Exercises — Distance formula — derivation using Pythagoras

Identify: $x_1=2,y_1=3,x_2=2,y_2=8$. $d=\sqrt{(2-2{)}^2+(8-3{)}^2}=\sqrt{0+25}=\sqrt{25}=5$ What this means: the two points share the same $x$-value, so they sit on a vertical line. The horizontal gap is $0$ and the distance is just the vertical gap. Answer: $5$ units.

$x_1=-1,y_1=4,x_2=6,y_2=4$. $d=\sqrt{(6-(-1){)}^2+(4-4{)}^2}=\sqrt{7^2+0}=\sqrt{49}=7$ Same $y$-value → a horizontal line; the distance is the pure horizontal gap. Answer: $7$ units.

$d=\sqrt{(5-0{)}^2+(12-0{)}^2}=\sqrt{25+144}=\sqrt{169}=13$ This is the famous $5\text{–}12\text{–}13$ right triangle. Answer: $13$ units.

$x_1=-3,y_1=-2,x_2=1,y_2=1$. $x_2-x_1=1-(-3)=4,y_2-y_1=1-(-2)=3$ Why the plus signs? Subtracting a negative adds: $1-(-3)=1+3$. $d=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$ Answer: $5$ units.

$x_2-x_1=-4-2=-6,y_2-y_1=3-(-5)=8$ $d=\sqrt{(-6{)}^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$ Note $(-6{)}^2=+36$: the minus disappears under the square. Answer: $10$ units.

$d=\sqrt{(4-1{)}^2+(5-1{)}^2}=\sqrt{9+16}=\sqrt{25}=5$ Here it happens to be whole. Compare with $A(1,1)$, $B(3,4)$: $d=\sqrt{2^2+3^2}=\sqrt{13}\approx 3.606$ Leave $\sqrt{13}$ as-is — it is exact; the decimal is only an approximation. Answer: $5$ units (first pair).

Compute all three side lengths (look at the figure above — the three coloured legs): $AB=\sqrt{(4-0{)}^2+(0-0{)}^2}=\sqrt{16}=4$ $BC=\sqrt{(4-4{)}^2+(3-0{)}^2}=\sqrt{9}=3$ $CA=\sqrt{(0-4{)}^2+(0-3{)}^2}=\sqrt{16+9}=\sqrt{25}=5$ Test Pythagoras: does the sum of the two smaller squares equal the largest square? $A{B}^2+B{C}^2=16+9=25=C{A}^2✓$ Because the equality holds, the angle opposite the longest side $CA$ is $90^{\circ }$ — that is the angle at $B$. Answer: right-angled at $B$.

Three points lie on a line exactly when the two shorter distances add up to the longest. $PQ=\sqrt{(3-1{)}^2+(4-2{)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$ $QR=\sqrt{(6-3{)}^2+(7-4{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$ $PR=\sqrt{(6-1{)}^2+(7-2{)}^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$ Check: $PQ+QR=2\sqrt{2}+3\sqrt{2}=5\sqrt{2}=PR$. ✓ Since the pieces add up exactly, there is no bend at $Q$ — the points are collinear.

Any point on the $x$-axis has the form $(x,0)$ — its $y$-value is $0$. Call it $P(x,0)$. Set the two distances equal. Squaring both sides first (to kill the roots): $P{A}^2=(x-2{)}^2+(0-(-5){)}^2=(x-2{)}^2+25$ $P{B}^2=(x-(-2){)}^2+(0-9{)}^2=(x+2{)}^2+81$ Set $P{A}^2=P{B}^2$: $(x-2{)}^2+25=(x+2{)}^2+81$ Expand: $(x^2-4x+4)+25=(x^2+4x+4)+81$. The $x^2$ and the $4$ cancel from both sides: $-4x+25=4x+81\Rightarrow -8x=56\Rightarrow x=-7$ Answer: $(-7,0)$.

Same $x$-value, so only the $y$'s differ: $5=\sqrt{(3-3{)}^2+(9-k{)}^2}=\sqrt{(9-k{)}^2}=|9-k|$ $|9-k|=5\Rightarrow 9-k=5\text{ or }9-k=-5$ $k=4\text{or}k=14$ Why two answers? A point $5$ above or $5$ below $B$ both sit at distance $5$. Answer: $k=4$ or $k=14$.

