This topic falls under “Coordinate Geometry“. We will learn the following topics today.

The Distance Formula (refer to the diagram as well)

Let the two given points be $A (x_1, y_1)$ and $B (x_2, y_2)$

In the diagram, you can see that the $\triangle ABC$ is a right angled triangle and $\angle C = 90^o$ . Which means that $AC^2+BC^2=AC^2$ (Pythagoras Theorem) Given $AC = x_2-x_1$ and $BC = y_2-y_1$

Hence $AB^2 = (x_2-x_1)^2 + (y_2-y_1)^2$ $\Rightarrow AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Therefore the distance between any two points $(x_1, y_1)$ and $(x_2, y_2)$ is $= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

Notes: If a point is on $x-axis$ , its ordinate is $0$ , therefore the point on $x-axis$ is taken as $(x,0)$ . Similarly, if the point is on $y-axis$ , its abscissa is $0$ , therefore the point on $y-axis$ is taken as $(0,y)$ .

Co-Linearity Three points are said to be co-linear if and only if $AC+CB=AB$ (as shown in the diagram). Which means that the distance from $A \ to \ C$ plus the distance from $C \ to \ B$ is equal to the distance from $A \ to \ B$ .

Circumcentre of  a Triangle It is a point that is equidistant from the three vertices of a triangle. i.e. if point $P$ is equidistant from the three vertices $A, B \ and \ C$ , then $AP = BP = CP = Circumradius$ of $\triangle ABC$ .

What this means is that if a circle is drawn with $P$ as the center and any of the vertices as the radius, the circle will touch all the three vertices of the $\triangle ABC$.

Section Formula

This is used to find a point that divides a line segment joining two points in a given ratio.

Let $AB$ be the line segment. Let coordinates of $A \ be \ (x_1,y_1)$ and $B \ be \ (x_2, y_2)$ .

Let $P$ be a point dividing $AB$ in the ratio of $m_1:m_2$.  $\displaystyle \text{i.e. } \frac{AP}{PB} = \frac{m_1}{m_2}$

We need to find $P(x,y)$ .

Refer to the diagram. $AR=LM=OM-OL=x-x_1$ $PR=PM-RM=PM-AL=y-y_1$ $PS=MN=ON-OM=x_2-x$ $BS=BN-SN=BN-PM=y_2-y$

Since $\triangle APR$ and $\triangle PBS$ are similar $\displaystyle \frac{AR}{PS} = \frac{PR}{BS} = \frac{AP}{PB}$ $\displaystyle \frac{AR}{PS} = \frac{AP}{PB} \Rightarrow \frac{x-x_1}{x_2-x} = \frac{m_1}{m_2}$ $\Rightarrow m_2x-m_2x_1=m_1x_2-m_1x$ $\Rightarrow m_1x+m_2x=m_1x_2+m_2x_1$ $\Rightarrow x(m_1+m_2)=m_1x_2+m_2x_1$ $\displaystyle \Rightarrow x = \frac{m_1x_2+m_2x_1}{m_1+m_2}$ $\displaystyle \frac{PR}{BS} = \frac{AP}{PB} \Rightarrow \frac{y-y_1}{y_2-y} = \frac{m_1}{m_2}$ $\displaystyle \Rightarrow y = \frac{m_1y_2+m_2y_1}{m_1+m_2}$ $\displaystyle \text{Therefore the co-ordinates of } P = \Big( \frac{m_1x_2+m_2x_1}{m_1+m_2} , \frac{m_1y_2+m_2y_1}{m_1+m_2} \Big)$

Note: If instead of $m_1:m_2$ you used $k:1$, then the formula would become as follows: $\displaystyle P = \Big( \frac{kx_2+x_1}{k+1} , \frac{ky_2+y_1}{k+1} \Big)$

Points of Trisection Let points $P \ and \ Q$ lie on a line segment $AB$ such that it divides the line in three equal parts i.e. $AP=PQ=QB$

Point $P$ divides the line segment $AB$ in the ratio of $1:2 \ (i.e. \ m_1 = 1, m_2 = 2)$ $\displaystyle \text{Hence } P = \Big( \frac{x_2+2x_1}{3} , \frac{y_2+2y_1}{3} \Big)$

Similarly, point $Q$ divides the line segment $AB$ in the ratio $2:1 \ (i.e. \ m_1=2, m_2=1)$ $\displaystyle \text{Hence } Q = \Big( \frac{2x_2+x_1}{3} , \frac{2y_2+y_1}{3} \Big)$

Midpoint Formula Let points $P$ lie on a line segment $AB$ such that it divides the line in two equal parts i.e. $AP=PB$

Point $P$ divides the line segment $AB$ in the ratio of $1:1 \ (i.e. \ m_1 = 1, m_2 = 1)$ $\displaystyle \text{Hence } P = \Big( \frac{x_2+x_1}{2} , \frac{y_2+y_1}{2} \Big)$

The centroid of a Triangle The centroid of a triangle is the point of intersection of its medians and the centroid divides each of the medians in the ratio of $2:1$.

To find the coordinates of the centroid:

1. First, find the coordinates of the midpoint of the sides of the triangle. $(m_1:m_2=1:1)$
2. The find out the coordinates of the centroid $(m_1:m_2=1:2)$ between the vertex and the opposite midpoint.