VIJNAN CLASSES-TWO DIMENSION CO-ORDINATE GEOMETRY -I PUCCONCEPTS, FORMULAE AND SHORTCUTS

Any point on the x-axis will have the co-ordinates of the form (x,0) i.e has ordinate=0Any point on the y-axis will have the co-ordinates of the form (y,0) i.e. has abscissa=0DISTANCE FORMULA:Distance between Two points: Let A and B are two points whose co-ordinates are given by A(x1 ,y1), B(x2 ,y2) respectivelythen AB= x2−x1

2 y2− y12= x1−x2

2 y1− y22

= Differenceof abscissa2Difference of ordinates 2Note: In order to prove that a given figure is a

a) square:Prove all sides and diagonals are equalb) A parallelogram: To prove that opposite sides are equalc) A parallelogram but not a rectangle: To prove its opposite sides are equal but diagonals are not equald) Rectangle: Prove opposite sides are equal and diagonals are equale) Rhombus: Prove all sides are equal and diagonals f) Rhobmus but not a squre: Only sides are equal but diagonals are not equal

Remark: i. Every square is a rectangleii. Every rhombus is a parallelogramiii. In a rhombus diagonals bisect each other.

SECTION FORMULA:

i)Internal division: If P(x,y) divides the line joining A(x1 ,y1) and B(x2 ,y2) in the ratio m:n then x=mx2nx1

mn

y=my2ny1

mni.e co-ordinates of the point P which divides AB internally in the ratio m:n is P

mx2nx1

mn,my2ny1

mn ii)External Division: co-ordinates of the point P which divides AB internally in the ratio m:n is

P mx2−nx1

m−n,my2−ny1

m−n Note: 1. If m/n is positive, the division is internal , but if m/n is negative, the division is external. 2.If R(x,y) divides the join of P(x1 ,y1) and Q(x2 ,y2) in the ration k:1(k>0) then

x=kx2± x1

k±1; y=

ky2± y1

k±1 +ve sign is taken for internal division and negative sign is taken for external

division 3.For finding ratio, use k:1. If k is positive, then divides internally and if k is negative, then divides internally MIDPOINT FORMULA: Co-ordinates of the midpoint of the line joining the points (x1 ,y1) and (x2 ,y2) is given by

x1x2

2,y1 y2

2 Co-ordinates of CENTROID, INCENTRE, CIRCUMCENTRE, ORTHOCENTRE AND EXCENTRE OF A TRAINGLE

CENTROID(G): Centroid of a traingle is the point of concurrency of medians. (The median is the line joining any vertex to midpoint of the opposite side.) The centroid G of the traingle ABC, divide median AD in the ratio 2:1 and Co-ordinate of centroid of the traingle formed by the vertices (x1 ,y1)

(x2 ,y2) and (x3,y3) are given by x1x2 x3

3,y1 y2 y3

3 INCENTRE(I): The incentre I of the traingle is the point of concurrency of the internal bisectors of the angles of the traingle. This is the centre of the circle which touches the sides of the given traingle. (The bisector of the angle is the line which passes through the vertex of the angle and divides into two equal angles). Co-ordinates of incentre I(x,y)=

I ax1bx2cx3

abc,ay1by 2cy3

abc CIRCUMCENTRE(O): Circumcnetre is the point of concurrency of perpendicular bisectors of the sides of traingle.(The lines through the mid point of a side and perpendicular to it ). This is a point which is equidistant from the three vertices of the traingle. This is also the centre O of the circle passing through the vertices of the given traingle. The co-ordinates of circumcentre is given by

O x1 sin2Ax2 sin2Bx3 sin2Csin2Asin2Bsin2C

,y1 sin2A y2 sin2B y3 sin2Csin2Asin2Bsin2C

ORTHOCENTRE(H):Orthocentre of the traingle ABC is the point of concurrency of altitudes of the traingle . Altitude of a traingle is the line drawn through a vertex perpendicular to the opposite side.

H x1 tanA x2tanB x3tanCtanAtanB tanC

,y1 tanA y2 tanB y3tanCsin2Asin2Bsin2C PAGE 1

IMPORTANT NOTE: a)Let ABC be any traingle, If the lengths of the perpendicular from I upon the three sides of the traingle are equal. then I is the Incentre of traingle ABC.b)Orthocentre of a right angled traingle ABC with right angle at the point A is the point A. c)If OA=OB=OC, then O is the circumcentre of traingle ABCd)Centroid G of traingle ABC divdes the median AD or BE or CF in the ratio 2:1

AREA OF A TRAINGLE:Area of traingle ABC with vertices A(x1 ,y1) B(x2 ,y2) and C(x3 ,y3) dentoed by Δ and given by

Δ =12 [x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]=

12 [(x1y2+x2y3 +x3y1)-(y1x2+y2x3+y3x1)]

or= ∣12∣x1 y1 1x2 y2 1x3 y3 1∣∣ =or ∣1

2∣ x1−x3 x2− x3

y1− y3 y2− y3∣∣Corollary: i)If one vertex (x1,y1) is at origin (0,0) then Δ =

12 [ x1 y2−x2 y1]

ii)If points A,B, C are collinear (lies in a straight line) then

Δ =0 i.e. ∣x1 y1 1x2 y2 1x3 y3 1∣ =0 or AB+BC=AC(using distance formula) or AC+AB=BC

iii)If arx+bry+Cr=0(r=1,2,3) are equations the sides of a traingle then area of traingle is given by

Δ =1

2k1 k 2 k 3∣a1 b1 C1

a2 b2 c2

a3 b3 c3∣ Where k1,k2,k3 are the cofactors of c1,c2,c3 in the determinant.

