Exercises — Slope (gradient) — definition, formula, interpretation

WHAT: run is horizontal change, rise is vertical change. WHY: displacement is always final − initial, in the same order for both. $\Delta x=5-2=3,\Delta y=7-1=6.$ $m=\frac{\Delta y}{\Delta x}=\frac{6}{3}=2.$ WHAT IT LOOKS LIKE: for every $1$ step right the line climbs $2$ up — a steep uphill line (see the red arrows in the figure above).

WHAT/WHY: we only need the shape of the ratio, not its value.

• (a) $\Delta y=4-4=0$ ⇒ rise is zero ⇒ $m=0$ (horizontal).
• (b) $\Delta x=1-1=0$ ⇒ run is zero ⇒ we would divide by zero ⇒ undefined (vertical).
• (c) $\Delta y=-6-0=-6$, $\Delta x=2-0=2$ ⇒ up-then-down ⇒ negative ($m=-3$).
• (d) $\Delta y=5-(-2)=7>0$, $\Delta x=3-(-1)=4>0$ ⇒ both positive ⇒ positive.

$m=\frac{-2-4}{5-(-3)}=\frac{-6}{8}=-\frac{3}{4}.$ WHY the double-negative care: $5-(-3)=5+3=8$. The result is negative ⇒ downhill to the right.

$m=\frac{4-(-2)}{-3-5}=\frac{6}{-8}=-\frac{3}{4}.✓$ WHY it must match: swapping the two points flips the sign of both top and bottom; two sign-flips cancel. So the slope is genuinely a property of the line, not of which point you call "first".

WHAT: set the slope formula equal to the given slope and solve. $\frac{k-3}{6-2}=\frac{5}{2}\Rightarrow \frac{k-3}{4}=\frac{5}{2}.$ Multiply both sides by $4$: $k-3=\frac{5}{2}\cdot 4=10$, so $k=13$. Check: $\frac{13-3}{6-2}=\frac{10}{4}=\frac{5}{2}.✓$

WHAT/WHY: three points are on one line exactly when the slope $A\to B$ equals the slope $B\to C$ — same steepness the whole way means no bend. $m_{AB}=\frac{6-2}{3-1}=\frac{4}{2}=2,m_{BC}=\frac{12-6}{6-3}=\frac{6}{3}=2.$ Equal ⇒ collinear. WHAT IT LOOKS LIKE: the little rise/run triangles are similar (same tilt), so they stack into one straight line.

WHY this tool: on the rise/run right triangle, $\Delta y$ is opposite the angle and $\Delta x$ is adjacent, and $\frac{\text{opposite}}{\text{adjacent}}=\tan \theta$. So $m=\tan \theta$ (see Angle of Inclination). To recover $\theta$ we ask "which angle has this tangent?" — that is $\arctan$. $m=1:\theta =\arctan 1=45^{\circ }.m=\sqrt{3}:\theta =\arctan \sqrt{3}=60^{\circ }.$

First $m_1=\frac{3-1}{4-0}=\frac{2}{4}=\frac{1}{2}$. WHY the rule: for perpendicular lines $m_1{m}_2=-1$ (see Parallel and Perpendicular Lines) — turning a line $90^{\circ }$ swaps rise and run and flips one sign. $m_2=-\frac{1}{m_1}=-\frac{1}{1/2}=-2.$

Step 1 — slope: $m=\frac{8-(-1)}{4-1}=\frac{9}{3}=3.$ Step 2 — use point-slope (see Equation of a Straight Line): $y-y_1=m(x-x_1)$ with $(1,-1)$: $y-(-1)=3(x-1)\Rightarrow y+1=3x-3\Rightarrow y=3x-4.$ Check other point: $x=4\Rightarrow y=3(4)-4=8.✓$ So $m=3,c=-4$.

Slope: $m=\frac{4-0}{3-0}=\frac{4}{3}.$ Distance (see Distance Formula): $\sqrt{(3-0{)}^2+(4-0{)}^2}=\sqrt{9+16}=\sqrt{25}=5.$ WHY both fit together: run $=3$, rise $=4$, and the straight-line segment joining them is the hypotenuse $\sqrt{3^2+4^2}=5$ of the same rise/run triangle. Slope reads the tilt; distance reads the length — one triangle, two questions.

Compute the three side-slopes: $m_{AB}=\frac{2-0}{4-0}=\frac{1}{2},m_{BC}=\frac{5-2}{1-4}=\frac{3}{-3}=-1,m_{AC}=\frac{5-0}{1-0}=5.$ WHY: two sides are perpendicular when the product of their slopes is $-1$.

• At $A$: sides $AB,AC$ ⇒ $\frac{1}{2}\cdot 5=\frac{5}{2}\ne -1$. Not right.
• At $B$: sides $AB,BC$ ⇒ $\frac{1}{2}\cdot (-1)=-\frac{1}{2}\ne -1$. Not right.
• At $C$: sides $BC,AC$ ⇒ $(-1)\cdot 5=-5\ne -1$. Not right. Conclusion: no product equals $-1$, so this triangle has no right angle — the premise "right-angled at $B$" is false. (Never trust a claim; test it.)

$m(h)=\frac{5-1}{(2+h)-2}=\frac{4}{h}.$

• As $h\to 0^{+}$ (approach from the right): $m=\frac{4}{\text{tiny }+}\to +\infty$.
• As $h\to 0^{-}$ (from the left): $m=\frac{4}{\text{tiny }-}\to -\infty$. WHAT IT LOOKS LIKE: the second point slides straight above the first; the line tips toward vertical. The slope grows without bound and the sign flips depending on the side — this is exactly why a truly vertical line ($h=0$) has an undefined, not infinite, slope: there is no single value the two sides agree on.

$m=\frac{46-10}{4-1}=\frac{36}{3}=12.$ WHY: on a $(t,s)$ graph, rise is metres and run is seconds, so slope has units $\frac{\text{m}}{\text{s}}$ — it is the average speed $=12\text{m/s}$ over that interval (see Rate of Change). Slope is "how fast $s$ changes per unit $t$."

Points: $(3,9)$ and $(3+h,(3+h{)}^2)$. $m(h)=\frac{(3+h{)}^2-9}{(3+h)-3}=\frac{9+6h+h^2-9}{h}=\frac{6h+h^2}{h}=6+h(h\ne 0).$ WHY cancel $h$: for $h\ne 0$ the run is nonzero, so we may divide top and bottom by $h$ — this removes the fake $\frac{0}{0}$ and reveals a clean formula. As $h\to 0$: $m(h)\to 6$. WHAT IT MEANS: the secant lines pivot toward the tangent at $x=3$; that limiting slope $6$ is the derivative there (see Derivative as a Slope).

Missing coordinate: line through $(2,3)$ & $(6,k)$ has $m=\frac{5}{2}$; $k=?$ ::: $k=13$ Perpendicular slope to $m_1=\frac{1}{2}$? ::: $m_2=-2$ Slope $=\tan \theta$ with $m=\sqrt{3}$ gives $\theta =?$ ::: $60^{\circ }$ Secant slope of $y=x^2$ at $x=3$ as $h\to 0$? ::: $6$ (the derivative) Why is vertical slope "undefined" not "infinite"? ::: from one side it $\to +\infty$, other side $\to -\infty$ — no single value, and run $=0$ divides by zero.