Coordinate Geometry: Question 4

Worked solution

Step 1: Find the gradient of $l_1$

Rearranging $2x + 3y = 12$ into the form $y = mx + c$:

$3y = -2x + 12 \implies y = -\frac{2}{3}x + 4.$

So $l_1$ has gradient $m_1 = -\dfrac{2}{3}$.

Step 2: Find the gradient a perpendicular line must have

Two lines with gradients $m_1$ and $m_2$ are perpendicular when $m_1 m_2 = -1$, so

$m_2 = -\frac{1}{m_1} = -\frac{1}{-2/3} = \frac{3}{2}.$

We need the option whose gradient is $\dfrac{3}{2}$.

Step 3: Rearrange each option and compare

• A: $3x - 2y = 7 \implies -2y = 7 - 3x \implies y = \dfrac{3}{2}x - \dfrac{7}{2}$. Gradient $= \dfrac{3}{2}$. $\checkmark$
• B: $2x - 3y = 5 \implies -3y = 5 - 2x \implies y = \dfrac{2}{3}x - \dfrac{5}{3}$. Gradient $= \dfrac{2}{3}$.
• C: $3x + 2y = 9 \implies 2y = 9 - 3x \implies y = -\dfrac{3}{2}x + \dfrac{9}{2}$. Gradient $= -\dfrac{3}{2}$.
• D: $2x + 3y = -6 \implies 3y = -6 - 2x \implies y = -\dfrac{2}{3}x - 2$. Gradient $= -\dfrac{2}{3}$.

Only option A has gradient $\dfrac{3}{2}$, matching $m_2$ from Step 2. As a check: $m_1 \times m_2 = \left(-\dfrac{2}{3}\right)\times\dfrac{3}{2} = -1$, confirming the perpendicular condition holds.

Why the other options are wrong

• B (gradient $\dfrac{2}{3}$): this is $-m_1$, the result of flipping the sign of $l_1$‘s gradient without taking the reciprocal.
• C (gradient $-\dfrac{3}{2}$): this is the reciprocal of $m_1$ without the sign changed.
• D (gradient $-\dfrac{2}{3}$): this is the same gradient as $l_1$ itself. This line is parallel to $l_1$, not perpendicular.

Final answer

$\boxed{\text{Option A: } 3x - 2y = 7}$