coordinategeometry1 1
TRANSCRIPT
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Rene DescartesA French mathematician who
discovered the co-ordinate geometry.
He was the first man who unified Algebra and Geometry.
According to him every point of the plane can be represented uniquely by
two numbers
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co-ordinate geometry
It is that branch of geometry in which two numbers called co-
ordinates , are used to locate the position of a point in a plane.
It is also called as Cartesian-geometry
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Axes of reference• The whole plane is divided in to the four
parts by two straight lines , which are perpendicular to each other.
• The line which is parallel to the horizontal line is called as the X - AXIS
• The line which is parallel to the vertical line is called as the Y - AXIS
• the point of intersection is called as ORIGIN
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CO-ORDINATES OF A POINT
• X-CO-ORDINATE -the distance of the point from the origin along X - axis is called X- co-ordinate ( abscissa)
• Y-CO-ORDINATE -the distance of the point from the origin along Y - axis is called Y- co-ordinate ( ordinate)
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REPRESENTATION OF A POINT
THE CO-ORDINATES OF A POINT IS ALWAYS REPRESENTED BY
ORDERED -PAIR ( )
FIRST PUT X-CO-ORDINATE THEN Y-CO-ORDINATE IN BRACKET
( X, Y )
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Representation of points on plane
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Co-ordinate of origin (0,0)
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Distance formulaTo find out the distance
between two points in the plane
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Let the two points are P(x1,y1) and Q(x2,y2)
P(x1,y1)
Q(x2,y2)
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P(x1,y1)
Q(x2,y2)
x1 x2
y2
y1 R(x2,y1)
Then distance QR = y2 - y1 and PR = x2 - x1
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• Since PRQ is a right triangle• therefore by using Pythagorus
theorem• PQ2 = PR2 + RQ2
• PQ2 = (x2 - x1) 2 + (y2 - y1) 2
• PQ = (x2 - x1) 2 + (y2 - y1) 2
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PROBLEMS ON COLLINEARITY OF THREE
POINTS
POINTS A , B and C ARE said to be collinear if AB +BC = AC
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PROBLEM 1Determine by distance formula , whether the points (2,5) , (-1,2)
and (4,7) are collinear.• Sol. We are given three points A (2,5)
• B (-1,2) and C (4,7)
AB= (-1-2) 2 +(2-5) 2
• AB= (-3) 2 +(-3) 2= 9+9= 18= 3 2
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SIMILARLY
BC= (4+1) 2 +(7-2) 2
= (5) 2 +(5) 2
= 25+25= 50 = 25X2= 5 2
AC= (2-4) 2 +(5-7) 2
= (-2) 2 +(-2) 2 = 4+4= 8 =2 2
THUS 3 2+ 2 2 =5 2
AB +AC = BCPOINTS B , A C are collinear
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Problem on equidistant
• Find the point on y-axis which is equidistant from (-5,-2) and (3,2)
• solution: let p(0,y) be a point on y-axis which is equidistant from
• A(-5,-2) and B (3,2)
PA = PB
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• PA2 = PB2
• (-5-0)2 + (-2-y)2 =(3-0)2 +(2-y)2
• 25+4+ y2 +4y = 9+4 + y2 -4y• 8y = -16• y= -2 • so the required point is p(0,-2)
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Note
• If we have to find the point which is on the x-axis and equidistant from the given two points then that point will be p(x,0) . Then find x by applying the similar procedure and you will get the required point.
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Note • To show that the given points are
vertices of an equilateral triangle .find the length of all sides using distance formula and check all sides are equal.
• To show that the given points are vertices of a right angle triangle check whether sides LENGTHS are following Pythagorus theorem
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For square1.all sides are equal and 2.diagonals are equal
For rhombus
1.all sides are equal but2. diagonals are not equal
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For rectangle
1. Opposite sides are equal2. Diagonals are equal
For parallelogram1. Opposite sides are equal2. Diagonals bisect each other
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It gives the co-ordinates of the point which divides the given line segment in the ratio m : n
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Let AB is a line segment joining the points A(x1,y1) and B(x2,y2)
LET P(x,y) be a point which divides line segment AB
in the ratio m : n internally
therefore
(x1,y1)
(x2,y2)
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(x1,y1)
(x2,y2)
(x,y1)(x-x1)
(x2,y)(x2 - x)
(y2 - y)
(y-y1)
x1 xx x2
Complete the figure as follows
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Now AQP PRB …..by AA similarity
AP/PB = AQ/PR = PQ/BR
BY CPST
m/n = x-x1/x2-x =y-y1/y2-y
m/n = x-x1/x2-xsolving this equation for x we will get
x = mx2+nx1
m+n
Similarly solving the equation m/n =y-y1/y2-y
we will gety = my2+ny1
m+n
(x1,y1)
(x-x1)
(y-y1)
(y2 - y)
(x2 - x)
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Thus if a point p(x,y) divides a line segment joining the points
A(x1,y1) and B(x2,y2) in the ratio m : n then the co-ordinates
of P are given by
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• The co-ordinates of A and B are (1,2) and (2,3). Find the co-ordinates of R so that AR/RB = 4/3.
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SOLUTION 1
Let co-ordinates of point R (X,Y)
X = 4[2]+3[1] 4+3
X= {8+3]/7 = 11/7
Here m=4 , n=3, x1=1 , y1=2 x2= 2 , y2 = 3
Y = 4[3]+3[2] 4+3
Y = [ 12+6]/7 = 18/7
Hence co-ordinates of R = (11/7 , 18/7)
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• Find the ratio in which the point (11,15) divides the line segment joining the points (15,5) and (9,20).
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Let k : 1 be the required ratio
Here x=11 , y=15 , x1= 15 , y1=5 ,x2=9 , y2=20
m= k , n= 1
Therefore using section formula
11 = 9k+15 k+1
and 15 = 20k+5 k+1
solve either of these two equation let’s take first equation
11k+11=9k+1511k-9k=15-11
2k=4
k=2So,the required ratio is 2:1
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• Find the ratio in which the line segment joining the points (6,4) and (1,-7) is divided internally by axis of x.
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Solution 3Let k:1 is the required ratio.
P(x,0) point on x-axis which divides AB in required ratio.
Here x=x , y=0 , x1= 6 , y1=4 ,x2=1 , y2= -7
Therefore using section formula
x = k+6 k+1
and0= -7k+4 k+1
solve either of these two equation let’s take second equation
0= -7k+4
7k=4
K= 4/7
So,the required ratio is 4:7.
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AREA OF TRIANGLE
Let vertices of triangle are ( x1,y1) (x2,y2) and (x3,y3).
Area of Triangle =1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))
A( x1,y1)
B(x2,y2) C(x3,y3).
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•Find the area of triangle with vertices A(6,4) B(1,-7) C(2,3).
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Here x1= 6 , y1=4 ,x2=1 , y2= -7, x3=2 , y3=3
Area of Triangle =1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))
Area of Triangle =1/2(6(-7-3)+1(3-4)+2(4+7)) =1/2(-60-1+22)
=1/2(-39) = -39/2
= -39/2
= 39/2
So,Area of Triangle is 19.5