Coordinate System and Coordinates
5.0 Distance formulae
5.0 Distance formulae
The distance between two points $A({x_1},{y_1})$ and $B({x_2},{y_2})$ is given by $$\left| {AB} \right| = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $$
Proof: Let $A({x_1},{y_1})$ and $B({x_2},{y_2})$ be any two points in the plane. From $A$ and $B$ draw $AE$ and $BD$ perpendicular to $X$-axis. From $A$ draw $AC \bot BD$. Then, as shown in figure, $$OE = {x_1},{\text{ }}OD = {x_2},{\text{ }}AE = {y_1}{\text{ and }}BD = {y_2}$$$$\therefore {\text{ }}ED = {x_2} - {x_1}{\text{ and }}BC = {y_2} - {y_1}$$
Apply pythagoras theorem in $\Delta ACD$, we get $$\begin{equation} \begin{aligned} A{B^2} = A{C^2} + B{C^2} \\ AB = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} \\\end{aligned} \end{equation} $$
Question 1. Find the value of $x$, if the distance between the points $(x,-1)$ and $(3,2)$ is $5$.
Solution: Let $P(x,-1)$ and $Q(3,2)$ be the given points. Then $PQ=5$(given). Therefore, using distance formulae we get, $$\sqrt {{{(x - 3)}^2} + {{( - 1 - 2)}^2}} = 5$$ Squaring both sides, we get$$\begin{equation} \begin{aligned} {(x - 3)^2} + 9 = 25 \\ x = 7{\text{ or }}x = - 1 \\\end{aligned} \end{equation} $$
Question 2. Find the distance between the points $(a\cos \alpha ,a\sin \alpha )$ and $(a\cos \beta ,a\sin \beta )$ where $a>0$.
Solution: Let $P \equiv (a\cos \alpha ,a\sin \alpha )$ and $Q \equiv (a\cos \beta ,a\sin \beta )$
then, using distance formulae,
$$\begin{equation} \begin{aligned} \left| {PQ} \right| = \sqrt {{{(a\cos \alpha - a\cos \beta )}^2} + {{(a\sin \alpha - a\sin \beta )}^2}} \\ = \sqrt {{a^2}\{ {{(\cos \alpha - \cos \beta )}^2} + {{(\sin \alpha - \sin \beta )}^2}\} } \\ = \sqrt {{a^2}({{\cos }^2}\alpha + {{\cos }^2}\beta - 2\cos \alpha \cos \beta + {{\sin }^2}\alpha + {{\sin }^2}\beta - 2\sin \alpha \sin \beta )} \\ = \sqrt {{a^2}(2 - 2\cos (\alpha - \beta ))} \\ = \sqrt {2{a^2}(1 - \cos (\alpha - \beta ))} \\ = \sqrt {2{a^2}.2{{\sin }^2}(\frac{{\alpha - \beta }}{2})} \\ = \sqrt {4{a^2}{{\sin }^2}(\frac{{\alpha - \beta }}{2})} \\ = \left| {2a\sin (\frac{{\alpha - \beta }}{2})} \right| \\ = 2a\left| {\sin (\frac{{\alpha - \beta }}{2})} \right|{\text{ (}}\because {\text{ a > 0)}} \\\end{aligned} \end{equation} $$
