Cartesian Coordinate Geometry

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Session
Cartesian Coordinate Geometry And Straight Lines
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Session Objectives

• Cartesian Coordinate system and Quadrants
• Distance formula
• Area of a triangle
• Collinearity of three points
• Section formula
• Special points in a triangle
• Locus and equation to a locus
• Translation of axes - shift of origin
• Translation of axes - rotation of axes

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René Descartes
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Coordinates
Y-axis YOY
X-axis XOX
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Coordinates
(2,1)
Abcissa
Ordinate
(-3,-2)
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Coordinates
(2,1)
Abcissa
Ordinate
(-3,-2)
(4,?)
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Coordinates
(2,1)
Abcissa
Ordinate
(-3,-2)
(4,-2.5)
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Quadrants
(,)
(-,)
I
II
III
IV
(,-)
(-,-)
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Quadrants
Ist? IInd?
Q (1,0) lies in which Quadrant?
A None. Points which lie on the axes do not lie
in any quadrant.
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Distance Formula
PQN is a right angled ?. ? PQ2 PN2 QN2
? PQ2 (x2-x1)2(y2-y1)2
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Distance From Origin
Distance of P(x, y) from the origin is
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Applications of Distance Formula
Parallelogram
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Applications of Distance Formula
Rhombus
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Applications of Distance Formula
Rectangle
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Applications of Distance Formula
Square
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Area of a Triangle
Area of ? ABC Area of trapezium ABML
Area of trapezium ALNC
- Area of trapezium BMNC
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Area of a Triangle
Sign of Area Points anticlockwise ?
ve Points clockwise ? -ve
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Area of Polygons
Area of polygon with points Ai ? (xi, yi) where i
1 to n
Can be used to calculate area of Quadrilateral,
Pentagon, Hexagon etc.
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Collinearity of Three Points
Method I Use Distance Formula
a
b
c
Show that ab c
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Collinearity of Three Points
Method II Use Area of Triangle
A ? (x1, y1) B ? (x2, y2) C ? (x3, y3) Show that
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Section Formula Internal Division
Clearly ?AHP ?PKB
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Midpoint
Midpoint of A(x1, y1) and B(x2,y2) mn ? 11
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Section Formula External Division
P divides AB externally in ratio mn
Clearly ?PAH ?PBK
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Centroid
Intersection of medians of a triangle is called
the centroid.
Centroid is always denoted by G.
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Centroid
Consider points L, M, N dividing AD, BE and CF
respectively in the ratio 21
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Centroid
Consider points L, M, N dividing AD, BE and CF
respectively in the ratio 21
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Centroid
Consider points L, M, N dividing AD, BE and CF
respectively in the ratio 21
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Centroid
Medians are concurrent at the centroid, centroid
divides medians in ratio 21
We see that L ? M ? N ? G
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Centroid
We see that L ? M ? N ? G
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Incentre
Intersection of angle bisectors of a triangle is
called the incentre
Incentre is the centre of the incircle
Let BC a, AC b, AB c AD, BE and CF are the
angle bisectors of A, B and C respectively.
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Incentre
Similarly I can be derived using E and F also
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Incentre
Angle bisectors are concurrent at the incentre
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Excentre
Intersection of external angle bisectors of a
triangle is called the excentre
Excentre is the centre of the excircle
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Excentre
Intersection of external angle bisectors of a
triangle is called the excentre
Excentre is the centre of the excircle
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Excentre
Intersection of external angle bisectors of a
triangle is called the excentre
Excentre is the centre of the excircle
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Cirumcentre
Intersection of perpendicular bisectors of the
sides of a triangle is called the circumcentre.
OA OB OC circumradius
The above relation gives two simultaneous linear
equations. Their solution gives the coordinates
of O.
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Orthocentre
Intersection of altitudes of a triangle is called
the orthocentre.
Orthocentre is always denoted by H
We will learn to find coordinates of Orthocentre
after we learn straight lines and their equations
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Cirumcentre, Centroid and Orthocentre
The circumcentre O, Centroid G and Orthocentre H
of a triangle are collinear.
G divides OH in the ratio 12
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Locus a Definition
The curve described by a point which moves under
