Mathematics

Cartesian Coordinate Geometry Session Cartesian Coordinate Geometry And Straight Lines

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

René Descartes Cartesian coordinate geometry is so named in honour of Rene Descartes. Rene Descartes, considered by many to be the father of modern philosophy was also an outstanding Mathematician. It is said that he invented cartesian coordinate geometry on observing a fly walk across the ceiling. Descartes used his method to apply principles of algebra to geometrical problems.

Coordinates Y Y’ 1 2 3 X X’ O Origin -1 -2 -3 -4 1 2 3 4 -1 -2 -3 +ve direction Y-axis : Y’OY X-axis : X’OX X X’ O Origin -1 -2 -3 -4 -ve direction 1 2 3 4 +ve direction -1 -2 -3 -ve direction Description of axes. The axes are arbitrarily chosen mutually perpendicular lines passing through an arbitrarily chosen point identified as the origin. Left to right and bottom to top are positive directions.

Coordinates X X’ Y Y’ O 1 2 3 4 -1 -2 -3 -4 (2,1) Abcissa Ordinate Any point on the plane can be uniquely identified by its x-coordinate and y-coordinate. The x-coordinate is also called the abcissa, the y coordinate the ordinate of the point. The point is identified as an ordered pair (abcissa, ordinate) or (x,y). Ordinate (-3,-2) (?,?)

Coordinates X X’ Y Y’ O 1 2 3 4 -1 -2 -3 -4 (2,1) Abcissa Ordinate (-3,-2) (4,?)

Coordinates X X’ Y Y’ O 1 2 3 4 -1 -2 -3 -4 (2,1) Abcissa Ordinate (-3,-2) (4,-2.5)

II I III IV (-,+) (+,+) (-,-) (+,-) Quadrants X X’ O Y Y’ The cartesian plane is divided into four quadrants. Quadrants are numbered in anticlockwise direction. All abcissae in a given quadrant will have the same sign and all ordinates in a given quadrant will have the same sign. (-,-) (+,-)

I II III IV (+,+) (-,+) (-,-) (+,-) Quadrants X X’ O Y Y’ Ist? IInd? Q : (1,0) lies in which Quadrant? A : None. Points which lie on the axes do not lie in any quadrant.

Distance Formula PQN is a right angled .  PQ2 = PN2 + QN2 X X’ Y’ O Y Q(x2, y2) N y2-y1 y2 P(x1, y1) PQN is a right angled .  PQ2 = PN2 + QN2 y1 x1 (x2-x1)  PQ2 = (x2-x1)2+(y2-y1)2 Explain the derivation step by step according to animation. x2

Distance From Origin Distance of P(x, y) from the origin is

Applications of Distance Formula Parallelogram To show that a figure is a parallelogram, prove that opposite sides are equal.

Applications of Distance Formula Rhombus To show that a figure is a rhombus, prove all sides are equal.

Applications of Distance Formula Rectangle To show that a figure is a rectangle, prove that opposite sides are equal and the diagonals are equal. Corollary : Parellelogram but not rectangle  diagonals are unequal

Applications of Distance Formula Square To show that a figure is a square, prove all sides are equal and the diagonals are equal. Corollary : Rhombus but not square  diagonals are unequal

Area of a Triangle Area of  ABC = Area of trapezium ABML A(x1, y1) C(x3, y3) B(x2, y2) X X’ Y’ O Y M L N Area of  ABC = Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC

Area of a Triangle Area of trapezium ABML + Area of trapezium ALNC X X’ Y’ O Y A(x1, y1) C(x3, y3) B(x2, y2) M L N Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC Sign of Area : Points anticlockwise  +ve Points clockwise  -ve

Can be used to calculate area of Quadrilateral, Pentagon, Hexagon etc. 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.

Collinearity of Three Points Method I : Use Distance Formula a b To show that three points are collinear, show that a+b = c c Show that a+b = c

Collinearity of Three Points Method II : Use Area of Triangle A  (x1, y1) B  (x2, y2) C  (x3, y3) Show that To show that three points are collinear, show that area of triangle is zero

Section Formula – Internal Division A(x1, y1) B(x2, y2) X X’ Y’ O Y P(x, y) m n : L N M H K Clearly AHP ~ PKB Explain the derivation step by step according to animation.

