SL 2.1—Equations of a line

The equation $y=\frac{3}{4}x-2$ is written in gradient-intercept form $y=mx+c$. We can read off the values directly.

The coefficient of $x$ is the gradient: $m=\frac{3}{4}$.

The constant term $c=-2$ gives the $y$-intercept as the point $(0,-2)$.

The gradient is $\frac{3}{4}$ and the $y$-intercept is $(0,-2)$.

We need to correctly identify both the gradient (coefficient of $x$) and the $y$-intercept (constant term) from the equation.

Option B states gradient $=-2$ and $y$-intercept $=\left(0,\frac{3}{4}\right)$. This confuses the roles of $m$ and $c$ — the values are swapped.

Option C gives gradient $=\frac{4}{3}$, which is the reciprocal of the actual gradient. This is a common error when reading the fraction.

Option D gives the correct gradient but describes $\left(-\frac{3}{4},0\right)$ as the $y$-intercept. That point has $y=0$, making it an $x$-intercept, not a $y$-intercept.

Option A correctly identifies $m=\frac{3}{4}$ and $y$-intercept $=(0,-2)$ directly from $y=mx+c$.

The gradient between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2-y_1}{x_2-x_1}$.

Using $P(2,1)$ and $Q(6,9)$: $m=\frac{9-1}{6-2}=\frac{8}{4}=2$

The gradient of the line is $m=2$.

We need the gradient (rise over run) between the two given points.

Option A gives $m=\frac{1}{2}$. This would result from inverting the formula: $\frac{x_2-x_1}{y_2-y_1}=\frac{4}{8}=\frac{1}{2}$, which is the reciprocal — an incorrect calculation.

Option C gives $m=-2$. Since $Q$ is above and to the right of $P$, the line slopes upward and the gradient must be positive.

Option D gives $m=4$, which would be the rise alone ($9-1=8$... actually not 4 either) — this does not equal $\frac{8}{4}$.

The correct calculation is $\frac{9-1}{6-2}=\frac{8}{4}=2$, confirming option B.