Straight-Line Graphs: Question 3

Worked solution

Step 1: Find the gradient

The gradient between two points $(x_1,y_1)$ and $(x_2,y_2)$ is

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{change in } y}{\text{change in } x}.$

Using $P(0,5)$ and $Q(4,-1)$:

$m = \frac{-1 - 5}{4 - 0} = \frac{-6}{4} = -\frac{3}{2}.$

The line falls as we move to the right, so a negative gradient is expected, a useful sanity check.

Step 2: Find the $y$-intercept

The point $P(0,5)$ has $x = 0$, so it lies on the $y$-axis. Its $y$-coordinate is therefore the intercept directly:

$c = 5.$

(If $P$ had not been on the axis, you would substitute a known point into $y = mx + c$ and solve. Using $Q(4,-1)$ as a check: $-1 = -\tfrac{3}{2}(4) + c \Rightarrow -1 = -6 + c \Rightarrow c = 5.$ ✓)

Step 3: Write the equation

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

Verify with point $Q(4,-1)$

$y = -\frac{3}{2}(4) + 5 = -6 + 5 = -1. \;\checkmark$

Both points satisfy the equation, so the answer is

$\boxed{y = -\tfrac{3}{2}x + 5} \quad \text{(Option B)}.$

Why the other options are wrong

• A $\left(y=\tfrac{3}{2}x+5\right)$: correct size of gradient but the sign was lost; this line rises, contradicting the points.
• C $\left(y=-\tfrac{2}{3}x+5\right)$: the gradient fraction was inverted ($\tfrac{\Delta x}{\Delta y}$ instead of $\tfrac{\Delta y}{\Delta x}$).
• D $\left(y=\tfrac{2}{3}x+5\right)$: both the inversion and the sign error were made.