Line Perpendicular to a Given Line Through a Point

The twist here is the starting point: the given line is in implicit form $3x - 2y = 6$, so the first job is to rearrange for its gradient. After that it is the standard perpendicular-line routine, flip the gradient, write the line through the point.

Find the equation of the line that passes through the point $(2, -1)$ and is perpendicular to the line $3x - 2y = 6$. Give your answer in the form $ax + by + c = 0$. [4]

The working

Step 1, gradient of the given line. Rearrange $3x - 2y = 6$ into $y = mx + c$: $2y = 3x - 6 \;\Rightarrow\; y = \frac{3}{2}x - 3 \;\Rightarrow\; m = \frac{3}{2} \quad \text{(M1)}$

Step 2, perpendicular gradient (negative reciprocal): $m_\perp = -\frac{2}{3} \quad \text{(A1)}$

Step 3, line through $(2, -1)$: $y - (-1) = -\frac{2}{3}(x - 2) \;\Rightarrow\; y + 1 = -\frac{2}{3}(x - 2) \quad \text{(M1)}$

Clear the fraction and rearrange: $3(y + 1) = -2(x - 2) \;\Rightarrow\; 3y + 3 = -2x + 4 \;\Rightarrow\; 2x + 3y - 1 = 0 \quad \text{(A1)}$

Where the marks are won and lost

• Rearrange first. The gradient of $3x - 2y = 6$ is $\frac32$, obtained only after solving for $y$, not $3$ or $-\frac32$ read off the coefficients.
• Watch the double negative: through $(2, -1)$ the $y$-term is $y - (-1) = y + 1$.
• Integer form requires clearing the $\frac23$; multiply through by $3$.

Common mistakes

• Taking the gradient as $3$ (coefficient of $x$ in the implicit form) without rearranging.
• Sign slip on $y - (-1)$.
• Forgetting to convert to the requested $ax + by + c = 0$ form.

Full method: Parallel & Perpendicular Lines notes. See also Equation of a Line. Topic home: Straight-Line Graphs pillar.