4
directions - match the equations for perpendicular lines. (note: you will not use all choices on the right side).
$y = -\frac{1}{2}x - 2$
$y = \frac{2}{1}x + 2$
$y = \frac{1}{2}x + 4$
$y = \frac{1}{2}x - 4$
$y = 2x - 4$
$y = -\frac{1}{2}x + 6$
$y = 5x + 2$
$y = -\frac{1}{5}x + 3$
$y = \frac{1}{5}x + 10$
$y = -5x - 1$
drag & drop the answer
$y = 3x - 4$
$y = 5x - 2$
$y = -2x$

Question

Accepted Answer

# Explanation:
## Step1: Recall Perpendicular Slopes Rule
Two lines are perpendicular if the product of their slopes is $-1$ (i.e., $m_1 \times m_2 = -1$ or $m_2 = -\frac{1}{m_1}$ where $m_1, m_2$ are slopes).

## Step2: Analyze $y = -\frac{1}{2}x - 2$
Slope $m_1 = -\frac{1}{2}$. Find $m_2$ such that $m_1 \times m_2 = -1$.
$-\frac{1}{2} \times m_2 = -1 \implies m_2 = 2$. Check right/drop options: $y = 2x - 4$ (slope $2$)? Wait, no—wait, the right-side initial options? Wait, the drag-drop options: $y = 3x - 4$, $y = 5x - 2$, $y = -2x$? Wait, no, original left and right boxes. Wait, let's re-express:

1. $y = -\frac{1}{2}x - 2$: slope $m = -\frac{1}{2}$. Perpendicular slope is $2$ (since $-\frac{1}{2} \times 2 = -1$). But in right-side initial, no $2x$ except $y = 2x - 4$ (left) and drag-drop $y = -2x$? Wait, no—wait, the problem's left equations:

- $y = -\frac{1}{2}x - 2$ (slope $-\frac{1}{2}$): perpendicular slope is $2$. But drag-drop has $y = -2x$? No, wait, $ -\frac{1}{2} \times 2 = -1$, but $y = -2x$ has slope $-2$, which is not. Wait, maybe I misread. Wait, the left equations:

1. $y = -\frac{1}{2}x - 2$: slope $m = -\frac{1}{2}$. Perpendicular slope is $2$ (since reciprocal and opposite sign). But in the drag-drop options, $y = 2x - 4$ is on the left? No, left has $y = 2x - 4$ (slope $2$). Wait, no—let's list all left slopes:

- $y = -\frac{1}{2}x - 2$: $m = -\frac{1}{2}$
- $y = \frac{1}{2}x + 4$: $m = \frac{1}{2}$
- $y = 2x - 4$: $m = 2$
- $y = 5x + 2$: $m = 5$
- $y = \frac{1}{5}x + 10$: $m = \frac{1}{5}$

Right-side initial (before drag-drop? No, the right-side boxes and drag-drop):

Wait, the drag-drop options are $y = 3x - 4$, $y = 5x - 2$, $y = -2x$. Wait, maybe the original matching is wrong, and we need to use drag-drop. Let's redo:

1. $y = -\frac{1}{2}x - 2$: slope $-\frac{1}{2}$. Perpendicular slope is $2$ (since $-\frac{1}{2} \times 2 = -1$). But drag-drop has no $2x$, but $y = -2x$ has slope $-2$ (no). Wait, no—wait, $ -\frac{1}{2} \times 2 = -1$, but maybe I made a mistake. Wait, $y = \frac{1}{2}x + 4$: slope $\frac{1}{2}$. Perpendicular slope is $-2$ (since $\frac{1}{2} \times (-2) = -1$). Ah! There we go. So $y = \frac{1}{2}x + 4$ (slope $\frac{1}{2}$) should match $y = -2x$ (slope $-2$), because $\frac{1}{2} \times (-2) = -1$.

2. $y = \frac{1}{2}x + 4$: slope $m = \frac{1}{2}$. Perpendicular slope $m_2 = -2$ (since $\frac{1}{2} \times (-2) = -1$). Drag-drop option $y = -2x$ has slope $-2$. So this matches.

3. $y = 2x - 4$: slope $m = 2$. Perpendicular slope $m_2 = -\frac{1}{2}$ (since $2 \times (-\frac{1}{2}) = -1$). Check right-side: $y = -\frac{1}{2}x + 6$ (slope $-\frac{1}{2}$)—yes, that's a match (left $y = 2x - 4$ and right $y = -\frac{1}{2}x + 6$ have slopes $2$ and $-\frac{1}{2}$, product $-1$).

4. $y = 5x + 2$: slope $m = 5$. Perpendicular slope $m_2 = -\frac{1}{5}$ (since $5 \times (-\frac{1}{5}) = -1$). Right-side: $y = -\frac{1}{5}x + 3$ (slope $-\frac{1}{5}$)—match.

5. $y = \frac{1}{5}x + 10$: slope $m = \frac{1}{5}$. Perpendicular slope $m_2 = -5$ (since $\frac{1}{5} \times (-5) = -1$). Drag-drop option $y = 5x - 2$? No, $y = -5x - 1$ (right-side) has slope $-5$? Wait, right-side: $y = -5x - 1$ (slope $-5$)—yes! $ \frac{1}{5} \times (-5) = -1$. So $y = \frac{1}{5}x + 10$ matches $y = -5x - 1$.

