Mathematics Session Cartesian Coordinate Geometry And Straight Lines Session Objectives 1. Cartesian Coordinate system and Quadrants 2. Distance formula 3. Area of a triangle 4. Collinearity of three points 5. Section formula 6. Special points in a triangle 7. Locus and equation to a locus 8. Translation of axes - shift of origin 9. Translation of axes - rotation of axes René Descartes Coordinates Y 3 Y-axis : Y’OY 2 X-axis : X’OX 1 +ve direction+ve X’ O X -4 -3 -2 -1 1 2 3 4 Origin -1 +ve direction -2 -ve direction ve directionve Y’ -3 - Coordinates Y 3 2 (2,1) Abcissa 1 X’ O X -4 -3 -2 -1 1 2 3 4 -1 Ordinate -2 (-3,-2) Y’ -3 (?,?) Coordinates Y 3 2 (2,1) Abcissa 1 X’ O X -4 -3 -2 -1 1 2 3 4 -1 Ordinate -2 (-3,-2) Y’ -3 (4,?) Coordinates Y 3 2 (2,1) Abcissa 1 X’ O X -4 -3 -2 -1 1 2 3 4 -1 Ordinate -2 (-3,-2) Y’ -3 (4,-2.5) Quadrants Y (-,+) (+,+) II I X’ O X III IV (-,-) (+,-) Y’ Quadrants Y (-,+) (+,+) II I X’ O X III IV (-,-) (+,-) 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 Y 1 y - 2 y 2 y N PQN is a right angled . 1 x1 y 2 2 2 (x2-x1) PQ = PN + QN 2 2 2 X’ O X PQ = (x2-x1) +(y2-y1) x2 Y’ 22 PQ =( x2 − x 1) +( y 2 − y 1 ) Distance From Origin Distance of P(x, y) from the origin is 22 =(x − 0) +( y − 0) =+xy22 Applications of Distance Formula Parallelogram Applications of Distance Formula Rhombus Applications of Distance Formula Rectangle Applications of Distance Formula Square Area of a Triangle Y A(x1, y1) ) 2 , y , C(x3, y3) 2 B(x X’ O M L N X Y’ Area of ABC = Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC Area of a Triangle A(x , y ) Y 1 1 ) 2 , y , C(x , y ) 2 3 3 B(x X’O M L N X Y’ Area of trapezium ABML + Area of trapezium ALNC - Area of trapezium BMNC 1 1 1 =(BM + AL) ( ML) +( AL + CN) ( LN) −( BM + CN) ( MN) 2 2 2 1 1 1 =(yyxx +) ( −) +( yyxx +) ( −) −( yyxx +) ( − ) 22112 2 1331 2 2332 x y 1 Sign of Area : Points anticlockwise ➔ +ve 1 11 = x y 1 2 22 Points clockwise ➔ -ve x33 y 1 Area of Polygons Area of polygon with points Ai (xi, yi) where i = 1 to n 1 x1 y 1xy22 x n−− 1 y n 1 x n y n = + +... + + 2 x2 y 2xy33 x n y n x 1 y 1 Can be used to calculate area of Quadrilateral, Pentagon, Hexagon etc. Collinearity of Three Points Method I : Use Distance Formula a b 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) x11 y 1 Show that x22 y 1= 0 x33 y 1 Section Formula – Internal Division Y K Clearly AHP ~ PKB H AP AH PH = = BP PK BK X’ O X m x−− x y y L N M =11 = n x−− x y y Y’ 22 mx2++ nx 1 my 2 ny 1 P, m++ n m n Midpoint Midpoint of A(x1, y1) and B(x2,y2) m:n 1:1 x1++ x 2 y 1 y 2 P, 22 Section Formula – External Division P divides AB externally in ratio m:n Y K Clearly PAH ~ PBK H AP AH PH = = BP BK PK X’ O X m x−− x y y L N M =11 = n x−− x y y Y’ 22 mx2−− nx 1 my 2 ny 1 P, m−− n m n Centroid Intersection of medians of a triangle is called the centroid. A(x1, y1) F G E B(x2, y2) D C(x3, y3) Centroid is always denoted x2++ x 3 y 2 y 3 D, by G. 22 x1++ x 3 y 1 y 3 x1++ x 2 y 1 y 2 E, F, 22 22 Centroid A(x1, y1) F G E B(x2, y2) D C(x3, y3) Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1 