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