Co-Ordinate Geometry
The geometry word is combination of two words 'Geo' and 'Metry'. Geo Means 'Land or Earth' and Metry means 'Measuring'.Geometry means that " Measuring of land or geo " .Geometry begins with a point and straight line. Uptil now, we have studied geometry without any use of algebra. Similarly, 'Co-ordinate geometry' is the branch of mathematics in which geometry is studied algebraically or we can say that geometric figures are studied with the help of equations.
For example, when any river flooded, then its bank would take away the boundaries between the fields of different people. Now, to re-make them, its area was calculated on the basis of geometry. Similarly, in order to make bricks they needed a proportion of their length, width and thickness or height. Their ratio was 4 : 2 : 1.
Cartesian System :-
When we use the method to show the position of any point in any given bottom plane , then that method is called the Cartesian method. It has two perpendicular lines, one horizontal line and the other vertical line. The horizontal line is called the X-axis and the vertical line is called the Y-axis . The two lines where each another intersects, it is called the origin point. It is denoted by O .There are four Quadrants Ⅰ, Ⅱ, Ⅲ and Ⅳ, Which are located in the anti - clockwise direction from X-axis. As shown in the given below.
In Ⅰ quadrant , the sign of ( x , y ) is ( + , + ) .
In Ⅱ quadrant , the sign of ( x , y ) is ( - , + ) .
In Ⅲ quadrant , the sign of ( x , y ) is ( + , - ) .
In Ⅳ quadrant , the sign of ( x , y ) is ( - , - ) .
Some Fundamental Formulae :-
1. Distance Formula Between Two Points :-
Let us consider two points P( x_{1} , y_{1} ) and Q( x_{2} , y_{2} ) are given , then
Example 1. Find the distance between any two points P( 2 , 3 ) and Q( -4 , 6 ) .
Sol :- Here , x_{1} = 2 , x_{2} = -4 , y_{1} = 3 and y_{2} = 6
PQ = √(-4 - 2)² + (6 - 3)²
= √(-6)² + (3)²
= √36 + 9 = √45
∴ PQ = 3√5
2. Division Formula :-
( a ) Internal Division Formula :-
Let us consider two points P( x_{1} , y_{1} ) and Q( x_{2} , y_{2} ) are given and point R( x , y ), which divides the join of two given points P and Q internally in the ratio m_{1} : m_{2} , then
coordinates of point R( x , y ) = | , | ||||
m_{1} + m_{2} | m_{1} + m_{2} |
Example 2 . Find the coordinates of the point which divides the join of points P( 3 , -7 ) and Q( 4 , 5 ) internally in the ratio 2 : 3 .
Sol :- Here , x_{1} = 3 , x_{2} = 4 , y_{1} = -7 , y_{2} = 5
m_{1} = 2 and m_{2} = 3
coordinates of point R( x , y ) = | , | ||||
2 + 3 | 2 + 3 |
= | , | ||||
5 | 5 |
Hence coordinates of the point of division are | , | ||||
5 | 5 |
( b ) External Division Formula :-
Let us consider two points P( x_{1} , y_{1} ) and Q( x_{2} , y_{2} ) are given and point R( x , y ), which divides the join of two given points P and Q externally in the ratio m_{1} : m_{2} , then
coordinates of point R( x , y ) = | , | ||||
m_{1} - m_{2} | m_{1} - m_{2} |
Example 3. Find the coordinates of the point which divides the join of points P( 3 , -7 ) and Q( 4 , 5 ) externally in the ratio 2 : 3 .
Sol :- Here , x_{1} = 3 , x_{2} = 4 , y_{1} = -7 , y_{2} = 5
m_{1} = 2 and m_{2} = 3
∴ the coordinates of point of division = | , | ||||
2 - 3 | 2 - 3 |
= | , | ||||
-1 | -1 |
( c ) Mid point Formula :-
If point R( x , y ) is the mid point of P( x_{1} , y_{1} ) and Q( x_{2} , y_{2} ), then
The coordinates of point R( x , y ) = | , | ||||
2 | 2 |
Example 4. What will be the coordinates of the mid point which divides the join of points P( 3 , -7 ) and Q( 4 , 5 ) .
Sol :- Here , x_{1} = 3 , x_{2} = 4 , y_{1} = -7 , y_{2} = 5
The coordinates of mid point = | , | ||||
2 | 2 |
= | , | ||||
2 | 2 |
Therefore the coordinates of the mid point are | , | ||||
2 | 2 |
3. Centroid of a triangle :-
When all the medians of a triangle intersect at each other, that point is called the centroid of the triangle. It divides the median in the ratio 2 : 1.
