The perpendicular bisector of a line segment passes through the midpoint of the segment and has a gradient that is the negative reciprocal of the gradient of the segment.

Let the coordinates of point \( B \) be \( (p, q) \).

The gradient of \( AB \) is given by: \[ \text{Gradient of } AB = \frac{q - 3}{p - 8} \] Since the perpendicular bisector has a gradient of -2, the gradient of \( AB \) must be the negative reciprocal of -2, which is \(\frac{1}{2}\). Therefore: \[ \frac{q - 3}{p - 8} = \frac{1}{2} \] Cross-multiplying gives: \[ 2(q - 3) = p - 8 \] Simplifying: \[ 2q - 6 = p - 8 \] \[ p = 2q + 2 \] The midpoint of \( AB \) is given by: \[ \left( \frac{8 + p}{2}, \frac{3 + q}{2} \right) \] Since the perpendicular bisector passes through this midpoint, it must satisfy the equation \( y = -2x + 4 \). Substituting the midpoint into the equation: \[ \frac{3 + q}{2} = -2 \left( \frac{8 + p}{2} \right) + 4 \] Simplifying: \[ \frac{3 + q}{2} = -2 \left( \frac{8 + 2q + 2}{2} \right) + 4 \] \[ \frac{3 + q}{2} = -2 \left( 5 + q \right) + 4 \] \[ \frac{3 + q}{2} = -10 - 2q + 4 \] \[ \frac{3 + q}{2} = -6 - 2q \] Multiplying both sides by 2: \[ 3 + q = -12 - 4q \] \[ 5q = -15 \] \[ q = -3 \] Substituting \( q = -3 \) back into \( p = 2q + 2 \): \[ p = 2(-3) + 2 \] \[ p = -6 + 2 \] \[ p = -4 \] So, \( p = -4\) and \(q = -3\).

\(4m=-1\), so \(m=-1/4\).

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\(y=(2/5)x-7/5\); parallel lines have gradient \(2/5\).

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The required horizontal line through the point is \(y=-4\).

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The given gradient is \(6/3=2\), so \(m=-1/2\).

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\(-3m=-1\), hence \(m=1/3\).

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\(y=-3x/4+3\), so a parallel line has gradient \(-3/4\).

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The required line is \(y=6\), with y-intercept \(6\).

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The original gradient is \(3/6=1/2\); the perpendicular gradient is \(-2\).

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