Class 9: Distance Formula

$\displaystyle \textbf{Distance Formula}$

$\displaystyle \textbf{Introduction}$ $\displaystyle \text{For any two given points in a co-ordinate (Cartesian) plane,}$
$\displaystyle \text{the knowledge of co-ordinate geometry may be used to find:}$
$\displaystyle \textbf{(i) }\text{the distance between the given points,}$
$\displaystyle \textbf{(ii) }\text{the co-ordinates of a point which divides the line joining}$
$\displaystyle \text{the given points in a given ratio,}$
$\displaystyle \textbf{(iii) }\text{the co-ordinates of the mid-point of the line segment}$
$\displaystyle \text{joining the two given points,}$
$\displaystyle \textbf{(iv) }\text{the equation of the straight line through the given points,}$
$\displaystyle \textbf{(v) }\text{the equation of the perpendicular bisector of a line segment, etc.}$

$\displaystyle \textbf{The Distance Formula}$
$\displaystyle \textbf{To find the distance between two given points:}$
$\displaystyle \text{Let the two given points be }A(x_1,y_1)\text{ and }B(x_2,y_2).$
$\displaystyle \text{Draw }AC\text{ parallel to the }x\text{-axis and }BC\text{ parallel to the }y\text{-axis.}$
$\displaystyle \text{Then }\triangle ABC\text{ is a right-angled triangle.}$
$\displaystyle AC=x_2-x_1=\text{difference between the abscissae of }A\text{ and }B.$
$\displaystyle BC=y_2-y_1=\text{difference between the ordinates of }A\text{ and }B.$
$\displaystyle \text{Using Pythagoras' Theorem,}$
$\displaystyle AB^2=AC^2+BC^2$
$\displaystyle =(x_2-x_1)^2+(y_2-y_1)^2$
$\displaystyle \therefore AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\displaystyle \text{Hence, the distance between }(x_1,y_1)\text{ and }(x_2,y_2)\text{ is}$
$\displaystyle \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.$
$\displaystyle \text{Since }(a-b)^2=(b-a)^2,$
$\displaystyle \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$\displaystyle =\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$
$\displaystyle \therefore\ \text{Distance}$
$\displaystyle =\sqrt{(\text{difference of abscissae})^2+(\text{difference of ordinates})^2}.$
$\displaystyle \textbf{Note: }\text{The distance formula is valid for points lying in any quadrant,}$
$\displaystyle \text{provided the proper signs of their co-ordinates are used.}$

$\displaystyle \textbf{Circumcentre of a Triangle}$ $\displaystyle \text{The circumcentre of a triangle is the point which is equidistant}$
$\displaystyle \text{from all the three vertices of the triangle.}$
$\displaystyle \text{If }P\text{ is the circumcentre of }\triangle ABC,\text{ then}$
$\displaystyle PA=PB=PC=\text{circumradius}.$
$\displaystyle \text{If a circle is drawn with }P\text{ as centre and }PA,\ PB\text{ or }PC$
$\displaystyle \text{as radius, it passes through all the three vertices of the triangle.}$