Question: Given $A(2, 1)$ and $B(8, 9)$, find: (a) the distance $AB$ (b) the midpoint $M$ of $AB$.

Step 1. Distance formula: $|AB| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.

$$|AB| = \sqrt{(8 - 2)^2 + (9 - 1)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

Step 2. Midpoint formula: $M = \left(\tfrac{x_1 + x_2}{2}, \tfrac{y_1 + y_2}{2}\right)$.

$$M = \left(\tfrac{2 + 8}{2}, \tfrac{1 + 9}{2}\right) = (5, 5)$$

Answer: $|AB| = 10$ units, $M = (5, 5)$.

Tip: Both formulas are symmetric in $A$ and $B$ — swap the order, get the same answer.

Question: Find the equation of the line through $A(-1, 4)$ and $B(3, -4)$.

Step 1. Gradient: $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-4 - 4}{3 - (-1)} = \dfrac{-8}{4} = -2$.

Step 2. Use point-slope form $y - y_1 = m(x - x_1)$ with $A(-1, 4)$:

$$\begin{aligned}y - 4 &= -2(x - (-1)) \\ y - 4 &= -2x - 2 \\ y &= -2x + 2\end{aligned}$$

Answer: $y = -2x + 2$ (or $2x + y = 2$).

Check: Sub $B(3, -4)$: $-2(3) + 2 = -4$ ✓.

Tip: Always verify by substituting the OTHER point. Catches arithmetic slips immediately.

Question: Find the equation of the perpendicular bisector of $A(1, 2)$ and $B(7, 6)$.

Step 1. Midpoint of $AB$:

$$M = \left(\tfrac{1 + 7}{2}, \tfrac{2 + 6}{2}\right) = (4, 4)$$

Step 2. Gradient of $AB$:

$$m_{AB} = \tfrac{6 - 2}{7 - 1} = \tfrac{4}{6} = \tfrac{2}{3}$$

Step 3. Perpendicular gradient (negative reciprocal):

$$m_\perp = -\tfrac{3}{2}$$

Step 4. Line through $M(4, 4)$ with gradient $-\tfrac{3}{2}$:

$$y - 4 = -\tfrac{3}{2}(x - 4) \;\Rightarrow\; y = -\tfrac{3}{2}x + 10$$

Answer: $y = -\tfrac{3}{2}x + 10$ (or $3x + 2y = 20$).

Why this works: The perpendicular bisector is the set of points equidistant from $A$ and $B$. It always passes through the midpoint with gradient $\perp$ to the original segment.

Question: Find the angle of inclination (the angle made with the positive $x$-axis) of (a) the line through $A(1, 2)$ and $B(3, 6)$; (b) the line through $P(0, 5)$ and $Q(4, 2)$. Give each answer correct to 1 decimal place.

Step 1. (a) Find the gradient:

$$m = \frac{6 - 2}{3 - 1} = 2$$

Step 2. The angle of inclination $\theta$ is measured anticlockwise from the positive $x$-axis to the line, and $m = \tan\theta$. Since $m > 0$, the angle is acute:

$$\theta = \tan^{-1} 2 = 63.4^\circ \;(1\text{ d.p.})$$

Step 3. (b) Find the gradient:

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

Step 4. Since $m < 0$, the angle is obtuse — take $180^\circ$ minus the basic angle:

$$\theta = 180^\circ - \tan^{-1}\tfrac{3}{4} = 180^\circ - 36.9^\circ = 143.1^\circ \;(1\text{ d.p.})$$

⚠ Watch out: Keying $\tan^{-1}(-0.75)$ into a calculator returns $-36.9^\circ$, which is not an inclination. The angle with the positive $x$-axis always satisfies $0^\circ \le \theta < 180^\circ$: for a negative gradient use $\theta = 180^\circ - \tan^{-1}|m|$.

Question: Find the equation of the line parallel to $2x - 3y + 5 = 0$ passing through $(4, -1)$.

