Given: Points: $P(1, 4)$, $Q(k, 3)$

Step 1: Find midpoint of PQ

Midpoint = $\left( \dfrac{1 + k}{2}, \dfrac{4 + 3}{2} \right) = \left( \dfrac{1 + k}{2}, \dfrac{7}{2} \right)$

Step 2: Find slope of PQ

Slope of PQ = $\dfrac{3 - 4}{k - 1} = \dfrac{-1}{k - 1}$

Step 3: Slope of perpendicular bisector = negative reciprocal = $k - 1$

Step 4: Use point-slope form for perpendicular bisector:

$y - \dfrac{7}{2} = (k - 1)\left(x - \dfrac{1 + k}{2}\right)$

Step 5: Find y-intercept (put $x = 0$)

$y = \dfrac{7}{2} + (k - 1)\left( -\dfrac{1 + k}{2} \right)$

$y = \dfrac{7}{2} - (k - 1)\left( \dfrac{1 + k}{2} \right)$

Given: y-intercept = -4, so:

$\dfrac{7}{2} - \dfrac{(k - 1)(k + 1)}{2} = -4$

Multiply both sides by 2:

$7 - (k^2 - 1) = -8 \Rightarrow 7 - k^2 + 1 = -8 \Rightarrow 8 - k^2 = -8$

$\Rightarrow k^2 = 16 \Rightarrow k = \pm4$

✅ Final Answer: $\boxed{k = -4 \text{ or } 4}$

Given:

Points: $A = (1, \frac{1}{2})$, $B = (3, -\frac{1}{2})$

Line: $2x + 3y = k$

Step 1: Evaluate $2x + 3y$

For A: $2(1) + 3\left(\frac{1}{2}\right) = \frac{7}{2}$
For B: $2(3) + 3\left(-\frac{1}{2}\right) = \frac{9}{2}$

✅ Option-wise Check:

• In between the lines $2x + 3y = -6$ and $2x + 3y = 6$: ✔️ True since $\frac{7}{2}, \frac{9}{2} \in (-6, 6)$
• On the same side of $2x + 3y = 6$: ✔️ True, both values are less than 6
• On the same side of $2x + 3y = -6$: ✔️ True, both values are greater than -6
• On the opposite side of $2x + 3y = -6$: ❌ False, both are on the same side

✅ Final Answer:

The correct statements are:

1. In between the lines $2x + 3y = -6$ and $2x + 3y = 6$
2. On the same side of the line $2x + 3y = 6$
3. On the same side of the line $2x + 3y = -6$