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The endpoints of one diagonal of a rhombus are (-7, -2) and (-1, -4). If the coordinates of the 3rd vertex are (-6, -9), what are the coordinates of the 4th vertex? (-8, 13) (-2, 3) (-6, 13) (7, -8)

Sagot :

The coordinates of the 4th vertex of the rhombus are (-2,3)

How to determine the coordinates of the 4th vertex?

The diagonals are given as:

(-7, -2) and (-1, -4)

The 3rd vertex is given as: (-6,-9)

Calculate the distance between the vertices of the diagonals and the 3rd vertex using:

[tex]d = \sqrt{(x_2 -x_1)^2 + (y_2 - y_1)^2}[/tex]

So, we have:

[tex]d_1 = \sqrt{(-7 + 6)^2 + (-2 + 9)^2} =\sqrt{50[/tex]

[tex]d_2 = \sqrt{(-1 + 6)^2 + (-4 + 9)^2} =\sqrt{50[/tex]

Let the 4th vertex be (x,y)

So, we have:

[tex]d_3 = \sqrt{(-7 - x)^2 + (-2 - y)^2} =\sqrt{50[/tex]

[tex]d_4 = \sqrt{(-1 - x)^2 + (-4 - y)^2} =\sqrt{50[/tex]

Equate d3 and d4

[tex]\sqrt{(-1 - x)^2 + (-4 - y)^2} = \sqrt{(-7 - x)^2 + (-2 - y)^2}[/tex]

Take the square of both sides

[tex](-1 - x)^2 + (-4 - y)^2 = (-7 - x)^2 + (-2 - y)^2[/tex]

Expand

[tex]1 + 2x + x^2 + 16 + 8y + y^2 = 49 + 14x + x^2 + 4 + 4y + y^2[/tex]

Evaluate the like terms

1 + 2x  + 16 + 8y  = 49 + 14x + 4 + 4y

Collect like terms

14x - 2x + 4y - 8y = 1 + 16 - 49 - 4

12x - 4y = -36

Divide through by 4

3x - y = -9

Next, we test the options in the above equation

Point (-8,13) means x = -8 and y = 13

So, we have:

3x - y = -9

3(-8) - 13 = -9

-37 = -9 --- this is false

Point (-2, 3) means x = -2 and y = 3

So, we have:

3x - y = -9

3(-2) - 3 = -9

-9 = -9 --- this is true.

Hence, the coordinates of the 4th vertex are (-2,3)

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