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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)

PD6

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

PD8

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

PD4It 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 which 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 .

PD7

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

\frac{AR}{PS}=\frac{PR}{BS}=\frac{AP}{PB}

\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

\Rightarrow x = \frac{m_1x_2+m_2x_1}{m_1+m_2}

\frac{PR}{BS}=\frac{AP}{PB} \Rightarrow \frac{y-y_1}{y_2-y}=\frac{m_1}{m_2}

\Rightarrow y = \frac{m_1y_2+m_2y_1}{m_1+m_2}

Therefore the co-ordinates of P = (\frac{m_1x_2+m_2x_1}{m_1+m_2},  \frac{m_1y_2+m_2y_1}{m_1+m_2})

Note: If instead of m_1:m_2 you used k:1 , then the formula would become as follows:

P = (\frac{kx_2+x_1}{k+1},  \frac{ky_2+y_1}{k+1})

Points of Trisection

PD3

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)

Hence P = (\frac{x_2+2x_1}{3},  \frac{y_2+2y_1}{3})

Similarly, point Q divides the line segment AB in the ratio 2:1 \ (i.e. \ m_1=2, m_2=1)

Hence Q = (\frac{2x_2+x_1}{3},  \frac{2y_2+y_1}{3})

Midpoint Formula

PD2

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)

Hence P = (\frac{x_2+x_1}{2},  \frac{y_2+y_1}{2})

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:PD1

  1. First find the coordinates of the mid point 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.

 

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