10 Coordinate Geometry Math Concepts .ppt

x
x
x’
x’
y
y
y’
y’
1
1st
Quadrant
Quadrant
2
2nd
Quadrant
Quadrant
3
3rd
Quadrant
Quadrant 4
4th
Quadrant
Quadrant
(x, y)
(-x, -y)
(-x, y)
(x, -y)
(0, 0)
RECTANGULAR COORDINATE SYSTEM
x-x’: x-axis
x-x’: x-axis
y-y’: y-axis
y-y’: y-axis

Answer the following:
In which quadrants do the following points lie?
In which quadrants do the following points lie?
1. (5, -3)
2. (-9, -7)
3. (3-, 6)
4. (3, (-5)2
)
5. (-15, 1-log525)
1. 4th
quadrant
2. 3rd
quadrant
3. 2nd
quadrant
4. 1st
quadrant
5. 3rd
quadrant
PROBLEM

Now, you can attempt question 1, 6 at page 211 and Q 1, 2, 3, 4 at page
213 .
Book : SAT Mathematics
NOW PRACTISE FROM YOUR BOOK!

SLOPE OF A LINE
The slope of the line , passing through the points (x1, y1) and
(x2, y2), where x1 ≠ x2
Slope = m = tan θ =
2 1
2 1
y y y
x x x
 

 


Rising as x moves Positive
from left to right
Falling as x moves Negative
from left to right
Horizontal 0
Vertical Not defined
Line
Line Slope
Slope Example
Example
GEOMETRIC INTERPRETATION OF SLOPE

CALCULATING SLOPE
What is the slope of the line passing through the points (2, 5) and (-1, -
7)?
Answer: m = 4

PARALLEL LINES
Two lines with slopes m1 and m2 are parallel if and only if
m1 = m2
(the slopes are the same)

INTERPRETING THE EQUATION
Ax + By + C = 0
Remember
Remember
Ax + By + C = 0 in coordinate plane represents a LINE.
x y
INTERCEPTS
INTERCEPTS
0
0 -2
-2
3
3 0
0
Plot
Plot
2x – 3y = 6

INTERCEPTS OF A LINE
x-intercept
y- i
n
t
e
r
c
e
p
t
(0,y)
(x,0)
Determine the x-intercept and y-intercept of the line 2x + 3y – 12 = 0.

EQUATION OF A STRAIGHT
LINE
If y–intercept (c) and the slope (m) of line are given.
The equation of the line in slope-intercept form is
y = mx + c
Where m is the slope and c is the y-intercept.
Slope-Intercept Form:
Slope-Intercept Form:

SLOPE OF A LINE (Ax + By + C = 0)
Find the slope of the line: 2x + 3y – 12 = 0
2x + 3y – 12 = 0
Always remember:
Always remember:
So, slope of the above line = –A/B

Find the slope of following lines:
L
L1
1:
: 3x + 4y = 7 L
L2
2:
: 4x + 3y = 15
L
L3
3:
: 6x + 8y = 8 L
L4
4:
: 6x – 8y = 17
1.
1. L
L1
1: -3/4
: -3/4
2.
2. L
L2
2: -4/3
: -4/3
3.
3. L
L3
3: -3/4
: -3/4
4.
4. L
L4
4 : 3/4
: 3/4
L1 and L3 are parallel as the slopes
are equal.
L2 and L4 are perpendicular as the
product of slopes is –1.
Solutions:
Solutions:
EXAMPLE

Book Reference
Now you can attempt Q 7, 14, 18 at page 211 – 212 and Q
19 at page 214.

Find the perimeter of ∆ABC.
C(-1, 4)
C(-1, 4)
A(-2, 0)
A(-2, 0)
B(2, -
B(2, -
1)
1)
y
y
x
x
Answer: 14.1
EXAMPLE

COORDINATES OF CENTROID
(x
(x1
1, y
, y1
1)
)
(x
(x3
3, y
, y3
3)
)
(x
(x2
2, y
, y2
2)
)
x-coordinate is given by
x1 + x2 + x3
Similarly, y-coordinate is given by
x-coordinate is given by: Arithmetic mean of x-coordinates
Arithmetic mean of x-coordinates
y-coordinate is given by: Arithmetic mean of y-coordinates
Arithmetic mean of y-coordinates
3
)
y
y
y
( 3
2
1 


3
3
The point of concurrence of the medians of a triangle is termed
as Centroid (G).

COORDINATES OF CENTROID
Find the coordinates of the centroid of the triangle formed with
the vertices (3, 4), (6, 7) and (9, 13).
Answer: (6, 8)

INCENTRE OF TRIANGLE
The point of concurrence of the angle bisectors of a triangle is termed as
incentre (I).

