Question

is $overrightarrow{ab}$ perpendicular to $overrightarrow{cd}$? explain
yes, because the slope of $overrightarrow{ab}$ is $\frac{5}{3}$ and the slope of $overrightarrow{cd}$ is $-\frac{3}{5}$
yes, because the slope of $overrightarrow{ab}$ is $\frac{4}{3}$ and the slope of $overrightarrow{cd}$ is $-\frac{3}{4}$
no, because the slope of $overrightarrow{ab}$ is $\frac{5}{3}$ and the slope of $overrightarrow{cd}$ is $-\frac{3}{4}$
no, because the slope of $overrightarrow{ab}$ is $\frac{4}{3}$ and the slope of $overrightarrow{cd}$ is $-\frac{3}{5}$

Explanation:

Step1: Calculate slope of $\overrightarrow{AB}$

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. For points $A(0,2)$ and $B(-3,-3)$, we have $m_{AB}=\frac{-3 - 2}{-3-0}=\frac{-5}{-3}=\frac{5}{3}$.

Step2: Calculate slope of $\overrightarrow{CD}$

For points $C(-4,1)$ and $D(0,-2)$, we have $m_{CD}=\frac{-2 - 1}{0+4}=\frac{-3}{4}$.

Step3: Check perpendicular - condition

Two lines are perpendicular if the product of their slopes is $- 1$. Here, $m_{AB}\times m_{CD}=\frac{5}{3}\times(-\frac{3}{4})=-\frac{5}{4}
eq - 1$. So they are not perpendicular.

Answer:

No, because the slope of $\overrightarrow{AB}$ is $\frac{5}{3}$ and the slope of $\overrightarrow{CD}$ is $-\frac{3}{4}$