Equation of parallel and perpendicular lines in general form

1) Which of the following statements are true?

Concepts:

• Two lines $Ax + By + C = 0$ and $A’x + B’y + C’ = 0$ are parallel if $\frac{A}{B} = \frac{A’}{B’}$.
• They are perpendicular if the product of their slopes ($m_1 \cdot m_2$) is $-1$, i.e., $-A/B \cdot -A’/B’ = -1 \implies \frac{A}{B} \cdot \frac{A’}{B’} = -1$.

Examine Each Option:

• a) $2x + 3y - 8 = 0$ and $3x - y - 2 = 0$ are parallel?
  • Slopes:
    • First: $m_1 = -2/3$
    • Second: $m_2 = -3/(-1) = 3$
    • Not equal, so not parallel.
• b) $3x + 5y - 10 = 0$ and $6x + 10y - 26 = 0$ are parallel?
  • Second equation is just 2 × first equation, so slopes are both $-3/5$.
  • True: They are parallel.
• c) $6x + 8y - 20 = 0$ and $4x - 3y = 0$ are perpendicular?
  • Slopes:
    • First: $m_1 = -6/8 = -3/4$
    • Second: $m_2 = -4/(-3) = 4/3$
    • Product: $(-3/4) \times (4/3) = -1$
    • True: They are perpendicular.
• d) $2x - 3y + 8 = 0$ and $3x + 2y - 18 = 0$ are not perpendicular to each other?
  • Slopes:
    • First: $m_1 = -2/(-3) = 2/3$
    • Second: $m_2 = -3/2$
    • Product: $(2/3) \times (-3/2) = -1$
    • But option says “are not perpendicular,” which is False (these are perpendicular).
• e) $4x + 5y + 10 = 0$ and $10x - 8y - 16 = 0$ are perpendicular?
  • Slopes:
    • First: $m_1 = -4/5$
    • Second: $m_2 = -10/(-8) = 5/4$
    • Product: $(-4/5) \times (5/4) = -1$
    • True: They are perpendicular.

Correct Statements:

• b) Lines $3x+5y-10=0$ and $6x+10y-26=0$ are parallel.
• c) Lines $6x+8y-20=0$ and $4x-3y=0$ are perpendicular.
• e) Lines $4x+5y+10=0$ and $10x-8y-16=0$ are perpendicular.

2) Given lines $4x + 2ky + 5 = 0$ and $12x + 6y + 15 = 0$ are parallel, find the value of $k$.

Concept:

• For two lines $A_1x + B_1y + C_1 = 0$ and $A_2x + B_2y + C_2 = 0$ to be parallel:

• So,

Final Answer: $\boxed{1}$