Precalculus Examples

$(1, 2)$ , $(- 3, 8)$

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 $x_{2}$ to get $x_{2} = 2 x_{M} - x_{1}$ .

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

Substitute the values for $x_{M}$ and $x_{1}$ in $x_{2} = 2 (x_{M}) - (x_{1})$ to get $x_{2} = 2 \cdot (- 3) - (1)$ .

$x_{2} = 2 \cdot (- 3) - (1)$

Simplify the right side of $x_{2} = 2 \cdot (- 3) - (1)$ .

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Simplify each term.

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Multiply $2$ by $- 3$ to get $2 (- 3)$ .

$x_{2} = 2 (- 3) - (1)$

Multiply $2$ by $- 3$ to get $- 6$ .

$x_{2} = - 6 - (1)$

Multiply $- 1$ by $1$ to get $- 1$ .

$x_{2} = - 6 - 1$

Subtract $1$ from $- 6$ to get $- 7$ .

$x_{2} = - 7$

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

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

Solve for $y_{2}$ to get $y_{2} = 2 y_{M} - y_{1}$ .

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

Substitute the values for $y_{M}$ and $y_{1}$ in $y_{2} = 2 (y_{M}) - (y_{1})$ to get $y_{1} = 2 \cdot (8) - (2)$ .

$y_{2} = 2 \cdot (8) - (2)$

Simplify the right side of $y_{2} = 2 \cdot (8) - (2)$ .

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Simplify each term.

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Multiply $2$ by $8$ to get $2 (8)$ .

$y_{2} = 2 (8) - (2)$

Multiply $2$ by $8$ to get $16$ .

$y_{2} = 16 - (2)$

Multiply $- 1$ by $2$ to get $- 2$ .

$y_{2} = 16 - 2$

Subtract $2$ from $16$ to get $14$ .

$y_{2} = 14$

The end point is $(- 7, 14)$ .

$(- 7, 14)$

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