$5=\sqrt{(x-2{)}^2+(3-(-1){)}^2}=\sqrt{(x-2{)}^2+16}$ Square both sides: $25=(x-2{)}^2+16\Rightarrow (x-2{)}^2=9\Rightarrow x-2=\pm 3$. $x=5\text{or}x=-1$ Answer: $x=5$ or $x=-1$. (Both make a right triangle with legs $3$ and $4$.)

$AB=\sqrt{(4-1{)}^2+(4-1{)}^2}=\sqrt{9+9}=\sqrt{18}=3\sqrt{2}$ $BC=\sqrt{(4-4{)}^2+(1-4{)}^2}=\sqrt{9}=3$ $CA=\sqrt{(4-1{)}^2+(1-1{)}^2}=\sqrt{9}=3$ Two sides equal ($BC=CA=3$) → isosceles. Right-angle test: $B{C}^2+C{A}^2=9+9=18=A{B}^2$ ✓, so the right angle is at $C$. Area $=\frac{1}{2}\times \text{leg}\times \text{leg}=\frac{1}{2}\times 3\times 3=\frac{9}{2}=4.5$. Answer: isosceles right triangle, area $=4.5$ square units.

A square needs four equal sides and two equal diagonals. Compute the four sides: $AB=\sqrt{(2-(-1){)}^2+(3-(-1){)}^2}=\sqrt{9+16}=5$ $BC=\sqrt{(6-2{)}^2+(0-3{)}^2}=\sqrt{16+9}=5$ $CD=\sqrt{(3-6{)}^2+(-4-0{)}^2}=\sqrt{9+16}=5$ $DA=\sqrt{(-1-3{)}^2+(-1-(-4){)}^2}=\sqrt{16+9}=5$ All four sides $=5$ → a rhombus. Now the diagonals: $AC=\sqrt{(6-(-1){)}^2+(0-(-1){)}^2}=\sqrt{49+1}=\sqrt{50}=5\sqrt{2}$ $BD=\sqrt{(3-2{)}^2+(-4-3{)}^2}=\sqrt{1+49}=\sqrt{50}=5\sqrt{2}$ Equal diagonals turn a rhombus into a square. Both diagonals $=5\sqrt{2}$ ✓. Proved.

The centre $O(x,y)$ is equidistant from all three points (that's the definition of a circle). Set $O{P}^2=O{Q}^2$ and $O{Q}^2=O{R}^2$.

$O{Q}^2=O{R}^2$ (same $x=3$, so this is the easy one): $(x-3{)}^2+(y+7{)}^2=(x-3{)}^2+(y-3{)}^2$ Cancel $(x-3{)}^2$: $(y+7{)}^2=(y-3{)}^2\Rightarrow y^2+14y+49=y^2-6y+9\Rightarrow 20y=-40\Rightarrow y=-2.$

$O{P}^2=O{Q}^2$ with $y=-2$: $(x-6{)}^2+(-2+6{)}^2=(x-3{)}^2+(-2+7{)}^2$ $(x-6{)}^2+16=(x-3{)}^2+25$ $x^2-12x+36+16=x^2-6x+9+25\Rightarrow -12x+52=-6x+34\Rightarrow -6x=-18\Rightarrow x=3.$ Centre $O(3,-2)$. Radius $=OR=\sqrt{(3-3{)}^2+(3-(-2){)}^2}=\sqrt{0+25}=5$. Answer: centre $(3,-2)$, radius $5$.

On $y=x$ every point looks like $P(t,t)$. Set $P{A}^2=P{B}^2$: $(t-1{)}^2+(t-2{)}^2=(t-4{)}^2+(t-1{)}^2$ Cancel $(t-1{)}^2$ from both sides: $(t-2{)}^2=(t-4{)}^2\Rightarrow t^2-4t+4=t^2-8t+16\Rightarrow 4t=12\Rightarrow t=3.$ Answer: $P(3,3)$. (Check: $PA=\sqrt{4+1}=\sqrt{5}$, $PB=\sqrt{1+4}=\sqrt{5}$ ✓.)