Coordinate Geometry Class 10 Notes CBSE Maths Chapter 7 (Free PDF Download)

The following is a list of all the important topics that are covered under the chapter on Coordinate Geometry:

1. Equation of a Straight Line

2. Distance Formula

3. Distance from the Origin

4. Section Formula

5. Midpoint Formula

6. Area of a Triangle

7. Area of a Polygon

8. The Centroid of a Triangle

Access Class 10 Maths Chapter 7 - Coordinate Geometry

Important Terms and Concepts:

• “The coordinate axes are of perpendicular lines " \[XOX`\] " and " \[YOY`\] " intersecting at \[O\] .” 

• The plane is divided into four quadrants by the axes.

• The Cartesian plane is the plane that contains the axes.

• The lines " \[XOX`\] " and " \[YOY`\] " are known as the  \[x\] -axis and  \[y\] -axis, respectively, and are commonly drawn horizontally and vertically as seen in the picture.

• \[O\], is the Point of Intersection of the axes which is known as origin.

• Abscissae are the values of \[x\] measured along the \[x\] -axis from \[O\]. \[x\] has positive values along \[OX\], but negative values along \[OX'\] .

• Similarly, ordinate refers to the values of  \[y\] measured along the \[y\] axis from \[O\].

• \[y\] has positive values along \[OY\], but negative values along \[OY'\].

• The coordinates of a point are the ordered pair containing the abscissa and ordinate of a point.

Distance Formula:

• To find the distance two points \[\text{A}\left( {{x}_{\text{1}}}\text{ , }{{\text{y}}_{\text{1}}} \right)\]and \[B\left( {{\text{x}}_{2}}\text{ , }{{\text{y}}_{2}} \right)\]

• From the Figure, 

$\text{AC = }{{\text{x}}_{\text{2}}}\text{- }{{\text{x}}_{\text{1}}}$

$\text{BC = }{{\text{y}}_{\text{2}}}\text{- }{{\text{y}}_{\text{1}}}$

Therefore, In \[\Delta ABC\],

By using the Pythagoras Theorem, we get

$\text{A}{{\text{B}}^{\text{2}}}\text{ = A}{{\text{C}}^{\text{2}}}\text{ + B}{{\text{C}}^{\text{2}}}$

$\text{A}{{\text{B}}^{\text{2}}}\text{ = }{{\left( {{\text{x}}_{\text{2}}}\text{- }{{\text{x}}_{\text{1}}} \right)}^{\text{2}}}\text{+ }{{\left( {{\text{y}}_{\text{2}}}\text{- }{{\text{y}}_{\text{1}}} \right)}^{\text{2}}}$

$\text{AB  = }\sqrt{{{\left( {{\text{x}}_{\text{2}}}\text{- }{{\text{x}}_{\text{1}}} \right)}^{\text{2}}}\text{+ }{{\left( {{\text{y}}_{\text{2}}}\text{- }{{\text{y}}_{\text{1}}} \right)}^{\text{2}}}}$

Section Formula:

• To find the coordinates of a point which divides the line segment joining two given points in a given ratio (internally).

• Let \[P\left( x\text{ , y} \right)\]divide the join of \[\text{A}\left( {{x}_{\text{1}}}\text{ , }{{\text{y}}_{\text{1}}} \right)\] and \[B\left( {{\text{x}}_{2}}\text{ , }{{\text{y}}_{2}} \right)\] in the Ratio \[\text{m:n}\].

Here,

$\angle \text{ PAC = }\angle \text{ BPD}$

$\angle \text{ PCA = }\angle \text{ BDP}={{90}^{0}}$

Therefore, by using \[AA\] similarity method, we get,

$AC=x-{{\text{x}}_{1}}$

$PD={{\text{x}}_{2}}-x$

From Similarity Property, we get,

$\dfrac{\text{AC}}{\text{PD}}\text{ = }\dfrac{\text{m}}{\text{n}}$

$\text{        = }\dfrac{\text{x-}{{\text{x}}_{\text{1}}}}{{{\text{x}}_{\text{2}}}\text{-x}}$ 

$\text{        =}\dfrac{\text{m}}{\text{n}}$

Now, make \[x\] the subject of the formula, 

$\text{   nx - n}{{\text{x}}_{\text{1}}}\text{ = m}{{\text{x}}_{\text{2}}}\text{ - mx}$

$\text{  Mx + nx = m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}$

$\text{X}\left( \text{m + n} \right)\text{ = m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}$

Therefore,

\[\text{X = }\dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}}\]

Similarly, we can show that

\[\text{Y = }\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{ + n}{{\text{y}}_{\text{1}}}}{\text{m + n}}\]

Thus, the Co-ordinates of \[P\] are \[\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}},\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{ + n}{{\text{y}}_{\text{1}}}}{\text{m + n}} \right)\] .

Mid−Point Formula:

• If \[P\] is the mid−point of \[AB\], then \[m:n\] , 

Therefore, the ratio becomes \[1:1\]

And hence,

$\text{X = }\dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}}$ 

$\text{X = }\dfrac{{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{\text{1}}}}{\text{1 + 1}}$. 

