Exercises — Area of triangle using coordinate formula

WHAT: Substitute into the formula. $S=0(0-8)+6(8-0)+0(0-0)=0+48+0=48$ WHY these numbers: $x_1(y_2-y_3)$ means "the $x$ of $A$" times "($y$ of $B$) minus ($y$ of $C$)". Each $x$ multiplies the difference of the other two $y$'s. $\text{Area}=\frac{1}{2}|48|=24\text{ square units}$ WHAT IT LOOKS LIKE: This is a right triangle with legs along the axes — base $6$, height $8$. Check: $\frac{1}{2}(6)(8)=24$. ✓ ($S>0$, so the corners are listed counterclockwise.)

WHAT: Substitute. $S=2(3-9)+7(9-3)+5(3-3)=2(-6)+7(6)+5(0)=-12+42+0=30$ $\text{Area}=\frac{1}{2}|30|=15\text{ square units}$ WHAT IT LOOKS LIKE: $A$ and $B$ share the height $y=3$, so $AB$ is horizontal with length $|7-2|=5$ (this is the base). $C$ sits at height $9$, so the perpendicular height is $9-3=6$. Check: $\frac{1}{2}(5)(6)=15$. ✓

WHAT: Substitute, keeping every sign. $S=(-4)(-1-5)+3(5-2)+2(2-(-1))$ WHY carefully: $y_2-y_3=-1-5=-6$; $y_3-y_1=5-2=3$; $y_1-y_2=2-(-1)=3$. $S=(-4)(-6)+3(3)+2(3)=24+9+6=39$ $\text{Area}=\frac{1}{2}|39|=19.5\text{ square units}$ WHAT IT LOOKS LIKE: All three corners are spread across quadrants; there is no axis-aligned base to eyeball, which is exactly why the coordinate formula earns its keep. ($S>0\Rightarrow$ counterclockwise.)

WHAT: Substitute. $S=1(-4-(-2))+(-2)((-2)-1)+4(1-(-4))$ $=1(-2)+(-2)(-3)+4(5)=-2+6+20=24$ $\text{Area}=\frac{1}{2}|24|=12\text{ square units}$ WHAT IT LOOKS LIKE: see the figure below — the three points and the enclosed triangle.

WHAT: Compute the area. $S=1(6-10)+3(10-2)+5(2-6)=1(-4)+3(8)+5(-4)=-4+24-20=0$ $\text{Area}=\frac{1}{2}|0|=0$ WHY this matters: zero area means the "triangle" has been squashed flat — the three points sit on a single straight line. They are collinear. This is the same detector used in the Colinearity Test. WHAT IT LOOKS LIKE: each step right by $2$ goes up by $4$, so all three points lie on the line of slope $2$: $y=2x$. No enclosed region exists.

WHAT: Substitute symbols the same way as numbers. $S=0(0-h)+b(h-0)+0(0-0)=bh$ $\text{Area}=\frac{1}{2}|bh|=\frac{1}{2}bh$ WHY this is the sanity anchor: it reproduces the classroom formula $\frac{1}{2}\times \text{base}\times \text{height}$ exactly, confirming the coordinate formula is the same law in disguise. WHAT IT LOOKS LIKE: listing $A\to B\to C$ walks along the $x$-axis then up — a counterclockwise loop, so $S=bh>0$ is positive, matching "$+$ means counterclockwise."

WHAT: Write the area in terms of $k$, then solve. $S=2(1-4)+6(4-1)+k(1-1)=2(-3)+6(3)+k(0)=-6+18=12$ WHY $k$ vanished: $A$ and $B$ have the same $y=1$, so $y_1-y_2=0$ kills the $k$-term. Geometrically the base $AB$ is fixed and horizontal; the area depends only on $C$'s height, not its horizontal position. $\text{Area}=\frac{1}{2}|12|=6\ne 9.$ The height of $C$ above the base is $4-1=3$, giving area $\frac{1}{2}(4)(3)=6$ for every $k$. Since $6\ne 9$, no value of $k$ works — the area can never reach $9$ while $C$ stays on the line $y=4$. WHAT IT LOOKS LIKE: sliding $C$ left or right along $y=4$ shears the triangle but keeps base and height fixed, so the area is locked at $6$. (This is the classic "constant area on a fixed base" picture.)

WHAT: A quadrilateral has no single "area formula" in the parent note, but we can split it into two triangles along the diagonal $PR$, then add. WHY split: the diagonal $PR$ cuts $PQRS$ into $△PQR$ and $△PRS$, and their areas add up to the whole because they share only the edge $PR$ (no overlap, no gap).

