How Three Points Can Tell You Whether They Lie on One Line

Placing a figure on coordinate axes lets you measure lengths, areas and angles with arithmetic instead of rulers and protractors. Straight lines are the simplest case, and their equations are the foundation for circles, parabolas and the rest of coordinate geometry.

This part covers the section formula, incentre, area and collinearity, slopes and angles between lines, slope-intercept and two-point forms, and intercept form.

**The point dividing the join of $(x_1,y_1)$ and $(x_2,y_2)$ internally in the ratio $m:n$ is $\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$, the incentre is the average of the vertices weighted by the lengths of the opposite sides, and three points are collinear exactly when the triangle they form has zero area.

Formulae:

- Incentre** — $\left(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c},\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c}\right)$, where a, b and c are the sides opposite the three vertices
- Area — $\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

Worked example (section). The point dividing $A(2,-3)$ and $B(8,6)$ in the ratio $1:2$ is $\left(\frac{8+4}{3},\frac{6-6}{3}\right)=(4,0)$.

Worked example (incentre). For $A(0,0)$, $B(6,0)$ and $C(0,8)$, the sides are $a=BC=10$, $b=CA=8$ and $c=AB=6$:

$$\left(\frac{10(0)+8(6)+6(0)}{24},\frac{10(0)+8(0)+6(8)}{24}\right)=(2,2)$$

Worked example (collinearity). For $(1,2)$, $(3,6)$ and $(5,10)$: $1(6-10)+3(10-2)+5(2-6)=-4+24-20=0$, so the points are collinear.

An everyday example. Checking whether three electricity poles along a village road stand in a straight line from their map coordinates is exactly the area test.

The substance. The incentre can be checked with the inradius — here the area is 24 and the semi-perimeter 12, so $r=2$, matching the point $(2,2)$ in a right angle at the origin.

**The slope of the line through $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\tan \theta =\frac{y_2-y_1}{x_2-x_1}$, the acute angle between lines with slopes $m_1$ and $m_2$ satisfies $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$, parallel lines have $m_1=m_2$, and perpendicular lines have $m_1{m}_2=-1$.

Worked example.** Find the acute angle between lines with slopes 3 and $-\frac{1}{2}$:

$$\tan \theta =\left|\frac{3+\frac{1}{2}}{1-\frac{3}{2}}\right|=\left|\frac{3.5}{-0.5}\right|=7$$

so $\theta \approx 81.9°$.

Worked example 2. The line through $(2,3)$ and $(4,7)$ has slope 2, and the line through $(1,5)$ and $(5,3)$ has slope $-\frac{1}{2}$. Since $2\times \left(-\frac{1}{2}\right)=-1$, the lines are perpendicular.

An everyday example. A hill road rising 1 m for every 20 m along the ground has slope $\frac{1}{20}=0.05$, shown on road signs as a 5 per cent gradient.

The substance. The product test fails for vertical lines — a vertical and a horizontal line are perpendicular even though no slope product can be formed.

**A line with slope m that cuts the y-axis at $(0,c)$ has equation $y=mx+c$, and the line through $(x_1,y_1)$ and $(x_2,y_2)$ has equation $y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$.

Forms:

- Slope-intercept** — $y=mx+c$
- Point-slope — $y-y_1=m(x-x_1)$
- Two-point — find the slope from the two points, then use point-slope form

Worked example. Find the line through $(1,-2)$ and $(4,7)$.

$$m=\frac{7-(-2)}{4-1}=3,y+2=3(x-1)\Rightarrow y=3x-5$$

Check with $(4,7)$: $3\times 4-5=7$. The y-intercept is $-5$.

Worked example 2. The line with slope $-\frac{2}{3}$ and y-intercept 4 is $y=-\frac{2}{3}x+4$, or $2x+3y-12=0$.

An everyday example. A taxi fare of ₹50 plus ₹12 per kilometre is the line $y=12x+50$ — the slope is the rate and the intercept is the fixed charge.

The substance. **Two-point form fails when $x_1=x_2$** — the line is then vertical, with equation $x=x_1$.

**A line that cuts the x-axis at $(a,0)$ and the y-axis at $(0,b)$ has equation $\frac{x}{a}+\frac{y}{b}=1$, which shows both intercepts at a glance and gives the triangle it forms with the axes an area of $\frac{1}{2}|ab|$.

Derivation.** The two-point form through $(a,0)$ and $(0,b)$ gives $y=-\frac{b}{a}(x-a)$, which rearranges to $\frac{x}{a}+\frac{y}{b}=1$.

Worked example. Write $3x-4y=12$ in intercept form. Dividing by 12 gives $\frac{x}{4}+\frac{y}{-3}=1$, so the intercepts are 4 and $-3$, and the triangle with the axes has area $\frac{1}{2}\times 4\times 3=6$ square units.

Worked example 2. A line through $(2,3)$ has equal positive intercepts. From $\frac{x}{a}+\frac{y}{a}=1$, substituting gives $\frac{5}{a}=1$, so $a=5$ and the line is $x+y=5$.

An everyday example. A ladder leaning against a wall meets the floor and the wall at its two intercepts, so the line of the ladder is naturally described in intercept form.

The substance. Intercept form cannot describe a line through the origin — both intercepts are zero and the fractions are undefined.