What is the 4th coordinate for the given coordinates (-1,2),(3,6), and (-2,4) to become a parallelogram?

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What is the 4th coordinate for the given coordinates (-1,2),(3,6), and (-2,4) to become a parallelogram?

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Answer 1 · 2017-12-12T04:36:24

$H (0, 0)$ $H(0, 0)$ and $G (- 5, - 2)$ $G(-5, -2)$

Explanation:

Given intersection of the diagonals $D (- 1, 2) = D (x_{D}, y_{D})$ $D(-1, 2)=D(x_D, y_D)$
All that is needed is to solve for the endpoints of the diagonals.
The formula

$x_{D} = \frac{x_{E} + x_{G}}{2}$ $x_D=(x_E+x_G)/2$

$- 1 = \frac{3 + x_{G}}{2}$ $-1=(3+x_G)/2$

$- 2 = 3 + x_{G}$ $-2=3+x_G$

$x_{G} = - 5$ $x_G=-5$

Also

$y_{D} = \frac{y_{E} + y_{G}}{2}$ $y_D=(y_E+y_G)/2$

$2 = \frac{6 + y_{G}}{2}$ $2=(6+y_G)/2$

$4 = 6 + y_{G}$ $4=6+y_G$

$y_{G} = - 2$ $y_G=-2$

and therefore $G (- 5, - 2)$ $G(-5, -2)$

For the other point $H (x_{H}, y_{H})$ $H(x_H, y_H)$

$x_{D} = \frac{x_{F} + x_{H}}{2}$ $x_D=(x_F+x_H)/2$

$- 1 = \frac{- 2 + x_{H}}{2}$ $-1=(-2+x_H)/2$

$- 2 = - 2 + x_{H}$ $-2=-2+x_H$

$x_{H} = 0$ $x_H=0$

Also

$y_{D} = \frac{y_{F} + y_{H}}{2}$ $y_D=(y_F+y_H)/2$

$2 = \frac{4 + y_{H}}{2}$ $2=(4+y_H)/2$

$4 = 4 + y_{H}$ $4=4+y_H$

$y_{H} = 0$ $y_H=0$

and therefore $H (0, 0)$ $H(0, 0)$

God bless.....I hope the explanation is useful.

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What is the 4th coordinate for the given coordinates (-1,2),(3,6), and (-2,4) to become a parallelogram?

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