Question: $A$ is $(-1, 3)$ and $B$ is $(5, -5)$. Find the length of $AB$ and the gradient of $AB$.

Step 1. Compute the run and the rise once, and reuse them:

$$\Delta x = 5 - (-1) = 6, \qquad \Delta y = -5 - 3 = -8$$

Step 2. Length — Pythagoras on the differences:

$$AB = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ units}$$

Step 3. Gradient — rise over run, keeping the coordinates in the same order top and bottom:

$$m = \dfrac{-5 - 3}{5 - (-1)} = \dfrac{-8}{6} = -\dfrac{4}{3}$$

Step 4. Read the sign: a negative gradient means the line slopes downwards from left to right — consistent with $B$ being lower and to the right of $A$. ✓

The same two numbers (run and rise) power both formulas — compute them once, use them twice.

⚠ Watch out: In the gradient, subtract in the same order in numerator and denominator ($y_B - y_A$ over $x_B - x_A$); mixing orders flips the sign. In the length, the squares erase the minus signs — a distance can never be negative, and $(-8)^2 = 64$, not $-64$.

Question: The distance between $A(2, 9)$ and $B(-4, k)$ is $10$ units. Find the possible values of $k$.

Step 1. Write the distance formula with the unknown left in:

$$\sqrt{\left(2-(-4)\right)^2 + (9-k)^2} = 10$$

Step 2. Square both sides to clear the root:

$$6^2 + (9-k)^2 = 100$$

Step 3. Isolate the bracket — no need to expand yet:

$$(9-k)^2 = 100 - 36 = 64$$

Step 4. Square-root with both signs, then solve each:

$$9 - k = \pm 8 \;\Rightarrow\; k = 1 \quad \text{or} \quad k = 17$$

Check. $k=1$: $\sqrt{36+64} = 10$ ✓. $k=17$: $\sqrt{36+64} = 10$ ✓ — geometrically, $B$ can sit above or below the level of $A$, which is exactly why two answers exist.

A quick sketch of the two candidate positions of the pair of answers feel inevitable rather than surprising.

⚠ Watch out: "Find the possible values" is the hint that a $\pm$ is coming — keep both roots. (Expanding to $k^2 - 18k + 17 = 0$ and factorising $(k-1)(k-17)$ works too, but isolating the bracket first is faster and safer.)

Question: Find the equation of the line passing through $A(2, 5)$ and $B(5, 11)$.

Step 1. Start from the general form and fill in the pieces one at a time:

$$y = mx + c$$

Step 2. Gradient from the two points:

$$m = \dfrac{11 - 5}{5 - 2} = \dfrac{6}{3} = 2$$

So far: $y = 2x + c$.

Step 3. Find $c$ by substituting either point — use $A(2,5)$:

$$5 = 2(2) + c \;\Rightarrow\; c = 1$$

Step 4. State the finished equation:

$$y = 2x + 1$$

Check. The unused point must also satisfy it: $B(5, 11)$: $2(5) + 1 = 11$ ✓ — a ten-second check that catches almost every slip.

For a line perpendicular to a given one, the recipe changes only at Step 2: take the negative reciprocal of the given gradient, then find the point exactly as in Step 3.

⚠ Watch out: Always find $m$ before $c$ — substituting a point into $y = mx + c$ with $m$ still unknown goes nowhere. The same recipe handles "through $(7,4)$ parallel to $y = 2x - 1$": parallel hands you $m = 2$ for free, then Step 3 finds $c$ as usual.

Question: $A(-1, 3)$, $B(5, -5)$, $C(1, 1)$ and $D(5, 4)$ are four points. Show that the line $AB$ is perpendicular to the line $CD$.

Step 1. The test: two lines are perpendicular exactly when the product of their gradients is $-1$:

$$m_1 \times m_2 = -1$$

Step 2. Gradient of $AB$:

$$m_{AB} = \dfrac{-5 - 3}{5 - (-1)} = \dfrac{-8}{6} = -\dfrac{4}{3}$$

Step 3. Gradient of $CD$:

$$m_{CD} = \dfrac{4 - 1}{5 - 1} = \dfrac{3}{4}$$

Step 4. Multiply and conclude in words:

$$m_{AB} \times m_{CD} = -\dfrac{4}{3} \times \dfrac{3}{4} = -1$$

Since the product of the gradients is $-1$, $AB$ is perpendicular to $CD$. (Shown)

Equivalently: a perpendicular gradient is the negative reciprocal — flip the fraction, flip the sign.

⚠ Watch out: Both moves are needed: $\frac{3}{4}$ is the flip of $\frac{4}{3}$ but without the sign change the product is $+1$, not $-1$. A "show that" answer must end with the sentence quoting $m_1 m_2 = -1$ — the arithmetic alone doesn't close the proof. (The rule fails only for horizontal/vertical pairs, where one gradient is undefined.)