Precalculus Examples

$(- 5, 9)$ , $(3, 0)$

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

$x_{2} = 2 \cdot (3) - (- 5)$

Simplify the right side of $x_{2} = 2 \cdot (3) - (- 5)$ .

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Simplify each term.

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Multiply $2$ by $3$ .

$x_{2} = 2 (3) - (- 5)$

Multiply $2$ by $3$ .

$x_{2} = 6 - (- 5)$

Multiply $- 1$ by $- 5$ .

$x_{2} = 6 + 5$

Add $6$ and $5$ .

$x_{2} = 11$

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

$y_{2} = 2 \cdot (0) - (9)$

Simplify the right side of $y_{2} = 2 \cdot (0) - (9)$ .

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Simplify each term.

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Multiply $2$ by $0$ .

$y_{2} = 2 (0) - (9)$

Multiply $2$ by $0$ .

$y_{2} = 0 - (9)$

Multiply $- 1$ by $9$ .

$y_{2} = 0 - 9$

Subtract $9$ from $0$ .

$y_{2} = - 9$

The end point is $(11, - 9)$ .

$(11, - 9)$

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