Midpoint Midpoint of A(x1, y1) and B(x2,y2) m:n  1:1

Section Formula – External Division P divides AB externally in ratio m:n X X’ Y’ O Y A(x1, y1) P(x, y) B(x2, y2) L N M H K Clearly PAH ~ PBK Explain the derivation step by step according to animation.

Intersection of medians of a triangle is called the centroid. A(x1, y1) B(x2, y2) C(x3, y3) D E F G Centroid is always denoted by G.

Centroid A(x1, y1) B(x2, y2) C(x3, y3) D E F G Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

Centroid A(x1, y1) B(x2, y2) C(x3, y3) D E F G Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

Centroid A(x1, y1) B(x2, y2) C(x3, y3) D E F G Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

Centroid A(x1, y1) B(x2, y2) C(x3, y3) D E F G Medians are concurrent at the centroid, centroid divides medians in ratio 2:1 We see that L  M  N  G

Centroid Centroid We see that L  M  N  G A(x1, y1) B(x2, y2) C(x3, y3) D E F G Centroid We see that L  M  N  G

Incentre is the centre of the incircle Intersection of angle bisectors of a triangle is called the incentre A(x1, y1) B(x2, y2) C(x3, y3) D E F I 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.

Similarly I can be derived using E and F also Incentre A(x1, y1) B(x2, y2) C(x3, y3) D E F I Similarly I can be derived using E and F also

Angle bisectors are concurrent at the incentre A(x1, y1) B(x2, y2) C(x3, y3) D E F I Angle bisectors are concurrent at the incentre

Excentre is the centre of the excircle A(x1, y1) B(x2, y2) C(x3, y3) D E F Intersection of external angle bisectors of a triangle is called the excentre Excentre is the centre of the excircle EA = Excentre opposite A

Excentre is the centre of the excircle A(x1, y1) B(x2, y2) C(x3, y3) D E F Intersection of external angle bisectors of a triangle is called the excentre Excentre is the centre of the excircle EB = Excentre opposite B

Excentre is the centre of the excircle A(x1, y1) B(x2, y2) C(x3, y3) D E F Intersection of external angle bisectors of a triangle is called the excentre Excentre is the centre of the excircle EC = Excentre opposite C

Cirumcentre Intersection of perpendicular bisectors of the sides of a triangle is called the circumcentre. A B C O OA = OB = OC = circumradius The above relation gives two simultaneous linear equations. Their solution gives the coordinates of O.

Orthocentre Intersection of altitudes of a triangle is called the orthocentre. Orthocentre is always denoted by H A B C H We will learn to find coordinates of Orthocentre after we learn straight lines and their equations Find orthocenter & circumcenter of a right and triangle??

Cirumcentre, Centroid and Orthocentre The circumcentre O, Centroid G and Orthocentre H of a triangle are collinear. O H G 1 : 2 G divides OH in the ratio 1:2

Circle!! 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!!

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!!

A Locus is NOT an equation. But it is associated with an equation 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

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

Class Exercise – 3.2 Three vertices of a parallelogram taken in order are (a+b,a-b), (2a+b,2a-b) and (a-b,a+b). Find the fourth vertex. Solution : Let the fourth vertex be (x,y). Diagonals bisect each other.  the required vertex is (-b,b)

EXERCISE 3.3 Four points A(6,3), B(-3,5), C(4,-2) and D(x,3x) are given such that Find x. Solution : Given that ABC = 2DBC

Class Exercise – 3.3 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 This method of proof is called reductio ad absurdum.

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. This method of proof is called reductio ad absurdum. Q.E.D.

Class Exercise – 3.4 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 = (h+2)2+(k-1)2  6h+4k = 5  equation of locus of (h,k) is 6x+4y = 5 No variable is involved

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, a2+b2 = l2  4h2+4k2 = l2  equation of locus of (h,k) is 4x2+4y2 = l2 B(0,b) A(a,0) O P(h,k) No variable is involved

Class Exercise - 7 A point moves so that the ratio of its distance from (-a,0) to (a,0) is 2:3. Find the equation of its locus. Solution : Let the point be P(h,k). Given that

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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