (-2, -4) = \sqrt{(3+2)^2+(1+4)^2}=5\sqrt{2}

    Distance between (3, 1) and (4, -4) = \sqrt{(3-4)^2+(1+4)^2}=\sqrt{26}

    Since none side of the triangle are not equal, triangle is not an isosceles triangle.

    3). For (-2,0),

    Distance between (-2, 0) and (-2, -4) = 4

    Distance between (-2, 0) and (4, -4) = \sqrt{(-2-4)^2+(0+4)^2} = 2\sqrt{13}

    None side is equal so the triangle formed is not an isosceles triangle.

    4). For a point (4, 2),

    Distance between (4, 2) and (-2, -4) = \sqrt{(4+2)^2+(2+4)^2}=6\sqrt{2}

    Distance between (4, 2) and (4, -4) = \sqrt{(4-4)^2+(2+4)^2}=6

    We find the two sides of this triangle measure 6 units therefore, its an isosceles triangle.

    Option (4) will be the third point.

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