The line L has equation $2x + 3y = 7$ - AQA - A-Level Maths Mechanics - Question 3 - 2018 - Paper 3

For a line to be perpendicular, its slope must be the negative reciprocal of the original line's slope. Therefore, the slope $m_P$ of the perpendicular line is:

$m_P = -\frac{1}{m_L} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2}.$ This means the perpendicular line has a slope of $\frac{3}{2}$.

Now, we need to check which equation among the options has a slope of $\frac{3}{2}$. Let's convert each equation to slope-intercept form:

1. Option A: $2x - 3y = 7$:

   • Rearranging gives $y = \frac{2}{3}x - \frac{7}{3}$, slope = $\frac{2}{3}$.

2. Option B: $3x + 2y = -7$:

   • Rearranging gives $y = -\frac{3}{2}x - \frac{7}{2}$, slope = $-\frac{3}{2}$.

3. Option C: $2x + 3y = 1$:

   • Rearranging gives $y = -\frac{2}{3}x + \frac{1}{3}$, slope = $-\frac{2}{3}$.

4. Option D: $3x - 2y = 7$:

   • Rearranging gives $y = \frac{3}{2}x - \frac{7}{2}$, slope = $\frac{3}{2}$. This matches our requirement for perpendicularity.

Thus, the correct equation that is perpendicular to line L is: Option D: $3x - 2y = 7$.