Exercises — Collinearity of three points

What it looks like: every point has $y=x$, so they climb the same $45^{\circ }$ diagonal. Look at the green line in the figure below — no bend at $B$.

Confirm with $\Delta$ (here $x_1,y_1=1,1;x_2,y_2=2,2;x_3,y_3=3,3$): $\Delta =1(2-3)+2(3-1)+3(1-2)=-1+4-3=0.$ Answer: collinear. ✓

Why suspect NO: $Q$ and $R$ share the same height $y=4$, but $P$ is at height $0$. To go $P\to Q$ you rise steeply; to go $Q\to R$ you rise not at all (flat). That is a turn. $\Delta =0(4-4)+1(4-0)+2(0-4)=0+4-8=-4\ne 0.$ Answer: NOT collinear. They enclose a triangle of area $\frac{1}{2}|-4|=2$.

A point $(x,y)$ lies on a line through the origin when $y/x$ is a fixed ratio (the slope).

• (a): $(1,2)$ gives $2/1=2$; $(3,6)$ gives $6/3=2$. Same slope through origin → collinear.
• (b): $(3,5)$ gives $5/3\approx 1.67\ne 2$. Not collinear.

Cross-check (a) with $\Delta$ using $(0,0),(1,2),(3,6)$: $\Delta =0(2-6)+1(6-0)+3(0-2)=6-6=0.✓$

Substitute $x_1,y_1=-2,-1;x_2,y_2=1,0;x_3,y_3=4,1$: $\Delta =(-2)(0-1)+1(1-(-1))+4((-1)-0)$ $=(-2)(-1)+1(2)+4(-1)=2+2-4=0.$ Collinear. Notice negative coordinates cause no trouble — $\Delta$ just tracks signed differences.

Why slope is banned here: slope needs $x_2-x_1$ in the denominator. Here $x_2-x_1=0$ — division by zero. This is a vertical line.

Correct test: all three $x$-coordinates equal $2$, so they all sit on the vertical line $x=2$ → collinear. Confirm with $\Delta$: $\Delta =2(7-(-5))+2((-5)-3)+2(3-7)=2(12)+2(-8)+2(-4)=24-16-8=0.✓$

$\Delta =1(1-5)+5(5-1)+3(1-1)=-4+20+0=16.$ Area $=\frac{1}{2}|16|=8$. Since $\Delta =16\ne 0$, the points are NOT collinear — a genuine triangle.

Set $\Delta =0$ with $x_1,y_1=2,3;x_2,y_2=4,k;x_3,y_3=6,7$: $2(k-7)+4(7-3)+6(3-k)=0$ $2k-14+16+18-6k=0$ $-4k+20=0⟹k=5.$ Check (slope): slope $AB=\frac{5-3}{4-2}=1$, slope $BC=\frac{7-5}{6-4}=1$. Equal ✓. So $k=5$.

Here $x_1,y_1=p,2;x_2,y_2=3,4;x_3,y_3=5,p$: $p(4-p)+3(p-2)+5(2-4)=0$ $4p-p^2+3p-6-10=0$ $-p^2+7p-16=0⟹p^2-7p+16=0.$ Discriminant: $(-7{)}^2-4(1)(16)=49-64=-15<0$. Answer: no real $p$ exists. Whatever real value you pick, these three points always bend — there is no way to line them up. The negative discriminant is the algebra telling you the geometry is impossible.

$\Delta =0(0-b)+a(b-0)+0(0-0)=ab.$ $\Delta =ab=0$ requires $a=0$ or $b=0$. But we assumed $a,b\ne 0$, so $\Delta =ab\ne 0$ — never collinear. This makes sense: $B$ sits on the $x$-axis, $C$ on the $y$-axis, and $A$ at the corner — a right triangle of area $\frac{1}{2}|ab|$.

By the Section Formula, the midpoint is $M=\left(\frac{1+3}{2},\frac{4+(-2)}{2}\right)=(2,1).$ Any midpoint lies on segment $AB$ by construction, so collinearity must hold. Verify with $\Delta$ on $A(1,4),B(3,-2),M(2,1)$: $\Delta =1(-2-1)+3(1-4)+2(4-(-2))=-3-9+12=0.✓$ General truth: any point produced by the section formula on $AB$ is collinear with $A$ and $B$, since it is defined as living on that segment.