a given condition or conditions is called its
locus
e.g. locus of a point having a constant distance
from a fixed point
Circle!!
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Locus a Definition
The curve described by a point which moves under
a given condition or conditions is called its
locus
e.g. locus of a point equidistant from two fixed
points
Perpendicular bisector!!
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Equation to a Locus
The equation to the locus of a point is that
relation which is satisfied by the coordinates of
every point on the locus of that point
Important A Locus is NOT an equation. But it is
associated with an equation
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Equation to a Locus
Algorithm to find the equation to a locus Step
I Assume the coordinates of the point whose
locus is to be found to be (h,k)
Step II Write the given conditions in
mathematical form using h, k Step III Eliminate
the variables, if any Step IV Replace h by x
and k by y in Step III. The equation thus
obtained is the required equation to locus
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Illustrative Example
Find the equation to the locus of the point
equidistant from A(1, 3) and B(-2, 1)
Solution
Let the point be P(h,k) PA PB (given) ?
PA2 PB2 ? (h-1)2(k-3)2 (h2)2(k-1)2 ?
6h4k 5 ? equation of locus of (h,k) is 6x4y
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Illustrative Example
A rod of length l slides with its ends on
perpendicular lines. Find the locus of its
midpoint.
Solution
Let the point be P(h,k) Let the ? lines be
the axes Let the rod meet the axes at A(a,0)
and B(0,b) ? h a/2, k b/2 Also, a2b2
l2 ? 4h24k2 l2 ? equation of locus of (h,k)
is 4x24y2 l2
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Shift of Origin
Consider a point P(x, y)
Let the origin be shifted to O with coordinates
(h, k) relative to old axes
Let new P ? (X, Y)
? x X h, y Y k
? X x - h, Y y - k
O ? (-h, -k) with reference to new axes
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Illustrative Problem
Show that the distance between two points is
invariant under translation of the axes
Solution
Let the points have vertices A(x1, y1), B(x2,
y2) Let the origin be shifted to (h, k) new
coordinates A(x1-h, y1-k), B(x2-h, y2-k)
Old dist.
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Rotation of Axes
R
Consider a point P(x, y)
?
Let the axes be rotated through an angle ?.
?
Let new P ? (X, Y) make an angle ? with the new
x-axis
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Rotation of Axes
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Class Exercise - 1
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Solution
Let O be the origin. ? OA2 a2b2, OB2 c2d2,
AB2 (c-a)2(d-b)2 Using Cosine formula in ?OAB,
we have AB2 OA2OB2-2OA.OBcos?
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Class Exercise - 2
Solution
Given that ?ABC 2?DBC
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Class Exercise - 3
If a ? b ? c, prove that (a,a2), (b,b2) and
(c,c2) can never be collinear.
Solution
Let, if possible, the three points be collinear.
R2 ? R2-R1, R3 ? R3- R2
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Solution Cont.
R2 ? R2-R3
This is possible only if a b or b c or c
a. But a ? b ? c. Thus the points can never be
collinear.
Q.E.D.
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Class Exercise - 4
Three vertices of a parallelogram taken in
order are (ab,a-b), (2ab,2a-b) and (a-b,ab).
Find the fourth vertex.
Solution
Let the fourth vertex be (x,y). Diagonals
bisect each other.
? the required vertex is (-b,b)
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Class Exercise - 5
If G be the centroid of ?ABC and P be any
point in the plane, prove that PA2PB2PC2GA2GB2
GC23GP2.
Solution
Choose a coordinate system such that G is the
origin and P lies along the X-axis.
Let A ? (x1,y1), B ? (x2,y2), C ? (x3,y3), P ?
(p,0) ? LHS (x1-p)2y12(x2-p)2y22(x3-p)2y32
(x12y12)(x22y22)(x32y32)3p2-2p(x1x2x3
) GA2GB2GC23GP2 RHS
Q.E.D.
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Class Exercise - 6
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Solution
Let the line intercept at the axes at A and B.
Let R(h,k) be the midpoint of AB.
? Ans (b)
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Class Exercise - 7
A point moves so that the ratio of its
distance from (-a,0) to (a,0) is 23. Find the
equation of its locus.
Solution
Let the point be P(h,k). Given that
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Class Exercise - 8
Find the locus of the point such that the
line segments having end points (2,0) and (-2,0)
subtend a right angle at that point.
Solution
Let A ? (2,0), B ? (-2,0) Let the point be
P(h,k). Given that
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Class Exercise - 9
Find the coordinates of a point where the
origin should be shifted so that the equation
x2y2-6x8y-9 0 will not contain terms in x and
y. Find the transformed equation.
Solution
Let the origin be shifted to (h,k). The given
equation becomes (Xh)2(Yk)2-6(Xh)8(Yk)-9
0 Or, X2Y2(2h-6)X(2k8)Y(h2k2-6h8k-9)
0 ? 2h-6 0 2k8 0 ? h 3, k -4. Thus
the origin is shifted to (3,-4). Transformed
equation is X2Y2(916-18-32-9) 0 Or, X2Y2
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Class Exercise - 10
Through what angle should the axes be rotated
so that the equation 11x24xy14y2 5 will not
have terms in xy?
Solution
Let the axes be rotated through an angle ?. Thus
equation becomes
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Solution Cont.
Therefore, the required angle is
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