Precalculus Examples

$(0, 0)$ , $(- 6, 6)$

The midpoint formula is $M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ .

$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Set the $x_{M}$ equal to the $x_{M}$ coordinate.

$x_{M} = \frac{x_{1} + x_{2}}{2}$

Solve for $x_{2}$ .

$x_{2} = 2 (x_{M}) - (x_{1})$

Substitute the values for $x_{M}$ and $x_{1}$ in $x_{2} = 2 (x_{M}) - (x_{1})$ .

$x_{2} = 2 \cdot (- 6) - (0)$

Simplify the right side of $x_{2} = 2 \cdot (- 6) - (0)$ .

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Multiply $2$ by $- 6$ .

$x_{2} = 2 (- 6) - (0)$

Multiply $2$ by $- 6$ .

$x_{2} = - 12 - (0)$

Subtract $0$ from $- 12$ .

$x_{2} = - 12$

Set the $y_{M}$ equal to the $y_{M}$ coordinate.

$y_{M} = \frac{y_{1} + y_{M}}{2}$

Solve for $y_{2}$ .

$y_{2} = 2 (y_{M}) - (y_{1})$

Substitute the values for $y_{M}$ and $y_{1}$ in $y_{2} = 2 (y_{M}) - (y_{1})$ .

$y_{2} = 2 \cdot (6) - (0)$

Simplify the right side of $y_{2} = 2 \cdot (6) - (0)$ .

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Multiply $2$ by $6$ .

$y_{2} = 2 (6) - (0)$

Multiply $2$ by $6$ .

$y_{2} = 12 - (0)$

Subtract $0$ from $12$ .

$y_{2} = 12$

The end point is $(- 12, 12)$ .

$(- 12, 12)$

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