To find the equation of the line $JK$, perpendicular to the given line $GH: y + 2x + 14 = 0$, follow these steps:

Step 1: Rewrite the given line in slope-intercept form.

The equation of $GH$ is:

$y + 2x + 14 = 0$

Rearranging for $y$:

$y = -2x - 14$

From this, the slope ($m$) of $GH$ is $m = -2$.

Step 2: Determine the slope of $JK$.

The slope of a line perpendicular to another is the negative reciprocal of the slope. Therefore, the slope of $JK$ ($m_{\text{JK}}$) is:

$m_{\text{JK}} = \frac{1}{2}$

Step 3: Write the equation of $JK$.

The general equation of a line in slope-intercept form is:

$y = mx + c$

Substituting the slope $\frac{1}{2}$:

$y = \frac{1}{2}x + c$

Step 4: Find $c$ if a point is given.

If you are given a point on $JK$ (e.g., $(x_1, y_1)$), substitute it into $y = \frac{1}{2}x + c$ to solve for $c$. If no point is given, the equation of $JK$ remains in the form:

$y = \frac{1}{2}x + c$

Would you like to clarify the point or proceed with any related details? 😊

---

Follow-Up Questions:

1. What point does line $JK$ pass through?
2. Can we find the intersection point of $GH$ and $JK$?
3. How would the graph of both lines look on the coordinate plane?
4. Can you derive the perpendicular slope formula from the slopes' product rule?
5. Would $JK$'s equation change if we shifted the origin?

---

Tip:

When dealing with perpendicular lines, always remember their slopes satisfy $m_1 \times m_2 = -1$.