Question

let $a,b$ be the points on the coordinate plane with coordinates $(t-4,-1)$ and $(-2,t 3)$, respectively. the square of the distance between the midpoint of $\overline{ab}$ and an endpoint of $\overline{ab}$ is equal to $t^2/2$. what is the value of $t$

Answer 1

The midpoint
`m` of
`\overline{ab}` has coordinates

`\left(\frac{(t-4) + (-2)}2, \frac{-1 + (t+3)}2\right) = \left(\frac{t-6}2, \frac{t+2}2\right)`

The distance from
`m` to
`a` is

`\sqrt{\left((t-4) - \frac{t-6}2\right)^2 + \left(-1 - \frac{t+2}2\right)^2} = \sqrt{\left(\frac{t-2}2\right)^2 + \left(-\frac{t+4}2\right)^2} = \sqrt{\frac{t^2}2+t+5}`

If this is equal to
`\frac{t^2}2`, then we solve for
`t`.

`\sqrt{\frac{t^2}2 + t + 5} = \frac{t^2}2`

`\left(\sqrt{\frac{t^2}2 + t + 5}\right)^2 = \left(\frac{t^2}2\right)^2`

`\frac{t^2}2 + t + 5 = \frac{t^4}4`

`\frac{t^4}4 - \frac{t^2}2 - t - 5 = 0`

`t^4 - 2t^2 - 4t - 20 = 0`

Use a calculator to solve the quartic; there are two real solutions for
`t` at
`t\approx-2.13` and
`t\approx2.57`.