Step 1. Rearrange the given line to $y = \tfrac{2}{3}x + \tfrac{5}{3}$. Gradient $m = \tfrac{2}{3}$.

Step 2. Parallel lines share the same gradient. So the new line has gradient $\tfrac{2}{3}$.

Step 3. Point-slope through $(4, -1)$:

$$y - (-1) = \tfrac{2}{3}(x - 4) \;\Rightarrow\; y + 1 = \tfrac{2}{3}x - \tfrac{8}{3}$$ $$y = \tfrac{2}{3}x - \tfrac{11}{3}$$

Answer: $y = \tfrac{2}{3}x - \tfrac{11}{3}$ (or $2x - 3y = 11$).

Tip: Convert the given line into $y = mx + c$ form FIRST to read off $m$ cleanly. Then build the new line by changing only the $c$.

Question: Find the foot of the perpendicular from $P(7, 2)$ to the line $y = 2x - 1$.

Step 1. Gradient of given line: $m_1 = 2$. Perpendicular gradient: $m_2 = -\tfrac{1}{2}$.

Step 2. Perpendicular through $P(7, 2)$:

$$y - 2 = -\tfrac{1}{2}(x - 7) \;\Rightarrow\; y = -\tfrac{1}{2}x + \tfrac{11}{2}$$

Step 3. Solve simultaneously with $y = 2x - 1$:

$$\begin{aligned}2x - 1 &= -\tfrac{1}{2}x + \tfrac{11}{2} \\ \tfrac{5}{2}x &= \tfrac{13}{2} \\ x &= \tfrac{13}{5}, \; y = 2(\tfrac{13}{5}) - 1 = \tfrac{21}{5}\end{aligned}$$

Answer: Foot $F = \left(\tfrac{13}{5}, \tfrac{21}{5}\right)$.

Tip: Foot of perpendicular = intersection of (the given line) and (the perpendicular through the external point). Two-step plan, always.

Question: Find the area of triangle with vertices $A(1, 2)$, $B(5, 1)$, $C(4, 6)$.

Step 1. Shoelace formula:

$$\text{Area} = \tfrac{1}{2}\bigl|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\bigr|$$

Step 2. Substitute:

$$\begin{aligned}&= \tfrac{1}{2}\bigl|1(1 - 6) + 5(6 - 2) + 4(2 - 1)\bigr| \\ &= \tfrac{1}{2}\bigl|-5 + 20 + 4\bigr| = \tfrac{1}{2}(19) = 9.5\end{aligned}$$

Answer: Area $= 9.5$ square units.

Tip: Always take the absolute value — direction of traversal (clockwise vs anticlockwise) flips the sign but area is positive.

Question: Show that $P(-1, 4)$, $Q(2, -2)$, $R(5, -8)$ are collinear.

Method: shoelace area $= 0$.

Three points are collinear iff the triangle they form has zero area.

Step 1. Apply shoelace:

$$\text{Area} = \tfrac{1}{2}\bigl|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)\bigr|$$

Step 2. Substitute:

$$\begin{aligned}&= \tfrac{1}{2}\bigl|(-1)(-2 - (-8)) + 2(-8 - 4) + 5(4 - (-2))\bigr| \\ &= \tfrac{1}{2}\bigl|-6 - 24 + 30\bigr| = 0\end{aligned}$$

Area $= 0 \Rightarrow$ collinear. ✓

Alternative method: Show $m_{PQ} = m_{QR}$: $m_{PQ} = \tfrac{-2 - 4}{2 - (-1)} = -2$ $m_{QR} = \tfrac{-8 - (-2)}{5 - 2} = -2$ ✓

Both methods work — pick whichever is faster.

Question: The line $y = x + 2$ cuts the $y$-axis at $A$ and meets the line $y = -2x + 8$ at $B$. The line $y = -2x + 8$ cuts the $x$-axis at $C$. With $O$ the origin, find the area of quadrilateral $OABC$.

