Linear Relationships and the Cartesian Plane

Gradient formula: $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{9 - 3}{5 - 2} = \dfrac{6}{3} = 2$. ($\dfrac{1}{2}$ took the reciprocal (run over rise); 6 gave only the rise; 3 gave only the run.)

$m = \dfrac{-2 - 4}{3 - (-1)} = \dfrac{-6}{4} = -\dfrac{3}{2}$. Watch the double negative: $3 - (-1) = 4$. ($\dfrac{3}{2}$ mishandled the signs; $\dfrac{2}{3}$ took the reciprocal; $-6$ gave only the rise.)

The $y$-coordinates are equal, so the rise is $5 - 5 = 0$: $m = \dfrac{0}{7 - 2} = \dfrac{0}{5} = 0$. This is a horizontal line, and horizontal lines have gradient 0. (Undefined describes a vertical line, where the run is zero; here the rise is zero and the run is non-zero.)

The $x$-coordinates are equal, so the run is $3 - 3 = 0$: $m = \dfrac{8 - 1}{0} = \dfrac{7}{0}$, which is undefined (division by zero). This is a vertical line. (0 describes a horizontal line; division by zero is undefined, not zero.)

Equal gradients mean the lines have the same steepness and direction, so they are parallel. Being distinct (different intercepts), they never meet. (If the intercepts were also equal they would be the same line.)

Midpoint formula: $M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right) = \left(\dfrac{2 + 8}{2}, \dfrac{4 + 10}{2}\right) = (5, 7)$. ($(3, 3)$ subtracted instead of adding; $(10, 14)$ added but did not halve.)

$M = \left(\dfrac{-3 + 5}{2}, \dfrac{2 + (-8)}{2}\right) = \left(\dfrac{2}{2}, \dfrac{-6}{2}\right) = (1, -3)$. ($(2, -6)$ added correctly but did not divide by 2; $(-4, 5)$ subtracted instead of adding.)

Distance formula (from Pythagoras): $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(4 - 1)^2 + (6 - 2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. (7 added the differences $3 + 4$ without squaring; 25 forgot the square root, the canonical error.)

$d = \sqrt{(3 - 0)^2 + (3 - 0)^2} = \sqrt{9 + 9} = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$. ($\sqrt{18}$ is correct but not in simplest surd form; 6 added $3 + 3$.)

The burn rate is the gradient of height against time: $\dfrac{9 - 18}{5 - 2} = \dfrac{-9}{3} = -3$, so the height falls by $3$ cm each hour. ($9$ cm/h used only the change in height, ignoring the 3 hours.)

In $y = mx + b$, $m$ is the gradient and $b$ is the y-intercept. For $y = 2x - 5$: gradient $2$, y-intercept $-5$. (Gradient $-5$, y-intercept $2$ swapped the two; the gradient is a number, not $2x$.)

A negative gradient means the line slopes downward from left to right, so as $x$ increases, $y$ decreases. (A positive gradient would make $y$ increase; a zero gradient keeps $y$ constant.)

The $x$-intercept (set $y = 0$): $0 = -3x + 12$, so $x = 4$. The $y$-intercept is $12$. The right-angled triangle has base $4$ and height $12$, so its area is $\dfrac{1}{2} \times 4 \times 12 = 24$ units$^2$.

Use $y = mx + b$ with $m = 4$ and $b = -2$: $y = 4x - 2$. ($y = -2x + 4$ swapped gradient and y-intercept; $y = 4x + 2$ has a sign error on the intercept.)

On a distance-time graph the speed is the gradient of the line: $\dfrac{60 - 0}{4 - 0} = 15$ km/h. ($60$ km/h ignored that the journey took 4 hours.)

Parallel lines have the same gradient. The given gradient is $3$, so $y = 3x - 1$ (gradient 3, different intercept) is parallel. ($y = -\dfrac{1}{3}x + 4$ is the negative reciprocal, so it is perpendicular; $y = -3x + 4$ has gradient $-3$.)

Substitute $F = 50$: $50 = 1.8C + 32$. Subtract 32: $1.8C = 18$; divide by 1.8: $C = 10^\circ\text{C}$. ($90^\circ$ used $50 \times 1.8$; the gradient $1.8$ is the Fahrenheit rise per $1^\circ$C.)

Parallel means the same gradient $m = -2$. Point-gradient form through $(1, 5)$: $y - 5 = -2(x - 1)$, so $y = -2x + 2 + 5 = -2x + 7$. ($y = -2x + 5$ used the point's $y$-value as the intercept directly; $y = -2x + 3$ kept the original intercept.)

The perpendicular gradient is the negative reciprocal of $3$, namely $-\dfrac{1}{3}$. Point-gradient form through $(6, 4)$: $y - 4 = -\dfrac{1}{3}(x - 6)$, so $y = -\dfrac{x}{3} + 2 + 4 = -\dfrac{x}{3} + 6$. ($y = -3x + 6$ only negated; $y = \dfrac{1}{3}x + 6$ only took the reciprocal.)

In $C = 3d + 4$ the constant term $4$ is the $y$-intercept: the cost when $d = 0$. That is the fixed flag-fall paid before any distance is travelled. (The $3$ is the gradient, the cost per kilometre.)

Gradient $m = \dfrac{9 - 3}{4 - 1} = \dfrac{6}{3} = 2$. Point-gradient form through $A(1, 3)$: $y - 3 = 2(x - 1)$, so $y = 2x - 2 + 3 = 2x + 1$. ($y = 2x + 3$ used the $y$-value of $A$ as the intercept; $y = 2x - 1$ has a sign error.)

Rearrange to $y = mx + b$ form: $2y = -3x + 6$, so $y = -\dfrac{3}{2}x + 3$. The gradient is $-\dfrac{3}{2}$. ($3$ read the coefficient directly; in $ax + by + c = 0$ the gradient is $-\dfrac{a}{b}$, not $a$.)

Midpoint $M = \left(\dfrac{1 + 7}{2}, \dfrac{2 + 4}{2}\right) = (4, 3)$. Gradient of $AB$: $\dfrac{4 - 2}{7 - 1} = \dfrac{2}{6} = \dfrac{1}{3}$, so the perpendicular gradient is $-3$. Through $M(4, 3)$: $y - 3 = -3(x - 4)$, so $y = -3x + 15$. ($y = -3x + 5$ used point $A$ instead of the midpoint; $y = 3x - 9$ did not negate the reciprocal.)

Substitute $(2, 7)$: $7 = 2m + 3$, so $2m = 4$ and $m = 2$. ($m = 4$ did not divide by 2; $m = 7$ used the $y$-value as the gradient.)

Use the midpoint formula in reverse: $3 = \dfrac{1 + B_x}{2}$ gives $B_x = 5$, and $-1 = \dfrac{2 + B_y}{2}$ gives $B_y = -4$. So $B = (5, -4)$. ($(5, 4)$ made a sign error on $B_y$; $(2, -3)$ computed $M - A$ instead of $2M - A$.)