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 =−+−(x0y0) ( )

=+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 yxyx 1n 1xy22 y n 1n n −− =++++... 2 x2 yxyx 2nn1xy33 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) xy111 Show that xy1022 = xy133 Section Formula – Internal Division

Y

K

Clearly AHP ~ PKB H APAHPH  = = BP PK BK X’ O X m xxyy−− L N M == 11 nxxyy −− 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

xxyy1212++ P, 22 Section Formula – External Division

P divides AB externally in ratio m:n Y

K

Clearly PAH ~ PBK H APAHPH  = = BPBKPK X’ O X m xxyy−− L N M == 11 nxxyy−− 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(x , y ) 1 1 xxyy2323++ D,  F G E 22

B(x2, y2) D C(x3, y3)

x1++ x 3 y 1 y 3 xxyy1212++ E, F,  22 22 Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

x++ xy y x++ 2y2 2 32 3 1122 L,  1++ 21 2   Centroid

A(x , y ) 1 1 xxyy2323++ D,  F G E 22

B(x2, y2) D C(x3, y3)

x1++ x 3 y 1 y 3 xxyy1212++ E, F,  22 22 Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

x++ xy y x++ 2y1 2 31 3 2222 M,  1++ 21 2   Centroid

A(x , y ) 1 1 xxyy2323++ D,  F G E 22

B(x2, y2) D C(x3, y3)

x1++ x 3 y 1 y 3 xxyy1212++ E, F,  22 22 Consider points L, M, N dividing AD, BE and CF respectively in the ratio 2:1

x++ xy y x++ 2y1 2 21 2 3322 N,  1++ 21 2   Centroid

A(x , y ) 1 1 xxyy2323++ D,  F G E 22

B(x2, y2) D C(x3, y3)

x1++ x 3 y 1 y 3 xxyy1212++ E, F,  22 22 x1+ x 2 + x 3 y 1 + y 2 + y 3 L,  33 Medians are x1+ x 2 + x 3 y 1 + y 2 + y 3 M, concurrent at the 33 centroid, centroid x1+ x 2 + x 3 y 1 + y 2 + y 3 divides medians in N,  33 ratio 2:1 We see that L  M  N  G Centroid

A(x , y ) 1 1 xxyy2323++ D,  F G E 22

B(x2, y2) D C(x3, y3)

x1++ x 3 y 1 y 3 xxyy1212++ E, F,  22 22 x1+ x 2 + x 3 y 1 + y 2 + y 3 L,  33

x1+ x 2 + x 3 y 1 + y 2 + y 3 Centroid M,  33xxxyyy123123++++ G,  x1+ x 2 + x 3 y 1 + y 2 + y 3 33 N,  33 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) BD AB b bxcxbycy2323++  = = D, DC AC c bcbc++ AIABACABACcb ++ Now, ==== IDBDDCBDDCa + bx++ cxby cy axb+( cayb ++) c +2323 ( ) 11b++ cb c I, a+( b ++ ca) + b c ( )   Similarly I can be derived ax1++ bx 2 cx 3 I using E and F also a++ b c Incentre

A(x1, y1)

F I E

B(x2, y2) D C(x3, y3) BD AB b bxcxbycy2323++  = = D, DC AC c bcbc++ AIABACABACcb ++ Now, ==== IDBDDCBDDCa + bx++ cxby cy axb+( cayb ++) c +2323 ( ) 11b++ cb c I, a+( b ++ ca) + b c ( )   Angle bisectors are ax1++ bx 2 cx 3 I concurrent at the incentre 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 EA = Excentre opposite A

−ax1 + bx 2 + cx 3 − ay 1 + by 2 + cy 3 E,A   −a + b +− ca + 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 +− ca + 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 −+ ca − 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.

A Orthocentre is always H denoted by H

B C We will learn to find coordinates of Orthocentre after we learn straight lines and their equations Cirumcentre, Centroid and Orthocentre The circumcentre O, Centroid G and Orthocentre H of a triangle are collinear.

H

G O G divides OH in the ratio 1:2 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!! 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 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 = (h+2)2+(k-1)2  6h+4k = 5  equation of locus of (h,k) is 6x+4y = 5 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 B(0,b) A(a,0) and B(0,b)

 h = a/2, k = b/2 P(h,k) Also, a2+b2 = l2  4h2+4k2 = l2 O A(a,0)  equation of locus of (h,k) is 4x2+4y2 = l2 Shift of Origin

Y P(x,y)

X Y y O’(h,k) Consider a point P(x, y) Let the origin be shifted to O’ with coordinates (h, k) X’ O X relative to old axes x Y’ 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 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)

22 Old dist. = (x1 − x 2 ) + (y 1 − y 2 )

22 & New dist.= (x1 − h − x 2 + h) + (y 1 − h − y 2 + h) = Old dist. Rotation of Axes

