Distance between 2 points Midpoint of 2 points Distance between two points 5 18 3 17 A53 B1817 18 5 13 units 17 3 14 units AB 2 13 2 14 2 Using Pythagoras Theorem ID: 531327
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Slide1
COORDINATE GEOMETRY
Distance between 2 points
Mid-point of 2 pointsSlide2
Distance between two points.
5
18
3
17
A(5,3)
B(18,17)
18 – 5 = 13 units
17 – 3 = 14 units
AB
2
= 13
2
+ 14
2
Using Pythagoras’ Theorem,
AB
2
= (18 - 5)
2
+ (17 - 3)
2
y
xSlide3
Distance between two points.In general,
x
1
x
2
y
1
y
2
A(x
1
,y
1
)
B(x
2
,y
2
)
Length = x
2
– x
1
Length = y
2
– y
1
AB
2
=
(x
2
-x
1
)
2
+
(y
2
-y
1
)
2
Hence, the formula for
Length of AB
or
Distance between A and B
is
y
xSlide4
Find the distance between the points (-1,3) and (2,-6)
Simply by using the formula:
(-1,3) and (2,-6)
(x
1
,y
1
) and (x
2,y2)
Since
= 9.49 units (3 sig. fig)Slide5
Given 3 points A,B and C, distance formula is used to check whether the points are
collinear.If
not we may check for an isosceles, equilateral or right angled triangle
.
Perform the check on the following sets of points :
(1,5), (2,3), (-2, -11)(1,-1),(-½, ½),(1,2)
(a,a
),(-a,-a), (-a ,a )(12,8),(-2,6),(6,0)
(2,5),(-1,2),(4,7)Slide6
Distance Formula can be used to check for special quadrilaterals !!
Given 4 points A,B,C,D
If AB=CD, AD =
BC,it is a PARALLELOGRAM.
(Opposite sides are equal)If AB = CD, AD = BC, AC = BD ,it is a
RECTANGLE.(Diagonals are also equal)If AB=BC=CD=DA, it is a RHOMBUS
.(All sides are equal)If AB=BC=CD=DA and AC=BD, it is a SQUARE
.Slide7
Find the perimeter of the quadrilateral ABCD. Is ABCD a special quadrilateral?Slide8
Applications of Distance Formula
ParallelogramSlide9
Applications of Distance Formula
RhombusSlide10
Applications of Distance Formula
RectangleSlide11
Applications of Distance Formula
SquareSlide12
SPECIAL QUADRILATERALS
Show that (1,1),(4,4),(4,8),(1,5) are the vertices of a parallelogram.
Show that A(2,-2),B(14,10),C(11,13) and D(-1,1) are the vertices of a rectangle.
Show that the points (1,2),(5,4),(3,8),(-1,6) are the vertices of a square.
Show that (1,-1) is the centre of the circle circumscribing the triangle whose angular points are (4,3),(-2,3) and (6,-1).Slide13
FINDING CO-ORDINATES
Find the point on x – axis which is equidistant from (2, -5) and (-2,9).
Find the point on y – axis which is equidistant from (2,-5) and ( -2, 9).
Find a relation between x and y so that the point (x, y) is equidistant from (2,-5) and ( -2, 9).
Find the value of k such that the distance between the points (2, -5) and (k, 7) is 13 units.Slide14
The mid-point of two points.
5
18
3
17
A(5,3)
B(18,17)
Look at it’s horizontal length
= 11.5
11.5
Look at it’s vertical length
= 10
10
(11.5,
10)
Mid-point of AB
y
xSlide15
The mid-point of two points.
x
1
x
2
y
1
A(x
1,
y
1
)
B(
,
x
2
,y
2
)
Look at it’s horizontal length
Look at it’s vertical length
Mid-point of AB
y
x
y
2
Formula for mid-point isSlide16
Section Formula – Internal Division
A(x
1
, y
1
)
B(x
2
, y
2
)
X
X’
Y’
O
Y
P(x, y)
m
n
:
L
N
M
H
K
Clearly
AHP ~ PKBSlide17
The co-ordinates of the point which divides the line segment joining (x
1
, y
1
) and (x2, y2) in the ratio m : n internally are
The ratio in which the point (x, y)divides the line segment joining (x
1, y1) and (x
2, y2) is Slide18
Find the co-ordinates of the point which divides the line segment joining the points (4, -3) and (8,5) in the ratio 3:1 internally.
Find the co-ordinates of the point which divides the line segment joining the points (-1,7) and (4,-3) in the ratio 2:3 internally.Slide19
In what ratio does the point (-4,6) divide the line segment joining the points
A
(-6,10) and B (3,-8)?Find the coordinates of the points of trisection of the line segment joining (4,-1) and (-2,-3).Find the coordinates of the points which divide the line segment joining A(-2,2) and B(2,8) in four equal parts.Slide20
In what ratio
is the join of the points
(-4,6)
and (3, -8) divided by the (
i) x- axis. (ii) y-axis. Also find the co-ordinates of the point of division.
Find the coordinates of the centroid of the triangle whose vertices are (12,8),(-2,6
) and (6,0).Find the coordinates of the
vertices of a triangle whose midpoints are (4,3),(-2,3) and (6,-1).Slide21
Area of a triangle
Area of a triangle with vertices (x
1
, y
1), (x2, y2
), (x3, y3) is given by
Slide22
Collinearity of points using area of triangles
Three points (x
1
, y
1), (x2, y2), (x
3, y3) are collinear if and only if the area of the triangle with these points as vertices is 0.Slide23
Find the area of the triangle formed by the following points:
(3,4),(2,-1),(4,-6)
Show that the following points are collinear
(-5,1),(5,5) and (10,7)
For what value(s) of x,the area of the triangle formed by the points (5,-1),(x,4) and (6,3) is 5.5 square units.For what value(s) of x, will the following lie on
a line : (x,-1),(5,7),(8,11)If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.Slide24
Find the ratio in which 2x + 3y – 30 =0, divides the join of A(3, 4) and B(7, 8) and also find the point of intersection.