Area of polygon whose vertices are (x1,y1) (x2,y2)(x3, y3)............(xn,yn) is

A=12 [(x1y2-y1x2)+(x2y3-x3y2)+..............+(xny1-ynx1)]

=12 [∣x1 y1

x 2 y2∣∣x2 y2

x3 y3∣.........∣xn−1 yn−1

xn yn ∣∣xn ynx1 y1∣]

NOTE: There is a possibility of getting negative value, then take it's modulus value to find area.LOCUS AND EQUATION OF LOCUS:

LOCUS: When a point moves in a plane under certain geometrical condition or conditions, the point traces out a path. This path of the moving point is known its locus. EQUATION OF LOCUS: The equation to the locus of a point is the relation which is satisfied by the co-ordinates of every point on the locus of the point. Procedure for finding the equation of the locus of a point:

a) Consider the point(Co-ordinates) (h,k) or (x,y) whose locus is to be found. b) Using the geometrical condition satisfied by the point or conditions given , Express the given condition as

equation in terms of 'h' and 'k' by eliminating various parameters givenc) Now generalise (h,k) by replacing h by x and k by y. The equation thus obtained is the locus of the required

point P.

SLOPE OF A STRAIGHT LINE:Inclination of a line: The angle which a line makes with the positive direction of x-axis measured in theanti-clockwise direction is called the inclination of the line. Slope (or gradient) of a straight line: The tangent of the angle made by the line with x axis in the positive direction is called slope of a straight line. If θ(#90°) is the inclination of a straight line, then tanθ is called its slope (or gradient). The slope of a line is usually denoted by m. i.e m=tan θ 0≤≤1800Remark. Since tan θ is not defined when θ = 90°, therefore, the slope of a vertical line i.e. perpendicular to x axis is not defined.

(a) If tan θ >o then 0< θ <900 i.e slope of the line is +ve if inclination of line is acute(b) If tan θ =0, then θ=00 , 1800 which means line is itself x axis or it is parallel to x axis. ie slope of a line

parallel to x axis is zero.(c) If tan θ <0 then 900 < θ <1800 or 2700< θ 3600

(d) The slope is independent of the sense of the line segment i.e. slope of PQ = slope of QP

∴ Slope of QP=tan( +θ )=tanθ =slope of PQSlope of the line joining two pointsThe slope m of a non-vertical line passing through the points P(x1 , y1) and Q(x1, y1) is given by

slope = m = y2− y1x2− x1

=Difference of y co−ordinates ordinatesDifference of x co−ordinates abscissae

Note: If three points A, B, C are collinear then slope of AB =slope of BC =slope of ACRemarks

• Two (non-vertical) lines are parallel iff their slopes are equal. • Two (non-vertical) lines are perpendicular iff the product of their slopes = -1. • Slope of a perpendicular line is the negative reciprocal of the slope of the given line. PAGE 2

π +θ

θ

INTERCEPTS ON X AND Y AXIS: If a line meets the x axis at A, then OA (with proper sign) is calleIf a line meets X axis at A, then OA with proper sign is called the X intercept and usually denoted by a. If a line meets the y axis at B, then OB with proper sign is called the y-intercept, usually denoted by b. If the intercepts are a and b, then the line passes through the points (a,0) and (o,b). Thes intercepts may be +ve or -ve depending upon the location of points A and B.

EQUATION OF A STRAIGHT LINE: Equation of a Straight Line in different forms

1. (i) Equation of a line parallel to x-axis is y = b.

2. ii)Equation of a line parallel to y-axis is x = a.

3. Equation to x axis is y=0

4. Equation to y axis is x=0

5. Slope-Intercept formThe equation of a straight line having slope m and making an intercept c on y-axis isy = m x +c

6. Point-slope formThe equation of a straight line passing through the fixed point (x1, y1) and having slope m is y -y1 = m (x -x1)Corollary. The equation of a line passing through origin and having slope m is y = m x.

An equation of a line passing through the origin and a)making an angle θ with the positive direction of the x axis is y=xtan θ b)having a slope m is y=mx c)passing through the point (x1,y1) is x1y=yx1

7. Two-point form The equation of the line passing through two fixed points A (x1, y1) and B (x2,y2) is

y - y1 = [ y2− y1 x2−x1 ] x−x1 or

y− y1

x−x1=y2− y1

x2− x1In determinant form it is given as:

∣ x y 1x1 y1 1x2 y2 1∣=0 is the equation of line.

8. Intercept formThe equation of the line cutting off intercepts a and b on the axes isxa yb=1 or x

x intercept yy intercept

=1

Note:1. If Line L is parallel to X axis, X

intercept is undefined. If Line L is parallel to Y axis , Y intercept is undefined.2. If x intercept is a then line passes through (a,0) and if y intercept is b then line passes through (b,0) and viceversa

9. Normal (or perpendicular) formThe equation of a straight line in terms of the length of perpendicular from the origin upon it and the angle which this perpendicular makes with the positive direction of x-axis is given by x cosα + y sinα = p where α be the angle made by the perpendicular drawn from origin to the line and p be the length of this perpendicular.