$\text{X = }\dfrac{{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{\text{1}}}}{\text{2}}$

Similarly, we get

\[\text{Y = }\dfrac{{{\text{y}}_{1}}\text{ + }{{\text{y}}_{2}}}{\text{2}}\]

Thus, the Co-ordinate of \[P\]are \[\left( \dfrac{{{\text{x}}_{1}}\text{ + }{{\text{x}}_{2}}}{\text{2}},\dfrac{{{\text{y}}_{1}}\text{ + }{{\text{y}}_{2}}}{2} \right)\]

• Note:

When the point \[P\] divides the line joining \[AB\]  in the ration \[m:n\]  externally then,

The Co-ordinates of \[P\] are \[\left( \dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ - n}{{\text{x}}_{\text{1}}}}{\text{m - n}},\dfrac{\text{m}{{\text{y}}_{\text{2}}}\text{ - n}{{\text{y}}_{\text{1}}}}{\text{m - n}} \right)\]

The Centroid of a Triangle:

• The centroid is the point of intersection of three medians. 

• It is the point of intersection of a median

• $\text{AG : GD}$  is \[2:1\]

To Find the Coordinates of the Centroid of a Triangle:

Let the coordinates of the vertices of \[\vartriangle ABC\] be \[\text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{, B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\] and \[\text{ C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\]

Let \[\text{G}\left( \text{x,y} \right)\]be the centroid of the \[\vartriangle ABC\]

Here, \[\text{D}\] is the midpoint of \[\text{BC}\] , and hence

By applying the mid−point formula, we get

\[\text{a = }\dfrac{{{\text{x}}_{2}}\text{+}{{\text{x}}_{3}}}{\text{2}}\] and \[\text{b = }\dfrac{{{\text{y}}_{2}}\text{+}{{\text{y}}_{3}}}{\text{2}}\]

We know that,

In a \[2:1\] ratio, point \[G\] splits the median.

Therefore, by applying the section formula, we get,

$\text{X = }\dfrac{\text{m}{{\text{x}}_{\text{2}}}\text{ + n}{{\text{x}}_{\text{1}}}}{\text{m + n}}$

$\text{X = }\dfrac{\text{2}\left( a \right)\text{ + 1}\left( {{\text{x}}_{\text{1}}} \right)}{\text{2 + 1}}$

$\text{X = }\dfrac{\text{2}\left( \dfrac{{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{3}}}{2} \right)\text{ + }{{\text{x}}_{\text{1}}}}{3}$

$\text{X = }\dfrac{{{\text{x}}_{\text{1}}}+{{\text{x}}_{\text{2}}}\text{ + }{{\text{x}}_{3}}\text{ }}{3}$

Similarly,

$Y=\dfrac{\text{2}\left( b \right)\text{ + 1}\left( {{y}_{\text{1}}} \right)}{\text{2 + 1}}$

$\text{Y = }\dfrac{\text{2}\left( \dfrac{{{y}_{\text{2}}}\text{ + }{{\text{y}}_{3}}}{2} \right)\text{ + }{{\text{y}}_{\text{1}}}}{3}$

$\text{Y = }\dfrac{{{y}_{1}}\text{ + }{{\text{y}}_{3}}\text{ + }{{\text{y}}_{3}}}{3}$

Therefore, Coordinates of the centroid are \[\left( \dfrac{{{\text{x}}_{1}}\text{ + }{{\text{x}}_{2}}+{{x}_{3}}}{3},\dfrac{{{\text{y}}_{1}}\text{ + }{{\text{y}}_{2}}+{{y}_{3}}}{3} \right)\]

Area of the Triangle:

To find the area of a triangle whose vertices are \[\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{, }\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\] and \[\text{ }\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\]

Let, \[\text{A}\left( {{\text{x}}_{\text{1}}}\text{,}{{\text{y}}_{\text{1}}} \right)\text{, B}\left( {{\text{x}}_{\text{2}}}\text{,}{{\text{y}}_{\text{2}}} \right)\] and \[\text{ C}\left( {{\text{x}}_{\text{3}}}\text{,}{{\text{y}}_{\text{3}}} \right)\]be the vertices of a triangle \[\vartriangle ABC\]

$\text{Area of }\vartriangle \text{ ABC }$

$\text{= Area of Trapezium ABML + Area of Trapezium ALNC}$

$\text{    - Area of Trapezium BMNC}$

$\text{= }\dfrac{1}{2}ML\left( MB+LA \right)+\dfrac{1}{2}LN\left( LA+NC \right)-\dfrac{1}{2}MN\left( MB+NC \right)$

$=\dfrac{1}{2}\left( {{x}_{1}}-{{x}_{2}} \right)\left( {{y}_{2}}+{{y}_{1}} \right)+\dfrac{1}{2}\left( {{x}_{3}}-{{x}_{1}} \right)\left( {{y}_{1}}+{{y}_{3}} \right)-\dfrac{1}{2}\left( {{x}_{3}}-{{x}_{2}} \right)\left( {{y}_{2}}+{{y}_{3}} \right)$ 

$=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Arrow Method:

It is to obtain the formula for the area of the triangle

$ \dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\\end{matrix} \right| $

$=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$

Note:

1. If the points \[\text{A, B}\] and \[\text{C}\] we take in the anticlockwise direction, then the area will be positive. 

2. If the points we take in a clockwise direction, the area will be negative.

3. So we always take the absolute value of the area calculated.

Therefore, 

Area of triangle

$=\dfrac{1}{2}\left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right]$.

4. If the area of a triangle is zero, then the three points are collinear.