INCENTRE OF TRIANGLE
If the coordinates of a triangle are (2, 2), (2, 10) and (8, 2). Find the
incentre of the triangle.
Answer: (4, 4)

AREA OF A TRIANGLE
1
y
x
1
y
x
1
y
x
2
1
3
3
2
2
1
1
(x
(x1
1, y
, y1
1)
)
(x
(x3
3, y
, y3
3)
)
(x
(x2
2, y
, y2
2)
)
Area =
Area =
     
y
y
x
y
y
x
y
y
x
2
1
2
1
3
1
3
2
3
2
1 




Area =
Area =

AREA OF A TRIANGLE
What is the area of the triangle with coordinates of vertices as (1, 5), (-4,
6) and (3, 3)?
Answer: 4 square units

Now, you can attempt Q 3, 4, 5, 9, 10 at page 211 and Q
6, 10 at page 213.

y – y1 = m(x – x1)
EQUATION OF A STRAIGHT LINE
point = x1,y1
slope = m
Example:
Example:
Find the equation of the line that passes through the
point (2, 3) and has a slope of ½.
y – 3 = ½(x – 2)
(A) When one point and slope are given:

2 1
2 1



 
y y
y
x x x
Example:
Example:
Find the equation of the line that passes through the points (3, 1)
and (6, 0).
(B)
(B) When two points are given:
When two points are given:
points: (x1, y1) and (x2, y2)
y – y
y – y1
1 = m(x – x
= m(x – x1
1), where m =
), where m =
Equation of line will be )
3
(
3
1
1 


 x
y

x-intercept
y- i
n
t
e
r
c
e
p
t
b
(a,0)
1
b
y
a
x


a
Example:
Example:
Find the equation of a line whose x-intercept is 6
and y-intercept is -3.
Intercept Form:
Intercept Form:
If intercepts are given, then the equation of the straight
line is , whose x-intercept is 'a' and y-intercept
is 'b’.
or x – 2y – 6 = 0
x – 2y – 6 = 0
(0,b)

EXAMPLE
Find the equation of the line perpendicular to the line y = 2x – 5 and
passing through (1, 2).
Answer: 2y + x = 5

EXAMPLE
Find the equation of the line parallel to the line y = 2x – 5 and passing
through (1, 2).
Answer: y – 2x = 0

Now, you can attempt Q 14, 15, 19, 20 at page 211-212 and Q 11,
15, 16, 18, 20 page 213-214 .

The Equations a circle, Centre(0, 0), Radius r
A circle is a set of points which are equidistant from
a fixed point called centre.
Using distance, we find the radius.
Radius = )²
y
y
(
)²
x
x
( 2
1
2
1 


)²
0
y
(
)²
0
x
(
r 



²
y
²
x
²
r 

²
r
²
y
²
x 

Thus, the equations a circle, centre(0, 0), radius r is
²
r
²
y
²
x 


Let’s practice
Find the equations of the following circles with each centre (0, 0).
(i) Radius = √5 and (ii) which has points ( 4, -2)
Solution : (i) The centre is (0, 0) and there the equation is of the form
x² + y² = r²
Substitute, r = √5
x² + y² = 5
(ii) The centre is (0, 0) and there the equation is of the form
x² + y² = r²
Substitute, x = 4 and y = -2
16 + 4 = r²
r² = 20 = 2√5
Thus, the required equation is
x² + y² = 20

The Equations a circle, Centre(h, k), Radius r
Distance between (h, k) and (x, y) is equal to
the radius.
Radius = )²
y
y
(
)²
x
x
( 2
1
2
1 


)²
k
y
(
)²
h
x
(
r 



The equation of the circle with centre (h, k) and radius r is
)²
k
y
(
)²
h
x
(
²
r 




General Equation of the circle
The general equation of the circle is written as
x² +y² + 2gx + 2fy + c = 0
When equation of the circle is given in this form, we do the
following t o get the centre and the radius.
1.Bring ever term on the left side of the equation and make
coefficient of x² and y² equal to 1.
2.Centre = (-g, -f)
3.Radius = )
c
²
f
²
g
( 


Find the centre and radius of the circle (x – 2)² + (y + 4)² =
16
Solution :
Comparing ( x – 2)² + (y + 4)² = 16 with (x – h)² + (y - k )²
= r²
We get centre (h, k) = (2, -4) and radius = 4
Let’s practice

Let’s practice
Find the equation of the circle at (3, -4) and radius √12.
Solution :
circle at (3, -4) and radius √12
(h, k) = (3, -4) and r = √12
Plug-in the values in (x – h)² + (y - k)² = r²
(x – 3)² + (y + 4)² = 12

Miscellaneous Useful Formulae
Coordinate Geometry

SECTION FORMULAE
Consider two points A (x1, y1) and B (x2, y2)
(x1, y1)A B (x2, y2)
m n
nx
nx1
1
mx2
Externally:
n
m
ny
my
,
n
m
nx
mx 1
2
1
2




C
C
2 1
2 1
mx nx
x
m n
my ny
y
m n







SECTION FORMULAE
Find the ratio in which y-axis divides the line segment joining the
points (-6, -5) and (4, 1).
Answer: 3 : 2

DISTANCE BETWEEN PARALLEL LINES
2 2 2 2
difference of const.terms C D
p
(coeff of X) (coeff of Y) A B

 
 
Ax + By + C = 0
Ax + By + D = 0
p
p

EXAMPLE
Find the distance between lines 3x + 4y = 7 and 6x + 8y = 10.
Answer:
2
5

DISTANCE OF POINT TO A LINE
1 1
2 2 2 2
Ax By C
Substitute thegivenpt. inequation
p
(coeff of x) (coeff of y) A B
 
 
 
Ax + By + C = 0
p
(x1, y1)

EXAMPLE
Find the distance of a point (2, 3) to a line 3x + 4y = 7.
Answer:
11
5

10 Coordinate Geometry Math Concepts .ppt

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