10.Parametric or symmetrical or Distance form of a line: The equation of the line passing through fixed point A(x1,y1) makes an angle θ with the +ve direction of x axis is given byx−x1

cos=y− y1

sin=r say where r is the

distance of any point P(x,y) from the point A(x1,y1) on the same line and x=x1 +r cos θ , y=y1 +r sin θ is the co-ordinate of general point A(x,y) on the line. NOTE: The co-ordinates of any point on the line lying at a distance r from point A(x1 ,y1) can be taken as (x1 +r cos θ , y=y1 +r sin θ)

11. General formEvery straight line can be represented by an equation of the first degree in x and y, and conversely every first degree equation in x, y represents a straight line.The equation a x +b y +c= 0 (where at least one of a and b is non zero)

12. a) Slope and intercepts on axes by the line ax +by +c=0

Slope of a line = −ab= coeff. of xcoeff. of y

x intercept: = − ca y intercept: = − c

bIntercepts: Put y=o in the given equation and

find x= − ca

and

then put y=o and find y= − cb which are

respectively the intercepts on the axes of x and y.

b)The equation ax+by+c=0 represents a line i)parallel to x axis A=0 i.e coeff of x is 0 ii)paralle to y axis B=0 i.e. coeff of y is 0

iii)passing through origin C=0 i.e. constant term is 0.

CONVERSION OF GENERAL EQUATION ax+by+c=0 OF STRAIGHT LINE IN DIFFERENT FORMS:i)Slope intercept form: ax+by+c=0 is by=-ax-c

y= −abx− cb ; y=mx+c; where m= −a

b =−coefficient of xcoefficient of y =Slope of ax+by+c=0

and y intercept(Intercept on y-axis)=−constant termcoefficient of y

PAGE 3

x = a.

y=0

x=0

θ

θA(x1,y1)

P(x,y)

M

(0,0)

y = b.

ii)Intercept form of ax+by+c=0 is ax+by=-c x

− ca

y

− cb

=1 where x intercept= − c

a; y intercept= − c

b

iii)Normal form of general equation ax+by+c=0 is given as -(ax+by)=c − axa2b2

− bya2b2

= c a2b2

xcos ysin=p where xcosα+ysinα=p where

cosα= − axa2b2 sinα= − by

a2b2 p=c

a2b2 p=LENGTH OF PERPENDICULAR FROM ORIGIN TO

STRAIGHT LINE ax+by+c=0 is c

a2b2

ANGLE BETWEEN TWO LINESThe angle between two non-vertical and non-perpendicular lines:

• Let l1 and l2 be the two non-vertical and non-perpendicular lines with slopes m1 and m2 respectively. Let θ 1 and θ 2 be their inclinations, then m1 = tan θ 1 and m2 = tan θ 2. There are two angles θ and π - θ between the lines l1 and l2, given by

tan θ = ± m1−m21m1m2

• If θ is the acute angle between the two lines then tanθ = ∣ m1−m2

1m1m2∣

• Remark: 1)If lines are parallel then θ =00 tanθ =tan00 =0 m1−m2

1m1m2=0 m1=m2

2)If lines are perpendicular to each other i.e. θ=900 then 1+m1m2 =0 i.e. m1m2=-1

3)If lines re perpendicular and slope of one of the line be m then slope of the other line be − 1m

PARALLEL AND PERPENDICULAR LINES: • Conditions of parallelism and perpendicularity.

Two lines with slopes m1 and m2 are(i) parallel if and only if m1 = m2 (ii) perpendicular if and only if m1m2 = -1

• Any line parallel to the line ax+by+c=0 is of the form ax+by+k=0 and any line perpendicular to it is of the form bx-ay+k=0

ANGLE BETWEEN TWO LINES WHOSE GENERAL EQUATIONS ARE GIVEN:

Let θ be the angle between two lines L1 : a1x +b1y +c1 =0 L2 : a2x+b2y+c2 =0 ; then tan θ = ∣a2b1−b2a1

a1a2b1b2∣CONDITONS FOR TWO GIVEN LINES TO BE PARALLEL AND PERPENDICULAR: The two lines a1x +b1y +c1 =0 , and a2x+b2y+c2 =0

i)are parallel if a1

a2=b1

b2≠c1

c2or a1b2 =b1a2

ii) and coincident if a1

a2=b1

b2=c1

c2

iii)These lines will be perpendicular if a1a2 +b1b2 =0 i.e. product of the coefficients of x + product of coefficients of y=0.

iv)Intersecting if a1b2 -a2b1#0 i.e. a1

a2≠b1

b2

POINT OF INTESECTION OF TWO LINES&EQUATION OF LINE PASSING THROUGH IT:1) The point of intersection of two lines a1x +b1y +c1 =0 , and a2x+b2y+c2 =0 is given by obtained by solving the two equations

xb1 c1

b2 c2

= ya1 c1

a2 c2

= 1a1 b1

a2 b2

i.e x= x=b1 c2−b2c1a1b2−a2b1

and y=a2 c1−a1 c2a1b2−a2b1

Provided a1b2#a2c1[cramer's rule]2) Equation of the line through the intersection of L1=0 and L2=0 is of the form L1 +kL2 =0 The equation of family of lines passing through the point of intersection of a1x +b1y +c1 =0 , and

a2x+b2y+c2 =0 is given by : a1x +b1y +c1 +k( a2x+b2y+c2) =0 where k is arbitrary real number.

