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

For a line to be perpendicular, the product of its slope and the slope of line L must equal -1. Therefore:

$m_1 \cdot m_2 = -1$

Where $m_1 = -\frac{2}{3}$ and $m_2$ is the slope of the line we are looking for. Thus, we have:

$-\frac{2}{3} \cdot m_2 = -1$

Solving for $m_2$ gives:

$m_2 = \frac{3}{2}$

Now we need to determine which of the provided options has a slope of $\frac{3}{2}$:

1. $2x - 3y = 7$ can be rearranged to:

   $3y = 2x - 7$ $y = \frac{2}{3}x - \frac{7}{3}$

   • Slope = $\frac{2}{3}$ (not perpendicular).

2. $3x + 2y = -7$ can be rearranged to:

   $2y = -3x - 7$ $y = -\frac{3}{2}x - \frac{7}{2}$

   • Slope = $-\frac{3}{2}$ (not perpendicular).

3. $2x + 3y = -7$ can be rearranged to:

   $3y = -2x - 7$ $y = -\frac{2}{3}x - \frac{7}{3}$

   • Slope = $-\frac{2}{3}$ (not perpendicular).

4. $3x - 2y = 7$ can be rearranged to:

   $2y = 3x - 7$ $y = \frac{3}{2}x - \frac{7}{2}$

   • Slope = $\frac{3}{2}$ (this is the perpendicular line!).