Let three vertices of a triangle ( x_{1} , y_{1} ) , ( x_{2} , y_{2} ) and ( x_{3} , y_{3} ) are given , then
= | , | ||||
3 | 3 |
Example 5. Find the coordinates of the centroid of a triangle whose vertices are ( 0 , 6 ) , ( 8 , 12 ) and ( 8 , 0 ) ?
Sol :- Here , x_{1} = 0 , x_{2} = 8 , x_{3} = 8 , y_{1} = 6 y_{2} = 12 , y_{3} = 0
∴ The coordinates of the centroid of a triangle = | , | ||||
3 | 3 |
= | , | ||||
3 | 3 |
Hence , the coordinates of the centroid of a triangle are | , 6 | |||
3 |
4. Incentre of a triangle :-
When all the angles of a triangle are subtended , then the intersection point of these three angles is called the incentre of the triangle.
Let three vertices of a triangle ( x_{1} , y_{1} ) , ( x_{2} , y_{2} ) and ( x_{3} , y_{3} ) are given , then
= | , | ||||
a + b + c | a + b + c |
Example 6. Find the coordinates of the incentre of a triangle whose vertices are ( 4 , -2 ) , ( 5 , 5 ) and ( -2 , 4 ) ?
Sol :- Here , x_{1} = 4 , x_{2} = 5 , x_{3} = -2 , y_{1} = -2 y_{2} = 5 , y_{3} = 4
a = BC = √(-2 - 5)^{2} + (4 - 5)^{2}
= √(-7)^{2} + (-1)^{2}
= √49 + 1 = √50
a = 5√2
b = CA = √(4 + 2)^{2} + (-2 - 4)^{2}
= √(6)^{2} + (-6)^{2}
= √36 + 36 = √72
b = 6√2
c = AB = √(5 - 4)^{2} + (5 + 2)^{2}
= √(1)^{2} + (7)^{2}
= √1 + 49 = √50
c = 5√2
∴ The coordinates of the incentre of a triangle = | , | ||||
5√2 + 6√2 + 5√2 | 5√2 + 6√2 + 5√2 |
= | , | ||||
16 | 16 |
= | , | ||||
16 | 16 |
Hence , the coordinates of the incentre of a triangle are | , | ||||
2 | 2 |
5. Area of triangle :-
Let any triangle whose coordinates ( x_{1} , y_{1} ) , ( x_{2} , y_{2} ) and ( x_{3} , y_{3} ) of vertices P , Q and R are given respectively , then
Area of triangle ( A ) = | x_{1}( y_{2} - y_{3} ) + x_{2}( y_{3} - y_{1} ) + x_{3}( y_{1} - y_{2} ) | |||
2 |
Example 7. Find the area of triangle whose the coordinates of vertices are A ( 6 , -6 ) , B ( 3 , -7 ) and C ( 3 , 3 ) .
Sol :- Here , x_{1} = 6 , y_{1} = -6 , x_{2} = 3 , y_{2} = -7 ,
x_{3} = 3 , y_{3} = 3
Area of triangle ( A ) = | 6( -7 - 3 ) + 3( 3 + 6 ) + 3( -6 + 7 ) | |||
2 |
= | 6( -10 ) + 3( 9 ) + 3( 1 ) | |||
2 |
= | -60 + 27 + 3 | |||
2 |
= | = -15 | |
2 |
A = 15 units { ∴ leaving the negative sign }
Collinearity of three Points :-
If the area of the triangle formed by joining any of the three points becomes zero, then the given point ( x_{1} , y_{1} ) , ( x_{2} , y_{2} ) and ( x_{3} , y_{3} ) will be collinear .i.e. will lie on a straight line .
Area of triangle ( A ) = | x_{1}( y_{2} - y_{3} ) + x_{2}( y_{3} - y_{1} ) + x_{3}( y_{1} - y_{2} ) | = 0 | |||
2 |
Example 8. What will be the value of K whose points A ( -6 , -5 ) , B ( 6 , K ) and C ( 12 , 4 ) are collinear ?
Sol :- Here , x_{1} = -6 , y_{1} = -5 , x_{2} = 6 , y_{2} = K ,
x_{3} = 12 , y_{3} = 4
If given points are collinear .
