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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,  {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  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 by OP = \sqrt {{x^2} + {y^2}}
For Example: 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 are \left( {{{m{x_2} + n{x_1}} \over {m + n}},{{m{y_2} + n{y_1}} \over {m + n}}} \right)
For Example: 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: 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             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             x = {{12 - 4} \over 7}, y = {{ - 21 + 12} \over 7}
 \Rightarrow             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: 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}) are \left( {{{{x_1} + {x_2} + {x_3}} \over 3},{{{y_1} + {y_2} + {y_3}} \over 3}} \right)
For Example: Find centroid of triangle whose vertices are (1,4), (-1,-1), (3,-2)
Solution: 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}) is {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| 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: Find area of triangle whose vertices are (6,3), (-3,5), (4, 2)
Solution: 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}  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: 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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