(a) $\Delta =0$ with $x_1,y_1=1,2;x_2,y_2=4,8;x_3,y_3=2,k$: $1(8-k)+4(k-2)+2(2-8)=0$ $8-k+4k-8-12=0⟹3k-12=0⟹k=4.$ So $C(2,4)$ makes them collinear.

(b) With $k=5$, $C(2,5)$: $\Delta =1(8-5)+4(5-2)+2(2-8)=3+12-12=3.$ Area $=\frac{1}{2}|3|=1.5$. Non-zero ⇒ a real (thin) triangle.

Vector test (Vectors and Cross Product): form direction vectors from $P$. $\overrightarrow{PQ}=(4-1,3-1)=(3,2),\overrightarrow{PR}=(7-1,5-1)=(6,4).$ The 2D cross product is $3\cdot 4-2\cdot 6=12-12=0$, so $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ are parallel ⇒ collinear (and linearly dependent).

Parametric form (Parametric Equations): points on line $PQ$ are $\mathbf{r}(t)=P+t\overrightarrow{PQ}=(1+3t,1+2t)$. We need $\mathbf{r}(t)=R=(7,5)$: from $1+3t=7$ we get $t=2$; check $1+2(2)=5$ ✓. So $R=\mathbf{r}(2)$ — it lies on line $PQ$ at parameter $t=2$.

By the Section Formula, $B$ dividing $AC$ in ratio $m:n$ means $x_2=\frac{m{x}_3+n{x}_1}{m+n},y_2=\frac{m{y}_3+n{y}_1}{m+n}.$ Plug into $\Delta =x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$. Multiply everything by $(m+n)$ to clear denominators; write $s=m+n$: $s\Delta =x_1(m{y}_3+n{y}_1-s{y}_3)+(m{x}_3+n{x}_1)(y_3-y_1)+x_3(s{y}_1-m{y}_3-n{y}_1).$ Simplify the bracketed $y$-terms using $s=m+n$: $m{y}_3+n{y}_1-(m+n)y_3=n(y_1-y_3)$, and $(m+n)y_1-m{y}_3-n{y}_1=m(y_1-y_3)$. So $s\Delta =n{x}_1(y_1-y_3)+(m{x}_3+n{x}_1)(y_3-y_1)+m{x}_3(y_1-y_3).$ Factor $(y_1-y_3)$ (note $y_3-y_1=-(y_1-y_3)$): $s\Delta =(y_1-y_3)[n{x}_1-(m{x}_3+n{x}_1)+m{x}_3]=(y_1-y_3)\cdot 0=0.$ Since $s=m+n\ne 0$, $\Delta =0$. QED — any section-formula point is collinear with its endpoints, exactly as Q10 hinted.

Set up $\Delta$ with $x_1,y_1=0,0;x_2,y_2=100,0.01;x_3,y_3=200,k$ and require it to equal $0.02$: $\Delta =0(0.01-k)+100(k-0)+200(0-0.01)=100k-2.$ Set $100k-2=0.02⟹100k=2.02⟹k=\mathrm{0.0202.}$

Comment: $\Delta =0.02$ is tiny but not zero, so strictly the points are not collinear — they bow by area $\frac{1}{2}(0.02)=0.01{\text{m}}^2$ over a $200$m span. In surveying that is negligible ("practically collinear"), but mathematically the test is binary: only $\Delta =0$ is collinear. This is the punchline of the whole topic — collinearity is an exact, all-or-nothing condition, and $\Delta$ measures precisely how far you are from it.

Q1 collinear · Q2 not ($\Delta =-4$) · Q3 (a) · Q4 collinear · Q5 collinear (vertical) · Q6 not, area $8$ · Q7 $k=5$ · Q8 no real $p$ · Q9 never · Q10 $M=(2,1)$ · Q11 $k=4$, area $1.5$ · Q12 $t=2$ · Q13 proof · Q14 $k=0.0202$, not collinear.