∴ Area of triangle = ∆ = 0
⇒ | -6( K - 4 ) + 6( 4 + 5 ) + 12( -5 - K ) | = 0 | |||
2 |
⇒ [ -6K + 24 + 54 - 60 - 12K ] = 0
⇒ -18K + 18 = 0 ⇒ 18K = 18
⇒ K = | = 1 | |
18 |
Therefore the value of K will be 1 .
Shortcut Method for Finding the area of triangle :-
To find out the area of the triangle , we follow the steps given below :-
1.Firstly , we write the coordinates of the given triangle taken in order in two columns . The order of their writing will be as follows :-
( x_{1} , y_{1} ) ( in first row )
( x_{2} , y_{2} ) ( in second row )
( x_{3} , y_{3} ) ( in third row )
( x_{1} , y_{1} ) ( in fourth row )
2. At the end, repeat the coordinates of the first vertex.
3. Now mark them upwards from top to bottom (↘) Similarly, its opposite arrows (↗) denote them from the bottom to the top. As shown in the figure below.
4. The sign of multiplication (+) ve for the downward arrow is the same, while it changes for the upward arrow marks the sign of multiplication or (-) ve becomes.
5. Finally, obtain result divided by 2.
Thus , area of triangle ( ∆ ) = | ( x_{1}y_{2} - x_{2}y_{1} ) + ( x_{2}y_{3} - x_{3}y_{2} ) + ( x_{3}y_{1} - x_{1}y_{3} ) | |||
2 |
Example 9. if A ( 0 , 0 ) , B ( 3 , 0 ) and C ( 0 , -8 ) are given , then find the area of triangle .
Sol :- Here , x_{1} = 0 , y_{1} = 0 , x_{2} = 3 , y_{2} = 0 ,
x_{3} = 0 , y_{3} = -8
Using the above given method , we get
Area of triangle ( ∆ ) = | ( 0 x 0 - 3 x 0 ) + ( 3 x -8 - 0 x 0 ) + ( 0 x 0 - 0 x -8 ) | |||
2 |
∆ = | 0 - 24 + 0 | |||
2 |
⇒ ∆ = | = -12 | |
2 |
Hence area of triangle will be 12 .
Equations of straight lines :-
When a straight line is given , then it has the following equations.which are as follows -
(ⅰ) The equation of x - axis of any given equation gets y = 0.
(ⅱ) The equation of the y - axis of any given equation is obtained x = 0.
(ⅱⅰ) If an equation is given parallel to the y - axis and the equation x = a is obtained at a distance from it.
(iv) If an equation is given parallel to the x - axis and at the distance b from it , y = b gets the equation of a line.
Note :- As we know that general form of equation of a straight line is ax + by + c = 0.
where , a , b and c are real variables ( known value ) and x and y are independent variables ( unknown ) .
1. Slope or gradient of a equation :-
A line , which an angle formed with a tangent drawn from the positive direction of a x - axis is called the gradient or slope of the line. It is generally represented by m. If a line makes an θ angle with the x - axis and the intercept c is cut on the y - axis , then
Note :-
⇒ If the line passes through the origin (0 , 0) then the equation of the line is y = mx.
1.If line is parallel to x - axis , then θ = 0°
Or m = tan0° = 0
2. If the line is parallel to the y - axis , then θ = 90°
Or m = tan90° = ∞
Example 10. Find the slope of a line whose inclination with x - axis is 60° .
Sol :- Here , θ = 60°
Slope or gradient of a line m = tan60° = √3
2. Intercept form of a line Equation :-
When a straight line , which intersects at the length a from the x - axis and length b from the y - axis , then
The equation of this line is | + | = 1 | ||
a | b |
Example 11. Find the equation of the line which cuts from the x - axis to 3 units and the y - axis to the 4 unit intercept .
Sol :- Here , a = 3 and b = 4
Using the above given formula ,
Equation of line of intercept = | + | = 1 | ||
3 | 4 |
⇒ | = 1 | |
12 |
Therefore Equation of line of intercept will be 4x + 3y = 12 .
3. Perpendicular form of a line equation :-
When a perpendicular is drawn from the origin at any straight line , its length is p and makes with an angle α to the x - axis by that perpendicular , then the equation of that straight line is obtained
Example 12. Find the equation of that line , the length of the perpendicular drawn from the origin is 13 units and the angle making between the perpendicular and x - axis is 45° .
Sol :- Here , p = 13 , α = 45°
The equation of a line xcos45° + ysin45° = 13
⇒ | + | = 13 | ||
√2 | √2 |
⇒ | = 13 | |
√2 |
Hence , The equation of a line will be x + y = 13 √2 .
