Hello, mathhomework!

Draw $\displaystyle \Delta LMN$ with vertices: .$\displaystyle L(2, -1),\;M(6, -2),\;N (4, 2)$

Find the coodinates of the vertices after a counterclockwise 90° rotation:

(a) about the origin. . (b) about $\displaystyle L.$. . {c} about $\displaystyle M.$ . . (d) about $\displaystyle N.$

Luckily for both us, there are 90° rotations; there's a simple routine.

Suppose we have a point $\displaystyle P(3,2)$ relative to a point $\displaystyle O$ Code:

P
* (3,2)
↑
↑ 2
↑
O * → → → → → *
3

Think of the slope of $\displaystyle OP\!:\;\;m_{_{OP}} \:=\:\frac{+2}{+3}$

. . That is, going from $\displaystyle O$ to $\displaystyle P$, we move: .$\displaystyle \frac{\text{ up 2}}{\text{right 3}} $

Let $\displaystyle P'$ be the new point after a 90° CCW rotation.

The slope of $\displaystyle PP'$ is perpendicular to $\displaystyle OP.$

. . Hence: .$\displaystyle m_{_{PP'}} \;=\;-\frac{3}{2} \quad\Rightarrow\quad \frac{-3}{2} \;=\;\frac{3}{-2}$

That is, going from $\displaystyle P$ to $\displaystyle P'$, we move: .$\displaystyle \frac{\text{down 3}}{\text{right 2}}\,\text{ or }\,\frac{\text{up 3}}{\text{left 2}} $

We know that $\displaystyle P'$ will be northwest of $\displaystyle P,$

. . so we use the second fraction: .$\displaystyle \frac{\text{up 3}}{\text{left 2}} $

From point $\displaystyle P$, we move: .left 2, up 3.

Code:

P'
* (1,5)
↑
↑
↑ 3
↑
↑ 2
* ← ← ← * (3,2)
:
: 2
:
O * - - - - - *
3

And we arrive at: .$\displaystyle P'(1,5)$

Do a few more of these and the steps will get smoother.

You'll come up with your own intuitive shortcuts.

Trust me on this . . .