Equation of a perpendicular line passing through a point

1) The equation of a line passing through the point $(3,4)$ and perpendicular to the line $3x + 4y - 8 = 0$ is

Concepts:

• The slope of the given line $3x + 4y - 8 = 0$: Rearranged to $y = -\frac{3}{4}x + 2$, slope $ m_1 = -\frac{3}{4} $.
• The perpendicular slope $ m_2 $ satisfies $ m_1 \cdot m_2 = -1 $:

$$ m_2 = \frac{4}{3} $$

• Using point-slope form for a line through $(3,4)$ with this slope:

$$ y - 4 = \frac{4}{3}(x - 3) $$$$ 3(y - 4) = 4(x - 3) \implies 3y - 12 = 4x - 12 \implies 3y - 4x = 0 \implies 4x - 3y = 0 $$

None of the options match this directly, so multiply both sides to create equivalent forms:

$$ 8x - 6y = 0 $$

• Among the options, $8x - 6y = 0$ matches.

Final Answer: $ \boxed{8x - 6y = 0} $

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Concept:

• When a line is perpendicular to another, its slope is the negative reciprocal.
• Use point-slope or two-point form to get the equation through a given point.

2) Given lines $18x - 8y + 30 = 0$ and $4x + ky + 5 = 0$ are perpendicular to each other, find the value of $k$.

Concepts:

• Slope of line 1: $18x - 8y + 30 = 0 \implies y = \frac{18}{8}x + \ldots$ so $ m_1 = \frac{18}{8} = \frac{9}{4} $
• Slope of line 2: $4x + ky + 5 = 0 \implies ky = -4x - 5 \implies y = -\frac{4}{k}x - \frac{5}{k} $, so $ m_2 = -\frac{4}{k} $
• For perpendicular lines: $m_1 \cdot m_2 = -1$:

$$ \frac{9}{4} \cdot \left(-\frac{4}{k}\right) = -1 $$$$ \frac{9}{4} \cdot -\frac{4}{k} = -1 \implies -\frac{9}{k} = -1 \implies k = 9 $$

Final Answer: $ \boxed{9} $

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Concept:

• The product of slopes of perpendicular lines is $-1$.
• From the general form $Ax + By + C = 0$, slope is $ -A/B $.