CONCURRENT STRAIGHT LINES: Three or more lines are said to be concurrent if they pass through the same point. The condition of concurrency of the three given lines a1x +b1y +c1 =0 , a2x+b2y+c2 =0

and a3x +b3y +c3 =0 is ∣a1 b1 c1

a2 b2 c2

a3 b3 c3∣ =0

SOME FACTS about CONCURRENT ST LINES IN THE CARTESIAN SYSTEM:1. Three lines ax+by+c=0, bx+cy+a=0, cx+ay+b=0 will be concurrent if

i)a3 +b3 +c3 =3abc or ii)a2 +b2 +c2 =ab+bc+ca or iii)a+b+c=02. The condition when lines y=m1x +c1 , y=m2x +c2, y=m3x +c3 are concurrent or meet at a point is

m1(c2 -c3) +m2(c3 -c1) +m3(c1-c2)=0 PAGE 4

THE RATIO IN WHICH A GIVEN LINE DIVIDES THE LINE SEGMENT JOINING TWO POINTS: Let the given line ax+by+c=0 divide the line segment joining A(x1, y1) and B(x2 ,y2) in the ratio m:n , then : mn=−

ax1by1cax2by2c

If A and B are on the same side of the given line , then mn is negative, but if A and B are

on opposite sides of the given line, then mn is positive.

LOCATION OF TWO POINTS W.R.T A LINE:Two points (x1,y1) and (x2,y2) are on the same side of a line ax+by+c=0 if ax1 +by1 +c1 =0 if ax1 +by1 +c1 and ax2 +by2

+c are of same sign(i.e. either both positive or both negative) and points are on the opposite side if ax1 +by1 +c and ax2

+by2 +c are of opposite sign. POSITION OF A POINT W.R.T A LINE: A point (x1,y1) lies on the origin side of the line ax+by+c=0 if ax1 +by1+c=0 if ax1 +by1 +c and c have the same sign and (x1,y1) lies on the opposite side of origin if ax1 +by1 +c and c have the opposite signs.

DISTANCE OF A POINT FROM A LINEi) The(Length of ) perpendicular distance d of a point P (x 1, y 1) from the line ax +by +c = 0 is given by

d = ∣ax1by1ca²b² ∣

ii)The distance between the parallel lines ax+by+c=0 and ax+by+c'=0 is ∣ c−c 'a2b2∣ e.g. The distance between the parallel lines 3x-4y+5=0 and 6x-8y+5=0 i.e. 3x-4y+5/2=0

is ∣5−5 /2916∣= 5

2×5=1

2Rule to find the distance between parallel lines:

i. Choose a point on one of the given parallel lines. ii. Find the perpendicular distance from this point to the other line

EQUATION OF THE BISECTORS OF THE ANGLES: between the lines a1x+b1y+c1 =0 and a2x+b2y+c2 =0 (provided a1b2 #b1a2 and c1 , c2 >0) are ax1by1c1

a²b² =±ax 2by2c2

a²b² ii)To discriminate between the acute angle bisector and obtuse angle bisector: Let θ be the angle between one of the lines and one of the bisectors, find tan θ . If tan θ <1 than 2 θ <900 so that this bisector is the acute nagle bisector. If tan θ >1, then we get the bisector to be the obtuse angle bisector. iii) To discriminate between the bisector of the angle containing the origin and that of

the angle not containing the origin, Rewrite the equations 1x +b1y +c1 =0 and a2x+b2y+c2 =0 such that the constant terms c, c' are positive. (if constant terms c1

and c2 are negative make them positive by multiplying (-1) to both sides of the given equations)

Then ax1by1c1

a²b² =

ax2by2c2

a²b² gives the equation of the bisector of the angle containing the origin.

and ax1by1c1

a²b² =−

ax2by2c2

a²b² gives the bisector of the angle not containing the origin.

iv)case 1:As constant terms are positive and If a1a2 +b1b2>0, then the bisector corresponding to '+' sign gives the obtuse angle bisector and bisector corresponding to '-' sign is the bisector of acute angle between the given lines

Equation of the obtuse angle bisector is : ax1by1c1

a²b² =

ax2by2c2

a²b² (origin lies in the obtuse angle)

Equation of the acute angle bisector is : ax1by1c1

a²b² =−

ax2by2c2

a²b² case 2: As constant terms are positive and If a1a2 +b1b2<0, then the

a)Equation of obtuse angle bisector is : ax1by1c1

a²b² =−

ax2by2c2

a²b² (origin lies in the acute angle)

b)Equation of acute angel bisecotor is : ax1by1c1

a²b² =

ax2by2c2

a²b²

ortHOCENTRE, circUMCENTRE, centroid:1. a)If a traingle is right angle, orthocentre lies on that vertex at which traingle is right angle.

b)In an equilateral traingle centroid, circumcentre, Incentre and orhocentre coincide with each other.c)Circumcentre of rightangle traingle is midpoint of its hypotenus.d)Orthocentre, centroid, cirucmcentre of a traingle lies in a straight line(collinear) and centroid divide the line joining the orthocentre and circumcentre in the ratio 2:1. Orthocentre divides the circumcentre and centroid in the ratio 3:2 externally.