x++ xx y++ x y y y D, x2++ 2 32 2 3 3 y 2 2 3 112222 L, 1++ 2 1 2 x1++ x 3 y 1 y 3 x1++ x 2 y 1 y 2 E, F, 22 22 Centroid A(x1, y1) F G E B(x2, y2) D C(x3, y3) Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1 x++ xx y++ x y y y D, x2++ 2 31 2 3 3 y 2 1 3 222222 M, 1++ 2 1 2 x1++ x 3 y 1 y 3 x1++ x 2 y 1 y 2 E, F, 22 22 Centroid A(x1, y1) F G E B(x2, y2) D C(x3, y3) Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1 x++ xx y++ x y y y D, x2++ 2 31 2 2 3 y 2 1 2 332222 N, 1++ 2 1 2 x1++ x 3 y 1 y 3 x1++ x 2 y 1 y 2 E, F, 22 22 Centroid A(x1, y1) F G E B(x2, y2) D C(x3, y3) x1+ x 2 + x 3 y 1 + y 2 + y 3 L, 33 x+ x + x y + y + y Medians are M, 1 2 3 1 2 3 x2++ x 3 y 2 y 3 concurrent at the D,33 22 centroid, centroid x1+ x 2 + x 3 y 1 + y 2 + y 3 divides medians in N, x++ x33 y y xratio++ x 2:1 y y E, 1 3 1 3 F, 1 2 1 2 22 22 We see that L M N G Centroid A(x1, y1) F G E B(x2, y2) D C(x3, y3) x1+ x 2 + x 3 y 1 + y 2 + y 3 L, 33 x+ x + x y + y + y Centroid M, 1 2 3 1 2 3 x2++ x 3 y 2 y 3 D,33x1+ x 2 + x 3 y 1 + y 2 + y 3 22 G, x1+ x 2 + x 3 y 1 + y 2 + y 3 33 N, x++ x33 y y x++ x y y E, 1 3 1 3 F, 1 2 1 2 22 22 We see that L M N G Incentre Intersection of angle bisectors of a triangle is called the incentre A(x1, y1) F I E B(x2, y2) D C(x3, y3) Incentre is Let BC = a, AC = b, AB = c the centre of AD, BE and CF are the angle the incircle bisectors of A, B and C respectively. BD AB b bx2++ cx 3 by 2 cy 3 = = D, DC AC c b++ c b c Incentre A(x1, y1) F I E B(x2, y2) D C(x3, y3) AI AB AC AB++ AC c b Now, = = = = ID BD DC BD+ DC a bx++ cx by cy ax+( b + c) 2 3 ay +( b + c) 2 3 11b++ c b c I, a+( b + c) a +( b + c) ax++ bx cx Similarly I can be derived BD AB1 b 2 3 bx2++ cx 3 by 2 cy 3 I = = D,using E and F also DC ACa++ b c c b++ c b c Incentre A(x1, y1) F I E B(x2, y2) D C(x3, y3) AI AB AC AB++ AC c b Now, = = = = ID BD DC BD+ DC a bx++ cx by cy ax+( b + c) 2 3 ay +( b + c) 2 3 11b++ c b c I, a+( b + c) a +( b + c) ax++ bx cx Angle bisectors are BD AB1 b 2 3 bx2++ cx 3 by 2 cy 3 I = = D,concurrent at the incentre DC ACa++ b c c b++ c b c Excentre Intersection of external angle bisectors of a triangle is called the excentre E A(x1, y1) F E Excentre is B(x2, y2) D C(x3, y3) the centre of the excircle EA = Excentre opposite A −ax1 + bx 2 + cx 3 − ay 1 + by 2 + cy 3 E,A −a + b + c − a + b + c Excentre Intersection of external angle bisectors of a triangle is called the excentre E A(x1, y1) F E Excentre is B(x2, y2) D C(x3, y3) the centre of the excircle EB = Excentre opposite B ax1− bx 2 + cx 3 ay 1 − by 2 + cy 3 E,B a− b + c a − b + c Excentre Intersection of external angle bisectors of a triangle is called the excentre E A(x1, y1) F E Excentre is B(x2, y2) D C(x3, y3) the centre of the excircle EC = Excentre opposite C ax1+ bx 2 − cx 3 ay 1 + by 2 − cy 3 E,C a+ b − c a + b − c Cirumcentre Intersection of perpendicular bisectors of the sides of a triangle is called the circumcentre. A C O OA = OB = OC = circumradius B 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.