Linear Relationships and the Cartesian Plane

Coordinates are written $(x, y)$: across first, then up. P is 3 right and 2 up, so $(3, 2)$. ($(2, 3)$ reverses the order.)

$x = -3$ is to the left of the y-axis and $y = 4$ is above the x-axis, so the point is in the top-left region. (It is not on the y-axis, since $x \ne 0$.)

The lines cross at $(2, 4)$: read across to $x = 2$, then up to $y = 4$. This point lies on both lines at once. ($(0, 2)$ is Line A's y-intercept, not the intersection.)

Every point on the y-axis has an x-coordinate of 0, for example $(0, 3)$ or $(0, -5)$. (A point on the x-axis has y-coordinate 0; only the origin has both coordinates 0.)

Each step adds 3 ($4, 7, 10, 13, \ldots$), so the next value is $13 + 3 = 16$.

Each time $x$ rises by 1, $y$ rises by 2, so the coefficient of $x$ is 2. Since $2(1) = 2$ but $y = 3$, add 1: $y = 2x + 1$. (Check: $2(3) + 1 = 7$.)

The horizontal gap is 5 and the vertical gap is 12, forming a right-angled triangle. By Pythagoras, $d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$. (Adding $5 + 12 = 17$ instead of using Pythagoras is the common error.)

A linear pattern adds the same amount each step. $3, 6, 9, 12, 15$ adds 3 each time. (The squares $1, 4, 9, 16, 25$ and the doubling $1, 2, 4, 8, 16$ are non-linear.)

Setting $x = 0$ gives $y = c$, so the constant term $c$ is the y-intercept. The coefficient $m$ is the gradient (steepness), which is different.

The line crosses the y-axis at $(0, 3)$, so the y-intercept is 3. (2 is the rate of change, not the intercept.)

From $(0, 3)$ to $(1, 5)$, $y$ rises by 2 as $x$ rises by 1, so the rate of change is 2. (3 is the y-intercept; 5 is the y-value at $x = 1$.)

The line has y-intercept 3 and rises 2 per unit $x$, so $y = 2x + 3$. ($y = 3x + 2$ swaps the two values.)

The candle starts at 20 cm (the y-intercept) and loses 4 cm each hour, so the rate of change is $-4$: $h = -4t + 20$. (Using $+4$ would make the candle grow taller over time.)

Substitute $y = 9$: $9 = 2x + 1$, so $2x = 8$ and $x = \dfrac{8}{2} = 4$.

The width is the gap in the x-values: $6 - 1 = 5$. The height is the gap in the y-values: $5 - 2 = 3$. Area $= 5 \times 3 = 15$ square units. (Adding the two side lengths, $5 + 3 = 8$, gives the half-perimeter, not the area.)

A horizontal line has the same $y$ everywhere: $y = 3$. ($x = 3$ is vertical.)

A vertical line has the same $x$ everywhere: $x = -2$. ($y = -2$ is horizontal.)

A line through the origin has y-intercept 0. $y = 3x$ gives $y = 0$ when $x = 0$. ($y = 3x + 1$ passes through $(0, 1)$.)

Parallel lines share the coefficient of $x$. $y = 2x + 1$ has coefficient 2, and so does $y = 2x + 5$. ($y = 3x + 1$ is steeper.)

In $y = (\text{rate})\,x + (\text{intercept})$: rate 3, intercept $-2$, so $y = 3x - 2$. ($y = -2x + 3$ swaps the two values.)

The coefficient of $t$ is the rate of change of distance with time, which is the speed. Every extra hour adds 60 km, so the car travels at 60 km/h. (The line passes through the origin, so the starting distance is 0, not 60.)

The $5 base is the constant and the $2 per km multiplies $k$: $C = 2k + 5$. ($C = 5k + 2$ swaps them.)

Substitute $x = 3$: $k = 2(3) - 1 = 6 - 1 = 5$. (Forgetting the $-1$ gives 6.)

$y$ rises by 3 per step, and $3(1) = 3$ but $y = 6$, so $y = 3x + 3$. At $x = 10$: $y = 3(10) + 3 = 33$. ($y = 3x$ alone gives 30.)

The y-intercept is 5 (at $x = 0$). From $(0, 5)$ to $(2, 1)$, $y$ falls 4 over a run of 2, so the rate is $\dfrac{-4}{2} = -2$. So $y = -2x + 5$. ($y = 2x + 5$ has the wrong sign on the rate.)