2. If LMN be given traingle and traingle DEF is obtained by joining the midpoints of the sides of the traingle LMN thencentroid of Δ LMN =centroid of Δ DEF

PAGE 5

L1

L2

B1

B2

θθ

L

D E

FM N

NOTE: In an isosceles traingle altitude is a median as well as perpendicular bisecotor of the sides(base). In an equilateral traingle, the perpendicular from one vertex onto the opposite side bisects the angle at the vertex and is the perpendicular bisector, altitude and median.

1. NOTE: In an equilateral traingle centroid, cirucumcentre, Incentre and orthocentre coincide with each other. 2. NOTE: In a right angled traingle ABC, ∟C=90 0 , the orthocentre H is the vertex C of the traingle. 3. In a traingle any lines out of three lines i.e. AB, BC, CA are perpendicular then orthocentre is the

point of intersection of two perpendicular lines.

OTHER FACTS:

Area of parallelogram ABCD=p1 P2

sinWhere p1 and p2 are the distances between parallel sides and θ is the angle

between two adjacent sides

Area of Rhombus :(in case of Rhombus) (p1 =p2) =p1

2

sin

Area of Rhombus: =12d 1d 2 Where d1 and d2 are the lengths of two perpendicular diagonals of a rhombus.

SHORT CUTS: 1.MIRROR IMAGE(REFLECTION) OF A POINT P(x1,y1) W.R.T A GIVEN LINE: Let (x1 ,y1 ) be the given point and say ax+by+c=0 be given line, the point (x2,y2) will be mirror image of (x1,y1) in the said line if x2−x1

a=y2− y1

b= −

2 ax1by1c a2b2

Mirror image of (α ,β) in the line ax+by+c=0 is given byx−a

=y−b

= −2 a bc

a2b2

e.g: The image of (2,3) in the line x+3y+4=0 is x−2

1 =y−3

3 = −2 1334

10 or x=-1, y=-6 i.e (-1, 6)

2. FOOT (x2 ,y2 ) OF PERPENDICULAR DRAWN FROM (x1,y1) TO THE LINE ax+by+c=0: We count down the following conditionx2−x1

a=y2− y1

b= −

ax1by1c a2b2

i.e. Foot of the perpendicular drawn from (α ,β) to the line ax+by+c=0 is given byx−a

==y−b

= −abca2b2 e.g: The co-ordinates of the foot of perpendicular from (6,-2) on the

line 4x-5y+7=0 is x−6

4 =y−−2−5

= −4.6−5−27

42−52 or x=2, y=3

3.x axis divides the line segment joining (x1,y1), (x2,y2) in the ratio -y1:y2 and y axis divides the same line segment in the ratio -x1:x2

e.g: x axis divides the line segment joining (2,-4) , (6,2) in the ratio -(-4):2 is 4:2 and y axis divides in the ratio -2:6

4.If ABCD is a parallelogram, then D=A-B+C. e.g: If (1,6), (2,5) , (5,1) are three vertices of a parallelogram, then fourth vertex (1-2+5, 6-5+1) i.e. (4,2)

5.If D, E, F are the midpoints of the sides of BC, CA, AB of ABC, then A=E+F-D; B=F+D-E; C=D+E-F.

e.g: If D(1,1), E(3,-3), F(3,-1) are the midpoints of the sides of BC, CA, AB then A=(3+3-1, -3-1-1)=(5,-5) B=(3+1-3, -1+1-(-3))=(1,3) C=(1+3-3, 1-3+1)=(1,-1)AREA-SHORT CUTS

1. Area of the traingel formed by (x1,y1), (x2,y2), (x3,y3) is ∣12∣ x1−x3 x2− x3

y1− y3 y2− y3∣∣ e.g: Area of traingel

formed by (3,2), (3,-4) (7,2)

= ∣12∣3−7 3−7

2−2 −4−2∣∣ = ∣12∣−4 −4

0 −6∣∣ =12sq units

2. Area of quadrilateral formed by the points (x1,y1), (x2,y2), (x3,y3),(x4,y4) is ∣12∣ x1−x3 x2− x4

y1− y3 y2− y4∣∣e.g. Area of quadrilateral formed by points (2, -1), (4,3), (-1,2) (-3,-2) is ∣1

2∣ 21 43−1−2 32∣∣ =

121521 =18 sq units

PAGE 6

3. Area formed by the lines y=m1x +c1 and y=m2x +c2 and y=m3x +c3 is equals =

∣12 [ c1−c2

2

m1−m2c2−c1

2

m2−m3c3−c1

2

m3−m1 ]∣e.g: The area of the traingle formed by lines y+2x=4, y+3x=5 and y+x+1=0 is

∣12 [ 4−5

2

2

−232

521

2

−32−1

−1−42

−12 ]∣ =12 [ 9

4×−2

1 49

4×−2

125] =

12 [−9

2−49

225] =

12

[−2925 ] =2 sq units. [lines are y=-2x+4; y=-3/2x +5/2; y=-x-1]

4. The area of quadrilateral formed by the lines |x|+|y|=1 is equal to 2 square units.5. Area of the parallelogram formed by the lines

a1x+b1y+c1 =0; a2x +b2y +c2 =0,

a1x+b1y+d1 =0; a2x +b2y +d2 =0, is ∣d 1−c1d 2−c2a1b2−a2b1 ∣

If sides of parallelogram be given by y=mx+a, y=mx+b and y=nx+c, y=nx+d then area will be given by a−bc−d

m−ne.g. Area of parallelogram formedby the lines 3x+4y+5=0; 2x+3y+4=0; 3x+4y+7=0; 2x+3y+5=0 is 7−55−4