Step 1. Find each vertex by solving the right pair of equations. $A$ (on the $y$-axis): put $x = 0$ into $y = x + 2$, so $A(0, 2)$. $C$ (on the $x$-axis): put $y = 0$ into $y = -2x + 8$, so $x = 4$ and $C(4, 0)$.

Step 2. $B$ is where the lines meet — equate them:

$$x + 2 = -2x + 8 \implies 3x = 6 \implies x = 2,\; y = 4$$

So $B(2, 4)$.

Step 3. Shoelace with the vertices in order round the shape, $O \to A \to B \to C$, closing back at $O$:

$$\text{Area} = \frac{1}{2}\begin{vmatrix} 0 & 0 & 2 & 4 & 0\\ 0 & 2 & 4 & 0 & 0 \end{vmatrix}$$

Step 4. Down-products minus up-products:

$$\frac{1}{2}\left|(0\cdot2 + 0\cdot4 + 2\cdot0 + 4\cdot0) - (0\cdot0 + 2\cdot2 + 4\cdot4 + 0\cdot0)\right| = \frac{1}{2}\left|0 - 20\right| = 10$$

Area of $OABC$ $= 10$ units$^2$.

⚠ Watch out: The shoelace only works when the vertices are listed in order around the quadrilateral (and the list closes back at the start). Writing $O, A, C, B$ traces a crossed shape and gives a wrong "area" with no warning — sketch first, then read the order off the sketch.

Question: Find the point $P$ that divides the segment from $A(2, 3)$ to $B(8, 9)$ internally in the ratio $1 : 2$.

Step 1. Section formula for INTERNAL division in ratio $m : n$:

$$P = \left(\tfrac{m x_B + n x_A}{m + n}, \tfrac{m y_B + n y_A}{m + n}\right)$$

Step 2. With $m = 1$, $n = 2$, $A(2, 3)$, $B(8, 9)$:

$$P = \left(\tfrac{1(8) + 2(2)}{3}, \tfrac{1(9) + 2(3)}{3}\right) = \left(\tfrac{12}{3}, \tfrac{15}{3}\right) = (4, 5)$$

Answer: $P = (4, 5)$.

Sanity check: $P$ is closer to $A$ than $B$ (ratio $1:2$ means $AP:PB = 1:2$, so $P$ is one-third of the way from $A$). The midpoint of $A$ and $B$ would be $(5, 6)$ — $(4, 5)$ is correctly closer to $A$. ✓

Tip: "Ratio $m : n$" means $\overline{AP} : \overline{PB} = m : n$. The formula puts $m$ next to $B$'s coordinates and $n$ next to $A$'s — easy to remember by "cross-multiply."

Question: Find the equation of the locus of a point $P(x, y)$ that is equidistant from $A(1, 3)$ and $B(5, -1)$.

Step 1. Equidistant means $|PA| = |PB|$. Squaring (both sides non-negative):

$$|PA|^2 = |PB|^2$$ $$(x - 1)^2 + (y - 3)^2 = (x - 5)^2 + (y + 1)^2$$

Step 2. Expand both sides:

$$\begin{aligned}\text{LHS}: \;&x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + y^2 - 2x - 6y + 10 \\ \text{RHS}: \;&x^2 - 10x + 25 + y^2 + 2y + 1 = x^2 + y^2 - 10x + 2y + 26\end{aligned}$$

Step 3. Subtract LHS from RHS, simplify:

$$-2x - 6y + 10 = -10x + 2y + 26$$ $$8x - 8y - 16 = 0 \;\Rightarrow\; x - y - 2 = 0$$

Answer: $x - y = 2$ — the perpendicular bisector of $AB$.

Why this works: The locus of points equidistant from $A$ and $B$ IS the perpendicular bisector of $AB$. Same line, two ways to find it.