Question

A line is perpendicular to the line 3x + 2y - 1 = 0 and passes through the point A(1, 1). Find its equation.

Answer 1

2x - 3y + 1 = 0

Answer 2

Solution:

1. Write the given line in slope-intercept form.

   Given: 3x + 2y – 1 = 0 → 2y = -3x + 1 → y = -3/2x + 1/2

   Thus, the slope is $m_1 = -\frac{3}{2}$.

2. For a line perpendicular to this, its slope $m_2$ satisfies: $m_1 \times m_2 = -1$ $\Rightarrow \left(-\frac{3}{2}\right) \times m_2 = -1$ $\Rightarrow m_2 = \frac{2}{3}$.

3. Use point-slope form for the line passing through $(1, 1)$: $y - 1 = \frac{2}{3}(x - 1)$.

4. To write in standard form, multiply through by 3: $3(y - 1) = 2(x - 1)$ $\Rightarrow 3y - 3 = 2x - 2$ $\Rightarrow 2x - 3y + 1 = 0$.

Minimal Explanation:

Convert the given line to slope-intercept form to find its slope $\left(-\frac{3}{2}\right)$. The slope of the perpendicular line is $\frac{2}{3}$. Use the point-slope form with point $(1, 1)$ to derive the equation, which simplifies to $2x - 3y + 1 = 0$.