Slope of a Line Joining Two Given Points And The Equation of a Line :-
Let two points ( x_{1} , y_{1} ) and ( x_{2} , y_{2} ) are given.
The slope of a line joining two these points is m = | ||
( x_{2} - x_{1} ) |
Example 13. Find the slope of the line joining two points (1 , 3) and (3 , 7) and also the equation of a line.
Sol :- Here , x_{1} = 1 , y_{1} = 3 , x_{2} = 3 , y_{2} = 7
∴ slope of equation m = | = 13 | |
( 3 - 1 ) |
⇒ m = | = 2 | |
2 |
⇒ y - 3 = 2x - 2 ⇒ 2x - y = - 3 + 2 ⇒ 2x - y = -1
Therefore , the equation of a line will be 2x - y = -1
Angle between two given straight lines :-
Let two straight lines y = m_{1}x + c_{1} and y = m_{2}x + c_{2} are given , then angle between of these both lines
angle between of these both lines tanθ = | ||
( 1 + m_{1}m_{2} ) |
θ = tan^{-1} | ||
( 1 + m_{1}m_{2} ) |
1.When the two given lines are parallel , then m_{1} will be equal to m_{2} i.e. m_{1} = m_{2}.
2. When the two given lines are mutually perpendicular to each other , then m_{1}m_{2} = - 1 will be.
Example 14. If the slopes of both lines are equal , then what will be the angle between them ?
Sol :- Given that , m_{1} = m_{2}
Then ,
Angle between two straight lines tanθ = | ||
( 1 + m_{1}m_{2} ) |
tanθ = | ||
( 1 + m_{1}m_{1} ) |
tanθ = | ||
( 1 + m_{1} × m_{1} ) |
∴ θ = 0°
Therefore , angle between two straight lines will be 0° .
Equation of parallel and perpendicular lines of straight line : -
and the equation of the given perpendicular line is bx - ay = λ
Example 15. Find the equation of parallel and perpendicular lines of the given straight line 2x + y + 3 = 0.
Sol :- On comparing straight line 2x + y + 3 = 0 with ax + by + c = 0 , we get
a = 2 , b = 1 and c = 3
∴ Equation of a line 2x + y = λ is parallel to given any line 2x + y + 3 = 0
And the equation of a line x - 2y = λ is perpendicular to given line 2x + y + 3 = 0 .
The length of the perpendicular drawn from a point on the straight line : -
1. Let any straight line ax + by + c = 0 is given . At whose
The length of the perpendicular from a point ( x_{1} , y_{1} ) | ||
√a² + b² |
Example 16. Find the length of the perpendicular drawn from a point ( 3 , 4 ) on the straight line 2x + y + 3 = 0 .
Sol :- Here , x_{1} = 3 , y_{1} = 4
Comparing the given line 2x + y + 3 = 0 with ax + by + c = 0 ,
a = 2 , b = 1 and c = 3
∴ The length of the perpendicular drawn from a point ( 3 , 4 ) on the straight line 2x + y + 3 = 0
= | ||
√2² + 1² |
= | ||
√4 + 1 |
The length of the perpendicular = | ||
√5 |
2x + y + 3 = 0 will be | . | |
√5 |
2. The length of the perpendicular drawn from the origin on a straight line ax + by + c = 0 is
√a² + b² |
Example 17. Find the length of the perpendicular drawn from a point ( 0 , 0 ) on the straight line 2x + y + 3 = 0 .
Sol :- Here , x_{1} = 0 , y_{1} = 0
Comparing the given line 2x + y + 3 = 0 with ax + by + c = 0 ,
a = 2 , b = 1 and c = 3
∴ the length of the perpendicular drawn from the origin on the straight line 2x + y + 3 = 0
= | ||
√2² + 1² |
= | ||
√4 + 1 |
The length of the perpendicular = | ||
√5 |
line 2x + y + 3 = 0 will be | . | |
√5 |
3. If any two parallel lines ax + by + c_{1} = 0 and ax + by + c_{2} = 0 are given , then
distance between them = | ||
√a² + b² |
Example 18. What will be the distance between two given parallel lines 3x + 4y - 1 = 0 and 3x + 4y - 6 = 0 ?
Sol :- Here , a = 3 , b = 4 , c_{1} = -1 and c_{2} = -6
∴ Distance between two parallel lines = | ||
√a² + b² |
= | ||
√3² + 4² |
= | ||
√9 + 16 |
= | ||
√25 |
= | = 1 unit | |
5 |