Lines, Slopes, and Perpendicularity: Practice

AA similarity fixes the rise-to-run ratio. Let the second signed rise be $h$.

The negative sign agrees with a fall to the right.

The given slope is $-3/5$, so the required slope is $5/3$.

The point satisfies the equation and the slopes multiply to $-1$.

Its midpoint is $(-2,3)$ and its segment slope is $1/2$. The perpendicular direction therefore has slope $-2$.

The midpoint is $(4,-3)$. A perpendicular to the vertical segment is horizontal, so it fixes the midpoint's second coordinate.

The midpoint of $AB$ is $(2,1)$. The median slope is $7/4$, so

Both $(2,1)$ and $C=(6,8)$ satisfy the result.

The opposite side has slope $1/2$. The altitude has slope $-2$ and contains $A$.

A common point must give the same height in both equations.

Substitution of $(3,-3)$ works in both lines.

The slopes are $-k/4$ and $-3/2$. Equating them gives

Their intercepts are $9/4$ and $1/2$, so they are distinct.

The perpendicular direction has slope $-1/3$. Through $B$,

The slope and point conditions both hold.

Reflect the endpoint's coordinates across the midpoint coordinatewise.

The resulting segment slope is $-1/2$ and the given bisector slope is 2, checking perpendicularity.

The rightward run becomes rise 8, and the downward rise becomes run 6 right.

The original slope is $-3/4$, the negative reciprocal of this value.

The midpoint of $BC$ is $(2,1)$, so the line joining it to $A$ is vertical. The side $BC$ is horizontal, so the altitude from $A$ is the same vertical line.

The two definitions can select one line in this configuration.