9−8=2 sq units

6. Area of Rhombus formed by the lines ax±by±c=0 is given by ∣2c2

2ab∣e.g: Area of Rhombus formed by 4x±3y±5=0 is ∣252

4×3 ∣ =256 sq units

7. Area of traingle made by the line ax+by+c=0 with the co-ordinate axes is ∣ c2

2ab∣e.g: Area of the traingle formed by 4x+y+1=0 with the co-ordinate axes is

12∣41∣ =

18 sq. units

oTHERS: SHORTCUTS1. If the two vertices of an equilateral traingle are (x1,y1), (x2 ,y2) then its third vertex is given y

x1x2±3 y1− y22,

y1 y2±3 x1−x22

e.g: Let A(2,0), B(-2,0) be the ends of an equilateral traingle, then its hird vertex is

2−2±3 0−02,

00±3 222 i.e 0,±23

2. If (x1,y1) and (x2,y2) are the endpoints of hypotenus of a rightangle isosceles traingle then the co-ordinate of the vertex at which traingle is rightangle is given by

A(X,Y) = x1x2± y1− y22

,y1 y2±x1−x2

2 e.g: Let A(2,0), B(0,2) be the end points of

hypotenus of rightangled isoscele traingel.

third vertex is 20±0−22

, 02±2−02 i.e (0,0) or (2,2)

3. Circumcentre of the rightangle traingle ABC, rightangled at A is BC

2e.g.Circumcentre of the traingle formed by A(3,2), B(-3,4), C(7,2)

= 372,−42

2 =(5,-1) Clearly traingle ABC is rt angled. Why?

Because Circumcentre of rightangle traingle is midpoint of its hypotenus.

4. Orthocentre of the rightangled traingel ABC , rightangled at A is A.e.g. Orthocentre of the rt. angled traingle formed by A(3,2), B(3,-4), C(7,2) is (3,2).( rt.angle is at A, If a traingle is right angle, orthocentre lies on that vertex at which traingle is right angle. )

5. If D, E, F are the midpoints of the sides BC, CA, AB of ABC, then the centroid of ABC= centroid of DEFe.g.:If D(1,1), E(3,-3), F(3,-1) are the midpoints of BC, CA, AB, then centroid of ABC is

1333

, 1−3−13 i.e. 7

2,−1 PAGE 7

6. Centroid divides the line joining orhocentre and circumcentre in the ration 2:1e.g. If the orthocentre of a traingle is (-3,5) and circumcentre is (6,2) , then centroid is

261−321

, 2 21521 i.e. (3,3)

7. In an equilateral traingle, orthocentre, centroid, circumcentre, incentre coincide . e.g. The orthocentre, circumcentre, incentre, centroid of equilateral traingle formed by points (a,a), (-a,a),

−a3,a 3 is a−a−a33 , a−aa3

3 = −3a3 , 3a

3 8. A traingle is isosceles if any tow of its medians are equal 9. If the vertices of a triangle ABC are (a, 1/a), (b,1/b), (c,1/c) then co-ordinate of orthocentre is

−1abc

,−abc e.g: Orhocentre of traingle formed by vertices (4,1/4), (3, 1/3), (1,1) is

−14.3 .1

,−4.3.1 = −112,

1210. Equation of straight lines through (x1,y1) making α with

y=mx+c are :(y-y1)=tan(θ-α)(x-x1) and (y-y1)=tan(θ +α)(x-x1)OR

Equation of the lines through (x1,y1) and making an angle Φ with the line ax+by+c=0, b#0 are

y-y1 =m1(x-x1) where m1=tan−tanΦ

1tan tanΦ

y-y1 =m2(x-x2) where m2 =tantanΦ

1−tan tanΦwhere tanθ = −a

b is the slope of the given line. Note that

m1 =tan(θ -Φ), m2=tan(θ +Φ) and when b=0, θ=2

e.g.: the two st. lines through (2,-1) making an angle 450 with the line 6x+5y-1=0;

y+1=m1(x-2) where m1=

−65−tan450

1−65 tan450

=11; y+1=11(x-2) i.e. 11x-y-23=0;

y+1=m2 (x-2) where m2=

−65tan450

1−−65 tan450

=−111 y+1=

−111 (x-2) i.e. x+11y+9=0;

11. If set of lines ax+by+c=0 satisfy the condition la+mb+nc=0(where l, m, and n are constants, the line passing

through a fixed point lm , mn . For example if 3a+2b+4c=0, then set of line ax+by+c=0 satisfy through

a fixed point 34,

12

12. Equation of a line parallel to the lines ax+by+c1=0 and ax+by+c2 =0 and dividing distance between them in the ratio m:n

a)Internally is ax+by+ mc2nc1

mn =0 b)Externally is ax+by+ mc2−nc1

m−n =0

13. Equation of the line parallel to lines ax+by+c1 =0 and ax+by+c2 =0 and equidistant from them is

ax+by+ c1c2

2 =0 as m:n=1:1

EX-CENTRE: Let A(x1,y1), B(x2,y2), C(x3,y3) then (FOR AIEEE)

excentre I1(x,y)= −ax1bx2cx3

−abc,−ay1by2cy3

−abc i.e(centre of excircle)touching the side BC

I2(x,y)= ax1−bx2cx3

a−bc,ay1−by2cy3

a−bc I3(x,y)= ax1bx2−cx3

ab−c,ay1by2−cy3

ab−c

PAGE 8

αα

α1800-α

θ-αθ

θ

A(x1,y1)

y=mx+c

L

A(x1,y2)

