# Co-ordinate Geometry - Class 10 : Notes

Notes for coordinate geometry chapter of class 10 Mathematics. Dronstudy provides free comprehensive chapterwise class 10 Mathematics notes with proper images & diagram.

(1) The abscissa and ordinate of a given point are the distances of the point from y-axis and x-axis respectively.

(2) The coordinates of any point on x-axis are of the form (x,0).

(3) The coordinates of any point on y-axis are of the form (0,y).

(4) The distance between points $P({x_1},{y_1})$ and $Q({x_2},{y_2})$ is given by,
$PQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{({y_2} - {y_1})}^2}}$
For Example:
If $P({-6},{7})$ and $Q({-1},{-5})$, then distance between these two point is given as follow:
Here, ${x_1} = - 6$, ${y_1} = 7$
${x_2} = -1$, Â ${y_2} = -5$
Let O be the distance between two points $(-6, 7)$ and $(-1, -5)$.
The distance between two point is given by
$D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$
$= \sqrt {{{\left( { - 1 - \left( { - 6} \right)} \right)}^2} + {{\left( { - 5 - 7} \right)}^2}}$
$= \sqrt {{{\left( { - 1 + 6} \right)}^2} + {{\left( { - 12} \right)}^2}}$
$= \sqrt {{{\left( 5 \right)}^2} + {{\left( { - 12} \right)}^2}}$
$= \sqrt {25 + 144}$
$= \sqrt {169}$
$= 13$ Â $units$
Hence the distance between two points is 13 units.

(5) Distance of a point $P(x,y)$ from the origin $O(0,0)$ is given byÂ $OP = \sqrt {{x^2} + {y^2}}$
For Example:Â $P({-6},{7})$ and $O({0},{0})$ is given then distance between them i.e. $OP$ is calculated as follow:
$OP = \sqrt {{{\left( {{x_2} - {0}} \right)}^2} + {{({y_2} - {0})}^2}}$
$OP = \sqrt {{x^2} + {y^2}}$
$OP = \sqrt {{6^2} + {7^2}}$
$OP = \sqrt {85}$

(6) The coordinates of the point which divides the join of points $P({x_1},{y_1})$ and $Q({x_2},{y_2})$ internally in the ration m:n areÂ $\left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)$
For Example:Â Find the coordinates of the point which is divided the line segment joining (-1,3) and (4,-7) internally in the ratio 3:4.
Solution:Â Let the end points of AB be A(-1, 3) and B(4, -7).
(${x_1} = - 1$, ${y_1} = 3$) and (${x_2} = 4$, ${y_2} = - 7$)
Also, $m = 3$ and $n = 4$
Let $P(x, y)$ be the required point, then by section formula, we have $x = {{m{x_2} + n{x_1}} \over {m + n}}$, $y = {{m{y_2} + n{y_1}} \over {m + n}}$
$\Rightarrow$Â Â Â Â Â Â Â Â Â Â Â  $x = {{3 \times 4 + 4 \times \left( { - 1} \right)} \over {3 + 4}}$, $y = {{3 \times \left( { - 7} \right) + 4 \times 3} \over {3 + 4}}$
$\Rightarrow$Â Â Â Â Â Â Â Â Â Â Â  $x = {{12 - 4} \over 7}$, $y = {{ - 21 + 12} \over 7}$
$\Rightarrow$Â Â Â Â Â Â Â Â Â Â Â  $x = {8 \over 7}$, $y = {{ - 9} \over 7}$
Hence, the required point is $P\left( {{8 \over 7},{{ - 9} \over 7}} \right)$

(7) The coordinates of the mid-point of the line segment joining the points $P({x_1},{y_1})$ and $Q({x_2},{y_2})$ are $\left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right)$.
For Example:Â $P({4},{8})$ and $Q({6},{10})$
$Mid-Point = \left( {{{{4} + {6}} \over 2},{{{8} + {10}} \over 2}} \right)$
So mid-point of $PQ$ is $M(5, 9)$.

(8) The coordinates of the centroid of triangle formed by the points $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$ areÂ $\left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)$
For Example:Â Find centroid of triangle whose vertices are $(1,4)$, $(-1,-1)$, $(3,-2)$
Solution:Â We know that the coordinates of the centroid of a triangle whose angular points are $({x_1},{y_1})$, $({x_2},{y_2})$ and $({x_3},{y_3})$ are
$\left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)$
So, the coordinates of the centroid of a triangle whose vertices are $(1, 4)$, $(-1,-1)$ and $(3, -2)$ are
$\left( {{{1 - 1 + 3} \over 3},{{4 - 1 - 2} \over 3}} \right) = \left( {1,{1 \over 3}} \right)$

(9) The are of triangle formed by the points $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$ isÂ ${1 \over 2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \right|$Â Or, ${1 \over 2}\left| {({x_1}{y_2} + {x_2}{y_3} + {x_3}{y_1}) - ({x_1}{y_3} + {x_2}{y_1} + {x_3}{y_2})} \right|$
For Example:Â Find area of triangle whose vertices are $(6,3)$, $(-3,5)$, $(4, 2)$
Solution:Â Let $A = \left( {{x_1},{y_1}} \right) = \left( {6,3} \right)$, $B = \left( {{x_2},{y_2}} \right) = \left( { - 3,5} \right)$ and $C = \left( {{x_3},{y_3}} \right) = \left( {4, - 2} \right)$ be the given points
Area of $\Delta ABC = {1 \over 2}\left| {\left\{ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right\}} \right|$
$= {1 \over 2}\left| {\left\{ {6\left( {5 - \left( { - 2} \right)} \right) - 3\left( { - 2 - 3} \right) + 4\left( {3 - 5} \right)} \right\}} \right|$
$= {1 \over 2}\left| {\left\{ {6 \times 7 + 15 - 8} \right\}} \right|$
$= {1 \over 2}\left| {57 - 8} \right|$
$= {{49} \over 2}$ Â $sq.units$

(10) If points $A({x_1},{y_1})$, $B({x_2},{y_2})$ and $C({x_3},{y_3})$ are collinear, then
${{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})}$
For Example:Â Let $A(2,5)$, $B(4, 6)$ and $C(8, 8)$ be the given points.
Three points are collinear if area enclosed by three points is zero.
Area of $\Delta ABC = {1 \over 2}\left| {\left\{ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right\}} \right|$
$= {1 \over 2}\left| {2(6 - 8) + 4(8 - 5) + 8(5 - 6)} \right|$
$= {1 \over 2}\left| {2 \times ( - 2) + 4 \times (3) + 8 \times ( - 1)} \right|$
$= {1 \over 2}\left| { - 4 + 12 - 8} \right|$
$= {1 \over 2}\left| { - 12 + 12} \right|$
$= 0$

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