Straight-Line Graphs: Question 2

Worked solution

Part (a): Gradient of $L$

To read off the gradient we must write the equation in the form $y = mx + c$, where $m$ is the gradient. Start from

$2x + 3y = 12.$

Subtract $2x$ from both sides:

$3y = -2x + 12.$

Divide every term by $3$:

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

Comparing with $y = mx + c$, the gradient is

$\boxed{m = -\frac{2}{3}}, \qquad \text{(with } y\text{-intercept } c = 4\text{)}.$

Key idea: You cannot simply read the gradient off $2x + 3y = 12$; you must rearrange first.

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Part (b): Line $P_1$ parallel to $L$ through $A(6,1)$

Parallel lines have equal gradients, so $P_1$ also has gradient $m = -\dfrac{2}{3}$.

Using the point–gradient form $y - y_1 = m(x - x_1)$ with $A(6,1)$:

$y - 1 = -\frac{2}{3}\big(x - 6\big).$

Expand the bracket. Note $-\dfrac{2}{3}\times(-6) = +4$:

$y - 1 = -\frac{2}{3}x + 4.$

Add $1$ to both sides:

$\boxed{y = -\frac{2}{3}x + 5}.$

Alternative (substitution into $y = mx + c$): Since $m = -\dfrac{2}{3}$, write $y = -\dfrac{2}{3}x + c$ and substitute $x = 6,\ y = 1$:

$1 = -\frac{2}{3}(6) + c = -4 + c \quad\Rightarrow\quad c = 5,$

giving the same answer $y = -\dfrac{2}{3}x + 5$.

Check: at $x = 6$, $\ y = -4 + 5 = 1$ ✓. The line passes through $A(6,1)$.

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Part (c): Line $P_2$ perpendicular to $L$ through $A(6,1)$

For perpendicular lines the product of the gradients is $-1$, so the new gradient is the negative reciprocal of $-\dfrac{2}{3}$:

$m_{P_2} = -\frac{1}{\left(-\tfrac{2}{3}\right)} = +\frac{3}{2}.$

(Quick check: $-\dfrac{2}{3} \times \dfrac{3}{2} = -1$ ✓.)

Use the point–gradient form through $A(6,1)$:

$y - 1 = \frac{3}{2}\big(x - 6\big) = \frac{3}{2}x - 9,$

$y = \frac{3}{2}x - 8.$

Now convert to the required integer form $ax + by = c$. Multiply every term by $2$ to clear the fraction:

$2y = 3x - 16.$

Rearrange so the coefficients are integers:

$\boxed{3x - 2y = 16}.$

Check: substitute $A(6,1)$: $\ 3(6) - 2(1) = 18 - 2 = 16$ ✓.

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Final Answers

• (a) $y = -\dfrac{2}{3}x + 4$; gradient $= -\dfrac{2}{3}$
• (b) $y = -\dfrac{2}{3}x + 5$
• (c) $3x - 2y = 16$