Question

If the equation of a straight line passing through the point (1,2) and perpendicular to the line $3x + 4y = 7$ is:

• A. $4x - 3y + 2 = 0$
• B. $3x + 4y - 11 = 0$
• C. $4x + 3y - 10 = 0$
• D. $3x - 4y + 5 = 0$

Hint:

\begin{itemize} \item Find the slope ($m_1$) of the given line $3x+4y=7$. (Slope $m_1 = -A/B = -3/4$). \item The slope ($m_2$) of a line perpendicular to it is the negative reciprocal: $m_2 = -1/m_1 = 4/3$. \item Use the point-slope form $y-y_1 = m_2(x-x_1)$ with the point $(1,2)$ and slope $m_2=4/3$. \item Alternatively, a line perpendicular to $Ax+By+C=0$ is $Bx-Ay+K=0$. Substitute the point to find K. \end{itemize}

Answer 1

The Correct Option is A

The given line is \( L_1: 3x + 4y = 7 \). First, find its slope by writing it in slope-intercept form (\( y = mx + c \)): \[ 3x + 4y = 7 \implies 4y = -3x + 7 \implies y = -\frac{3}{4}x + \frac{7}{4} \] So, the slope of \( L_1 \) is \( m_1 = -\frac{3}{4} \). The required line, say \( L_2 \), is perpendicular to \( L_1 \). If two lines are perpendicular, the product of their slopes is \(-1\): \( m_1 m_2 = -1 \). So, the slope of \( L_2 \) is: \[ m_2 = -\frac{1}{m_1} = -\frac{1}{-\frac{3}{4}} = \frac{4}{3} \] Now, \( L_2 \) passes through the point \( (1, 2) \) and has slope \( m_2 = \frac{4}{3} \). Using the point-slope form: \[ y - y_0 = m(x - x_0) \] \[ y - 2 = \frac{4}{3}(x - 1) \] Multiply both sides by 3 to clear denominators: \[ 3(y - 2) = 4(x - 1) \] \[ 3y - 6 = 4x - 4 \] Bring all terms to one side (general form): \[ 4x - 3y + 2 = 0 \] This matches option (a). Alternatively, recall: A line perpendicular to \( Ax + By + C = 0 \) is of the form \( Bx - Ay + K = 0 \). Given \( 3x + 4y - 7 = 0 \), a perpendicular line has form \( 4x - 3y + K = 0 \). Substitute the point \( (1, 2) \): \[ 4(1) - 3(2) + K = 0 \implies 4 - 6 + K = 0 \implies -2 + K = 0 \implies K = 2 \] So, the equation is: \[ 4x - 3y + 2 = 0 \] \[ \boxed{4x - 3y + 2 = 0} \]