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

I1

I2

I3

NINE POINT CIRCLE:A circle which passes through the feet of perpendicular drawn from the vertices of a traingle to the opposite side, the middle points of the sides and the midpoints of the sides joining the angular points to the orthocentre is called nine point circle. (FOR IIT/JEE)In the plane of any traingle ABC, there are nine points through which one and only one circle can passThese nine points are:-

i. Three points, mid points D,E,F of traiangle ABC. ii. Three points, Altitudes G,H,I drawn from vertices A,B,Ciii. Three points, mid points J,K,L of line segments OA,OB, OC where O is the orthocentre (i.e. the point where

the three altitudes AG, BH and CI intersect)The centre of the circle is N which is the midpoint of the line segment OT where T is the circumcentre of the line segment OT where T is the circumcentre of Δ ABC.NOTE: 1)The centre of the nine point circle lies on the line segment joining orthocentre and circumcentre of the traingle. 2)The radius of the nine point circle of a traingle is half the radius of circumcircle3)The centroid, the nine point centre, circumcentre and orhocentre, lies on a straight line.4)Centroid, circumcentre and orthocentre in a traingle are collinear, and Centroid(G) divide the line segment joining the orthocentre(H) and circumcentre(M) internally in the ratio 2:1

Let M be the circum centre and H be the orthocentre andG be the centroid of the traingel in ABC. If OD is perpendicular to BC then D will be its midpoint. Let the median AD meet OP in G. Then it is clear that Δ OGD and Δ PGA are similar.

DGAG

=OGHG

=ODHA

DGAG

=OGHG

=ODHA

= RCOSA2RCOSA

=12

It follows that G is centroid of the traingle ABC and is situated on the line OH and divides it in the ratio 1:2

PAGE 9

A

HOG

D LB C

VIJNAN CLASSES -PAIR OF LINES-SYNOPSISHOMOGENEOUS EQUATION OF SECOND DEGREE:

(1) The homogeneous equation of degree 2 in x and y i.e. ax2 + 2hxy + by2 = 0 represents a pair of lines passing through the origin.These straight lines are

i)real and distinct if h2 -ab>0 i.e. h2ab ii)are coincident if h2-ab=0 i.e h2=ab iii) are imaginary if h2 -ab<0 . The point of intersection of the pair of lines ax2 +2hxy+by2 =0 is origin.

(2) If the individual equations of the above lines are y = m1x and y = m2x where m1 and m2 are the slopes of the lines then m1 + m2 = -2h/b and m1m2 = a/b

(3) Acute angle between these lines is given by tan−1∣2h2– abab ∣

(4) These lines are perpendicular if a + b = 0 and coincident if h2 = ab(5) If the individual equations of the above lines are y = m1x and y = m2x then

m1 + m2 = -2h/b and m1m2 = a/b

(6) The equation of a pair of lines perpendicular to the pair ax2 + 2hxy + by2 = 0 is(7) bx2 – 2hxy + ay2 = 0.(Interchange the co-efficients of x2 and y2 and change the sign of the term

containing xy.(If the lines represented by ax2 +2hxy+by2 =0 be perpendicular, then a+b=0 or b= -a. Hence the equation

becomes ax2 +2hxy-ay2=0 or x2 +2ha x-y2 =0 or x2 +pxy -y2 =0 where p is any constant.)

SECOND DEGREE EQUATION IN X AND Y: 1. The condition for the general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 to represent a pair of lines

is = 0 or abc + 2fgh – af2 – bg2 – ch2 = 0

2. Equation S= ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents a two degree curve, if 3. i)abc + 2fgh – af2 – bg2 – ch2 = 0, then it represents a pair of straight lines.

ii)abc + 2fgh – af2 – bg2 – ch2 # 0, then a) h2 =ab (it's a parabola) b)h2 >ab ( a hyperbola)

c)h2 <ab( an ellipse)

4. The point of intersection of these lines is [ hf – bgab – h2 ,gh– afab – h2 ]

5. Angle between these lines is the same as the angle between the pair of lines through the origin and hence it

is tan−1∣2h2– abab ∣

6. Condition for parallel lines is a/h = h/b = g/f7. The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will represent a pair of parallel lines if

abc + 2fgh – af2 – bg2 – ch2 = 0 and h2 =ab. Also ah=hb= gf

8. Distance between parallel lines = 2 g2 – aca ab

NOTE: Remember that a,b,c stand for the coefficients of x2 and y2 and constant term respectively and f, g, h stand for half the coefficients of y,x and xy.

SHORT CUTS: 1). Equation of the bisectors of angle between the lines represented by a)y=m1x and y=m2x is given by (m1 +m2)(x2 -y2) +2(m1. m2 -1)xy=0

b) ax2 + 2hxy + by2 = 0: are x2− y2

a−b=xyh

or x 2− y2

coeff.of x2−coeff.of.y2=xy

half of coeff. of xy

c) ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 are x− x1

2− y− y12

a−b=x−x1y− y1

h where (x1,y1) is the point of intersection of lines represented by above.