Y P(x,y)

y Consider a point P(x, y)   Let the axes be rotated through an angle . X’ O X x Let new P  (X, Y) make Y’ an angle  with the new x-axis x y Y X cos(  + ) = , sin( + ) = , sin= , cos = R R R R Rotation of Axes

x y Y X cos(  + ) = , sin( + ) = , sin= , cos = R R R R x cos  cos  − sin  sin  = R y sin cos  + cos  sin  = R X Y x cos  − sin  = RRR X Y y sin + cos  = RRR =xX − cosY sin yX= sinY + cos X = x cos  + y sin  Y= y cos  − x sin  Class Exercise Class Exercise - 1

If the segments joining the points A(a,b) and B(c,d) subtend an angle  at the origin, prove that

acbd+ cos = (abcd2222++) ( ) Solution

Let O be the origin.  OA2 = a2+b2, OB2 = c2+d2, AB2 = (c-a)2+(d-b)2 Using Cosine formula in OAB, we have AB2 = OA2+OB2-2OA.OBcos

22 −(ca) +−( db) = abcd2abcdcos2 + 2 + 2 + 22 −++ 2( 2) 2( )

acbd+ On simplifying, cos = (abcd2222++) ( ) Class Exercise - 2 Four points A(6,3), B(-3,5), C(4,-2) and D(x,3x) are given such that DBC1 = Find x. ABC2 Solution : Given that ABC = 2DBC

631x3x 1  −=−3512 351 42−− 142 1

652( ++) 343( ++) 1620( −) =++++ 2x52( ) 3x43 −( ) 1620( ) 2 28x − 14 = 49 113 xor == x − 49 28x − 14 =  88 2 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.

a a2 1 1 =b b2 1 0 2 c c2 1

R2 → R2-R1, R3 → R3- R2 a a2 1 a a2 1 b − a c − b 1 b + a 0 = 0 b − a b22 − a 0 = 0 ( ) ( ) 1 c+ b 0 c−− b c22 b 0 Solution Cont.

R2 → R2-R3

aa1 2 −−−=(bacb0ac00) ( ) 1cb0+

−−−=(bacbca0) ( ) ( )

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. Class Exercise - 4 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. abab+ + −+ +− 2abxabab + +− + 2aby == and 2222  the required vertex is (-b,b) Class Exercise - 5 If G be the centroid of ABC and P be any point in the plane, prove that PA2+PB2+PC2=GA2+GB2+GC2+3GP2.

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) 2 2 2 2 2 2  LHS = (x1-p) +y1 +(x2-p) +y2 +(x3-p) +y3 2 2 2 2 2 2 2 = (x1 +y1 )+(x2 +y2 )+(x3 +y3 )+3p -2p(x1+x2+x3) =GA2+GB2+GC2+3GP2 Q.E.D. =RHS Class Exercise - 6 The locus of the midpoint of the portion intercepted between the axes by the line xcos+ysin = p, where p is a constant, is 114 (a)xy4p(b)222+=+= xyp222 4112 (c)xy(d)22+=+= pxyp2222 Solution Let the line intercept at the axes at A and B. Let R(h,k) be the midpoint of AB.

pp Rh,k,( )  2cos2sin

pp sin  = , cos  = 2k 2h

pp22 114 += 1 +=Locus 4k4h22 xyp222

 Ans : (b) 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

2 2 2 (hak++) 2 (hak++) 2 4 = = 2 2 2 (hak−+) 2 3 (hak−+) 9 h2ahak4222+++ = h2ahak222−++ 9 +++=5h26ah222 5k5a0

 the required locus is 5x2+ 26ax + 5y 2 + 5a 2 = 0 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 PAPBAB222+= 222 −++++=+(h 2kh) 2k222( 2 ) ( ) ++=2h2k81622  the required locus is x22+= y4 Class Exercise - 9 Find the coordinates of a point where the origin should be shifted so that the equation x2+y2-6x+8y-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 (X+h)2+(Y+k)2-6(X+h)+8(Y+k)-9 = 0 Or, X2+Y2+(2h-6)X+(2k+8)Y+(h2+k2-6h+8k-9) = 0  2h-6 = 0; 2k+8 = 0  h = 3, k = -4. Thus the origin is shifted to (3,-4). Transformed equation is X2+Y2+(9+16-18-32-9) = 0 Or, X2+Y2 = 34 Class Exercise - 10 Through what angle should the axes be rotated so that the equation 11x2+4xy+14y2 = 5 will not have terms in xy? Solution :

Let the axes be rotated through an angle . Thus equation becomes

2 11( X cos − Y sin4 + −) X cos +( Y sin X sin) ( Y cos ) 2 +14 +( X sin = Y cos5) Solution Cont.

Or,( 11cos422 sincos14 + + sinX ) +(4 + cos6 −22 sincos4 sinXY ) +(11sin4 − +=22 sincos14 cos5 )

2cos3sincos2sin0 +22 − =

(cos2sin2cossin0 + −= ) ( ) 1 tanor = − = tan2 2 Therefore, the required angle is 1 − tanor−−11 tan 2 2 Thank you