$△PQR$ with $P(1,1),Q(7,2),R(6,6)$: $S=1(2-6)+7(6-1)+6(1-2)=-4+35-6=25,\text{Area}=\frac{25}{2}=12.5$ $△PRS$ with $P(1,1),R(6,6),S(2,5)$: $S=1(6-5)+6(5-1)+2(1-6)=1+24-10=15,\text{Area}=\frac{15}{2}=7.5$ WHAT: Add. $\text{Area}(PQRS)=12.5+7.5=20\text{ square units}$ WHAT IT LOOKS LIKE: the diagonal from $P$ to $R$ runs through the middle of the figure, splitting it into a lower-right and an upper-left triangle that tile the whole shape. (This is exactly the idea the Shoelace Theorem generalises to any polygon.)

WHAT: First find $M$ using the midpoint idea from the Section Formula (a midpoint is the section point at ratio $1:1$): $M=\left(\frac{8+4}{2},\frac{0+6}{2}\right)=(6,3).$ Area of $△ABC$ with $A(0,0),B(8,0),C(4,6)$: $S=0(0-6)+8(6-0)+4(0-0)=48,\text{Area}=24.$ Area of $△ABM$ with $A(0,0),B(8,0),M(6,3)$: $S=0(0-3)+8(3-0)+6(0-0)=24,\text{Area}=12.$ WHY the ratio is exactly a half: a median (here $AM$... note: $M$ is on $BC$, and $A$–$M$ splits nothing, but the segment $BM$ is half of $BC$). Triangles $ABM$ and $ABC$ share the base $AB$; since $M$ is the midpoint of $BC$, its height above line $AB$ is exactly half of $C$'s height. Half the height, same base $\Rightarrow$ half the area. $\frac{\text{Area}(△ABM)}{\text{Area}(△ABC)}=\frac{12}{24}=\frac{1}{2}.$ WHAT IT LOOKS LIKE: $M$ hangs at height $3$, halfway up to $C$'s height $6$; same base $AB$, half the height, half the paper.

WHAT — step 1, translate. Moving all three points by the same shift does not change the shape or its area, so slide $A$ to the origin. New corners: $A^{\prime }(0,0)$, $B^{\prime }(u_1,u_2)$, $C^{\prime }(v_1,v_2)$, where $u_1=x_2-x_1$, $u_2=y_2-y_1$, $v_1=x_3-x_1$, $v_2=y_3-y_1$. WHY translate: it kills all the $x_1,y_1$ clutter and lets $A^{\prime }$'s term vanish (since $x=y=0$ there), exposing the core quantity. WHAT — step 2, apply the formula to $A^{\prime }(0,0),B^{\prime }(u_1,u_2),C^{\prime }(v_1,v_2)$: $S=0(u_2-v_2)+u_1(v_2-0)+v_1(0-u_2)=u_1{v}_2-v_1{u}_2.$ Therefore $\text{Area}=\frac{1}{2}|S|=\frac{1}{2}|u_1{v}_2-u_2{v}_1|.■$ WHAT IT LOOKS LIKE / WHY this is the cross product: the quantity $u_1{v}_2-u_2{v}_1$ is precisely the $2\times 2$ determinant $\det \left(\begin{array}{cc}{u}_1& v_1\\ u_2& v_2\end{array}\right)$, which is the (signed) area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$. It is also the $z$-component of the 3D Vector Cross Product of $\vec{u}$ and $\vec{v}$. The triangle is exactly half that parallelogram — cut it along its diagonal — which is where the $\frac{1}{2}$ comes from.

Coordinate formula: $S=1(6-1)+4(1-2)+5(2-6)=5-4-20=-19,\text{Area}=\frac{19}{2}=\mathrm{9.5.}$ Cross-product method: $\vec{u}=B-A=(3,4)$, $\vec{v}=C-A=(4,-1)$. $u_1{v}_2-u_2{v}_1=(3)(-1)-(4)(4)=-3-16=-19,\text{Area}=\frac{1}{2}|-19|=\mathrm{9.5.}$ Agreement: both give $9.5$ square units. ✓ The two $S$-values are even identical ($-19$), because 5.1 proved the coordinate $S$ is the cross product — translating $A$ to the origin doesn't change $S$ at all.

WHAT: Write the area with $C=(x,y)$ general. $S=0(0-y)+4(y-0)+x(0-0)=4y,\text{Area}=\frac{1}{2}|4y|=2|y|.$ Set equal to $6$: $2|y|=6\Rightarrow |y|=3\Rightarrow y=3\text{ or }y=-3.$ WHY two lines: $|y|=3$ splits into $C$ being $3$ units above the base or $3$ units below it — both give the same area because area cares about distance, not direction. WHAT IT LOOKS LIKE: the locus is the pair of horizontal lines $y=3$ and $y=-3$. Slide $C$ anywhere along either line: the base $AB$ (length $4$) and the perpendicular height ($3$) never change, so the area stays $\frac{1}{2}(4)(3)=6$. ($x$ can be anything — the $x$-term dropped out because $A,B$ share $y=0$.)

The sign of $S$ (before $|\cdot |$) tells you what? ::: The walking direction around the vertices — positive $=$ counterclockwise, negative $=$ clockwise; the area itself is $\frac{1}{2}|S|$ either way.