2)Equation of the st. line joining origin to the point of intersection of a second degree curve ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 and a st. line lx+my+n=0 is (i.e.If line lx+my+n=0 intersect the pair of st.lines represented by above at twopoints A and B, then equation of pair of st.lines passing through origin and intersection points of curve and line AB i.e. OA and OB is )

ax2 + 2hxy + by2 + 2gx lxmy−n + 2fy lxmy−n + c lxmy−n 2

= 0

Method: step 1: Take all the terms of x and y in the equation of the st.line on L.H.S and constant term on R.H.S.. Then divide both sides by the constant term on R.H.S. to get 1 on R.H.S. PAGE 10

A

B

X

Y

O

Curve

Line

Step 2: We write 2gx as 2gx.1 ; 2fy as 2fy.1 and c as c(1)2 Putting the value of 1, in the equation of the curve we get the required joint equation e.g.: the joint equation of the pair of st.lines joining the origin to the point of intersection of the

line y=3x+2 and the circle x2 +y2 =a2 is x2 +y2 =a2.(1)2 =a2 y−3x2

2

[ y−3x=2⇒ y−3x2

=1 ]

⇒ 4(x2 +y2)=a2(y-3x)2 Note: If the pair of lines given by the above equation are perpendicular, then coeff.of x2 +coeff. of y2 =0 i.e. 4-9a2 +4-a2 =0 i.e. 8-10a2 =0 i.e. 4-5a2 =0. e.g.2: Find the equation of the st.line oining the origin to the point of intersection of the lne x-y=2 and the curve 5x2 +12xy-8y2 +8x-4y+12=0

Sol: From the equation of the line we have x-y=2 or x− y

2 =1; Equation of the curve

5x2 +12xy-8y2 +4(2x-y)+12=0 i.e 5x2 +12xy-8y2 +4(2x-y) .1 +12.(1)2 =0[Make it homogeneous]

Equation of the required line is 5x2 +12xy-8y2 +4(2x-y) . x− y2 +12. x− y2 2

=0[by(1)]

or (5x2 +12xy -8y2)+2(2x2 -3xy+y2) +3(x2 -2xy +y2)=0 or 12x2 +0.xy-3y2 =0 or y= ± 2x.

3)Point of Intersection of Lines: Let S=ax2 +by2 +2hxy+2gx +2fy+c=0 reprsents a pair of st.lines , then differentiate w.r.t x(assuming y constnat)

S x

=2ax+2hy+2g=0.............(i)

Now differentiate it w.r.t y (assuming x constant)

S y

=2by+2hx+2f=0............(ii)

Solution of equation (i) and (ii) gives the Intersection of lines e.g:The point of intersection of lines given by the equation 3x2 +10xy+3y2 -15x-21y+18=0 is

S x

=6x+10y-15=0..........(i) S y

=10x+6y-21=0............(ii)

From equation (i) and (ii) x=11548 y=

116

4)If the slope of one of the lines represented by ax2 +2hxy+by2 =0 be k times the other, then to determine the

value of k use the result 1k2h = k

ab e.g.: If the slope of the lines 3x2 +2hxy+4y2 =0 are in the ratio 3:1 then h equals____

Here a=3 b=4 k=3 ∴ 132h = 3

ab ⇒ h2 =4ab3 =

4 343

=42 ∴ h=4,-4

OR Let slopes be m and 3m m+3m=4m= −2hb ⇒ m= − h

2b = −h8

m.3m=ab ⇒ 3m2 =

34 ⇒ m2 =

14 ⇒ −h8

2

=14 ⇒ h2 =16 ∴ h=4,-4

5)Area of traingle formed by the lines ax2 +2hxy+by2 =0 and lx+my+n=0 is given by ∣ n2h2−abam2−2hlmbl 2∣

e.g.: Area of the traingle formed by the lines x2 +4xy+y2 =0 and x+y=1 is

∣ −12 22−1.1112−2.2 .1.11.12∣ = 3

2(By putting n=-1, h=2, a=b=l=m=1)

6)If the orhocentre of the traingle formed by the lines ax2 +2hxy+by2 =0 and lx+my+1=0(n=1) be denoted by

(x', y') then (x',y') be calculated as x 'l= y 'm= abam2−2hlmbl 2

7)If two sides of a traingle be represented by the pair of st. lins ax2 +2hxy+by2 =0 and co-ordinate of orthocentre be (c,d) then equation of third side of the traingle be given by (a+b)(cx+dy) =ad2 -2hcd +bc2 8)Product of perpendiculars drawn from the point (x1 ,y1) to the pair of st. lines ax2 +2hxy+by2 =0 is P1P2 =Product of ┴ 's drawn from P(x1,y1) to the given pair to st. lines

= ax122hx1 y1by1

2

a−b24h2

9)If the equation ax2 +by2 +2hxy+2gx +2fy+c=0 represent pair of st.lines then product of length of perpenduculars drawn from (0,0) to these lines is given by

P1 x P2 = c

a−b24h2

10)Joint equation of pair of st.lines through (x1,y1) and ┴ to the lines ax2 +2hxy+by2 =0 is given by b(x-x1)2 -2h(x-x1)(y-y1) +a(y-y1)2 =0 and parallel to ax2 +2hxy+by2 =0 is given by a(x-x1)2 +2h(x-x1)(y-y1) +b(y-y1)2 =0

PAGE 11

P(x1,y1)

P1

P2

L1

L2