Now, the drag-drop options: $y = 3x - 4$, $y = 5x - 2$, $y = -2x$. Let's see which left equation needs a drag-drop:

- $y = -\frac{1}{2}x - 2$ (slope $-\frac{1}{2}$): perpendicular slope $2$. But drag-drop has no $2x$, but wait—wait, $y = \frac{1}{2}x + 4$ (slope $\frac{1}{2}$) needs perpendicular slope $-2$, which is $y = -2x$ (drag-drop). So:

- $y = -\frac{1}{2}x - 2$: no, wait, $y = \frac{1}{2}x + 4$ (slope $\frac{1}{2}$) → perpendicular slope $-2$ (drag-drop $y = -2x$).

- $y = \frac{1}{2}x + 4$ → drag-drop $y = -2x$ (slope $-2$).

- $y = 2x - 4$ → right-side $y = -\frac{1}{2}x + 6$ (slope $-\frac{1}{2}$).

- $y = 5x + 2$ → right-side $y = -\frac{1}{5}x + 3$ (slope $-\frac{1}{5}$).

- $y = \frac{1}{5}x + 10$ → right-side $y = -5x - 1$ (slope $-5$).

Wait, the first left equation: $y = -\frac{1}{2}x - 2$ (slope $-\frac{1}{2}$): perpendicular slope $2$. But drag-drop has $y = 3x - 4$ (slope $3$)—no, $y = 5x - 2$ (slope $5$)—no, $y = -2x$ (slope $-2$)—no. Wait, maybe I messed up. Wait, the problem says "you will not use all choices on the right side" (drag-drop). Let's re-express each left equation:

1. $y = -\frac{1}{2}x - 2$ (slope $m = -\frac{1}{2}$):
   Perpendicular slope $m_2 = 2$ (since $-\frac{1}{2} \times 2 = -1$). But in drag-drop, no $2x$, but maybe the right-side initial has no, so drag-drop $y = -2x$ is for another? Wait, no—$y = \frac{1}{2}x + 4$ (slope $\frac{1}{2}$): perpendicular slope is $-2$ (since $\frac{1}{2} \times (-2) = -1$), so $y = -2x$ (drag-drop) has slope $-2$, so that's the match for $y = \frac{1}{2}x + 4$.

2. $y = \frac{1}{2}x + 4$ → drag-drop $y = -2x$ (slope $-2$).

3. $y = 2x - 4$ → right-side $y = -\frac{1}{2}x + 6$ (slope $-\frac{1}{2}$): match (product $-1$).

4. $y = 5x + 2$ → right-side $y = -\frac{1}{5}x + 3$ (slope $-\frac{1}{5}$): match (product $-1$).

5. $y = \frac{1}{5}x + 10$ → right-side $y = -5x - 1$ (slope $-5$): match (product $-1$).

Now, the first left equation $y = -\frac{1}{2}x - 2$ (slope $-\frac{1}{2}$): perpendicular slope is $2$, but drag-drop has no $2x$, but maybe the right-side initial has a typo? Wait, no—wait, the drag-drop options are $y = 3x - 4$, $y = 5x - 2$, $y = -2x$. Wait, $y = -\frac{1}{2}x - 2$ (slope $-\frac{1}{2}$): perpendicular slope is $2$, but none of drag-drop have slope $2$. Wait, maybe I made a mistake. Wait, $ -\frac{1}{2} \times 2 = -1$, but $y = -2x$ has slope $-2$, which is not. Wait, no—$y = \frac{1}{2}x + 4$ (slope $\frac{1}{2}$): perpendicular slope is $-2$ (since $\frac{1}{2} \times (-2) = -1$), so $y = -2x$ (drag-drop) is correct for $y = \frac{1}{2}x + 4$.

So the matches are:

- $y = -\frac{1}{2}x - 2$: no drag-drop (but maybe the problem's initial right-side is wrong, but the drag-drop has $y = -2x$ for $y = \frac{1}{2}x + 4$).

- $y = \frac{1}{2}x + 4$ → drag-drop $y = -2x$ (slope $-2$).

- $y = 2x - 4$ → $y = -\frac{1}{2}x + 6$ (right-side).

- $y = 5x + 2$ → $y = -\frac{1}{5}x + 3$ (right-side).

- $y = \frac{1}{5}x + 10$ → $y = -5x - 1$ (right-side).

So the drag-drop for $y = \frac{1}{2}x + 4$ is $y = -2x$.
# Answer:
For $y = \frac{1}{2}x + 4$, the perpendicular line is $y = -2x$ (drag-drop option).
Other matches:
- $y = -\frac{1}{2}x - 2$ → No drag-drop (or maybe error, but per slope rule, needs $2x$ slope, not in drag-drop).
- $y = 2x - 4$ → $y = -\frac{1}{2}x + 6$
- $y = 5x + 2$ → $y = -\frac{1}{5}x + 3$
- $y = \frac{1}{5}x + 10$ → $y = -5x - 1$

(But focusing on drag-drop: $y = \frac{1}{2}x + 4$ matches $y = -2x$ (drag-drop) as their slopes $\frac{1}{2}$ and $-2$ multiply to $-1$.)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Recall Perpendicular Slopes Rule

Step2: Analyze \(y = -\frac{1}{2}x - 2\)

Answer: