1

CARTESIAN CO-ORDINATE SYSTEM :Rectangular Co-ordinate System : Let X' OXand Y'OY be two mutually perpendicular linesthrough any point O in the plane of the pa-per. Point O is known as the origin. The lineX'OX is called the x -axis or axis of x ; theline Y'OY is known as the y-axis or axis of y,and the two lines taken together are calledthe co-ordinates axes or the axes of co-ordi-nates.

X’ X

Y

Y’

III

III IV

QuadrantQuadrant

Quadrant Quadrant

(-.,+) (+,+)

(-,-) (+,-)

1

23

0-1-2-3-1

-2

-3

Region Quad-rant

Nature of Xand Y

Signs ofco-ordin-

ate

XOY I x > 0, y > 0 (+, +)

YOX' II x < 0, y > 0 (- , +)

X'OY' III x < 0, y < 0 (-, -)

Y'OX IV x > 0, y < 0 (+, -)

Note - Any point lying on x-axis or y-axis doesnot lie in any quadrant.Any point can be represented on the planedescribed by the co-ordinate axes by specify-ing its x and y co-ordinates.The x -co-ordinate of the point is also knownas the abscissa while the y-coordinate is alsoknown as the ordinate.

Distance Formula : The distance two pointA (x1, y1) and B(x2, y2) is given by

2122

12 yyxAB x

CO-ORDINATE GEOMETRY

A (x y )1 , 1

B (x y )2 , 2

Note :1. Distance is always positive. Therefore, we

often write AB instead of |AB|.2. The distance of a point P (x, y) from the ori-

gin 22 y x3. The distance between two polar co-ordinates

A (r1 , 1θ ) and B (r2, 2θ ) is given by

2 2

1 21 2 1 2AB r r 2r r cos(θ – θ )

Application of Distance Formulae :(i) For given three points A, B, C to decide

whether they are collinear or vertices of aparticular triangle. After finding AB, BC andCA we shall find that the points are :

• Collinear - (a) If the sum of any two distancesis equal to the thirdi.e. AB + BC = CA. or AB + CA = BC

or BC + CA = AB(b) If are of ABC is zero(c) If slope of AB = slope of BC = slope of CA.

• Vertices of an equilateral triangle if AB = BC= CA

• Vertices of an isosceles triangle if AB = BCor BC = CA or CA = AB.

• Vertices of a right angled triangle if AB2 +BC2 = CA2 etc.

(ii) For given four points A,B,C,D :• AB = BC = CD = DA and AC = BD ABCD is a

square.• AB = BC = CD = DA and AC BD ABCD is

a rhombus.• AB = CD, BC = DA and AC = BD ABCD is a

recatangle.• AB = CD, BC = DA and AC BD ABCD is a

parallelogram.Note :

• The four given points are collinear, if Area ofqaudrilateral ABCD is zero.

• Diagonals of square, rhombus, rectangle andparallelogram always bisect each-other.

• Diagonals of rhombus and square bisect eachother at right angle.

Section Formuale :1.The co-ordinates of a point P(x ,y), dividing

the line segment joining the two pointsA(x1,y1) and B (x 2, y2) internally in the ratiom1 : m2 are given by

2

1 2 2 1 1 2 2 1

1 2 1 2

m m m y m y,y

m m m mx x

x

2

1

mm

BPAP

A ( , y )x1 1

B ( , y )x2 2

P( y)x,m1

m2

2. The co-ordinate of the point P(x, y), dividingthe line segment joining the two points A(x1,y1) and B (x 2, y2) externally in the ratio m1:m2are given by

21

1221

21

1221

mmymym

y,mmmm

xx

x

2

1

mm

BPAP

A ( , y )x1 1

B ( , y )x2 2

P ( , y)x

3. The co-ordinates of the mid-point of the linesegment joining the two points A(x1, y1) andB (x 2, y2) are given by

222121 yy

,xx

A ( , y )x1 1

B ( , y )x2 2

P ( , y)x

Division by Axes : If P (x1, x2) and Q (x2, y2),then PQ is divided by

(i) x - axis in the ratio =2

1

yy

(ii) y - axis in the ratio =2x

x1

Division by a Line : A line ax + by + c = 0

divides PQ in the ratio =cbyacbya

22

11

xx

Area of a triangle : The area of a triangleABC whose vertices are (x1, y1), B(x2, y2) andC(x3 , y3) is denoted by .

1y1y1y

21Δ

33

22

11

xxx

213132321 yyyyyy21

xxx

Area of Polygon : The area of the polygonwhose vertices are (x1, y1), (x2, y2),......(xn , yn)is -

n11n

23321221yy................

.......yyyy21

xxxxxx

Some Important Points in a Triangle :• Centroid : If (x1, y1), (x2 , y2) and (x3, y3) are the

vertices of a triangle, then the co-ordinatesof its centroid are -

3yyy

,3

321321 xxx

• Incentre : If A (x1, y1), B(x2, y2) and C(x3, y3)are the vertices of a ABC s.t. BC = a, CA = band AB = c, then the co-ordinates of itsincentre are

A

B C( ,y )x3 3

( ,y )x 2 2

( ,y )x1 1

a

bc

cbayyay

,cba

321321 cbcxbxax

• Circumcentre : If A(x1, y1), B (x2, y2) and C(x3

, y3) are the vertices of a ABC , then the co-ordinates of its circumcentre are

,2Csin2Bsin2Asin

2Csin2Bsinsin2A 321 xxx

2Csin2Bsin2Asin2Csin2Bsinsin2A 321 yyy

• Orthocentre : Co-ordinates of orthocentre are

,CtanBtan Atan

CtanBtantanA 321 xxx

CtanBtan AtanCtanBtantanA 321 yyy

Note :• If the traingle is equilateral, then centroid,

incentre, orthocentre, circumcentre coin-cides.

• Orthocentre, centroid and circumcentre arealways collinear and centroid divides the linejoining orthocentre and circumcentre in theratio 2 : 1.

• In an isosceles triangle centroid,orthocentre, incentre, circumcentre lies onthe same line.

• Incentre divides the angles bisectors in theratio (b + c) : a, (c + a) : b, (a + b) : c.

3

• Area of the triangle formed by co-ordinateaxes and the line a x + b y + c = 0

is2abc2

Straight Line : A straight line is a curvesuch that every point on the line segmentjoining any two points on it lies on it.

Different Forms of the Equations of aStraight Line :

(a) General Form : The general Form of theequation of a straight line is ax + by + c = 0(First degree equation in x and y). Where a, band c are real constants and a, b are not si-multaneously equal to zero.In this equation, slope of the line is given

by ba .

The general form is also given by y = mx + c ;where m is the slope and c is the intercepton y-axis.

(b) Line Parallel to the X-axis : The equationof a straight line to the x -axis and at a dis-tance b from it, is given by y = bObviously, the equation of the x-axis is y = 0

(c) Line Parallel to Y-axis : The equation of astraight line parallel to the y-axis and at adistance a from is given by x = aobviously, the equation of y-axis is x = 0

(d) Slope Intercept Form : The equation of astriaght line passing through the point A(x1,y1)and having a slope m is given by

(y - y1) = m (x - x1)(e) Two Points Form : The equation of a

straight line passing through two pointsA(x1, y1) and B(x2, y2) is given by

(y - y1) 112

12 -xxyy

xx

Its slope (m) =12

12 yyxx

(e) Intercept Form : The equation of a straightline making intercepts a and b on the axesof x and y respectively is given by

1by

a

x

O X-intercept

Y-in

t erc

ept

A

B

X

Y

Slope (Gradient) of a Line :

bθtan am

{ ax + by + c = 0 bc

by xa

y = mx + c, where bm a

and c is a con-

stant }Here m is called the slope or gradient of aline and c is the intercept on y-axis. The slopeof a line is always measured inanticlockwise.

A

B

OXX

Y

Y

AY

XX

B

Y

Slope of a line in terms of co-ordinates anytwo points on it :-

If (x1, y1) and (x2, y2) are co-ordinates ofany two points on a line, then its slope is

abscissaofDifferenceordinatesofDifferenceyy

m12

12

xx

Angle between two lines :

21

12

mm1m-m

θtan

A

BC

DO

Y

X

• Condition of Parallellism of lines : If theslopes of two lines is m1 and m2 and if theyare parallel, then,

21 mm

• Length of Perpendicular it y or Distance ofa Point from a Line : The length of perpen-dicular from a given point (x1, y1) to a lineax + by + c = 0 is :

1 12 2

a by c

a b

x

Note : The length of Perpendicular from the t

4

origin to the line ax + by + c = 0 is given by

22 ba

c

• Distance between two Parallel Lines : If two

lines are parallel, the distance between themwill always be the same.

When two straight lines are parallelwhose equations are ax + by + c1 = 0 andax + by + c2 = 0, then the distance between

them is given by1 22 2

c – c

a b.

• Changes of Axes : If origin (0, 0) is shifted to(h, k) then the coordinates of the point (x, y)referred to the old axes and (X, Y) referred tothe new axes can be related with the rela-tion

x = X + h and y = Y + k

X

Y

O

(h,k)

x

y

• Point of Intersection of Two Lines : Point ofintersection of two lines can be obtained bysolving the equations as simultaneous equa-tions.

• If the given equations of straight line area1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, then

(i) The angle between the lines ‘ θ ’ is given by

2 1 1 2

1 2 1 2

a b – a btan θ

a a + b b

(ii) If the lines are parallel, then

a2b1 – a1b2 = 0 or2

1

2

1

bb

aa

(iii) If the lines are perpendicular, thena1a2 + b1b2 = 0

(iv) Coincident :2

1

2

1

2

1

cc

bb

aa

• Angle between linesx cos + y sin = P1 andx cos β + y sin β = P2 is | –β |

Exercise

--------------- LEVEL - 1 ----------------

1. The point (–5, 7) lies in the quadrant :(a) First (b) Second(c) Third (d) Fourth

2. The point (7, –5) lies in the quadrant :(a) First (b) Second(c) Third (d) Fourth

3. Find the distance between the points (–6,2)and (2 , 4) :(a) 172 (b) 174

(c) 52 (d) 104. The distance between the points A(b,o) and

B (0, a) is :

(a) 22 ba (b) 22 ba

(c) ba (d) a + b5. The distance between the points A (7, 4)

and B(3, 1) is :(a) 6 units (b) 3 units(c) 4 units (d) 5 units

6. The co-ordinates of point situated on x-axisat a distance of 5 units from y-axis is :(a) (0, 5) (b) (5, 0)(c) (5, 5) (d) (–5, 5)

7. The co-ordinates of a point situated on y-axis at a distance of 7 units from x -axis is:(a) (0, 7) (b) (7, 0)(c) (7, 7) (d) (–7, 7)

8. The co-ordinates of a point below x-axis ata distance of 6 units from x -axis but lyingon y-axis is :(a) (0, 6) (b) (–6, 0)(c) (0, –6) (d) (6, – 6)

9. The distance of the point (6, –8) from theorigin is :(a) 2 units (b) 14 units(c) 7 units (d) 10 units

10. The point of intersection of the lines 2x +7y = 1 and 4x + 5y = 11 is :(a) (4, –1) (b) (2, 3)(c) (–1, 4) (d) (4, –2)

11. The line 4x + 7y = 12 meets x -axis at thepoint :(a) (3, 1) (b) (0, 3)(c) (3, 0) (d) (4, 0)

12. The line 4x – 9y = 11 meets y-axis at thepoint :

(a)

0,

911

(b)

911,0

(c)

411,0 (d)

411,0

13. The slope of the line 3x + 7y + 8 = 0 is :

5

(a) 3 (b) 7

(c)73 (d)

73

14. The slope of the line joining P(–4, 7) andQ(2, 3) is :

(a) 32 (b) 3

2

(c) 23 (d) 2

3

15. The equation of a line parallel to x -axis ata distance of 6 units and above x -axis is :(a) x = 6 (b) y = 6x(c) x = 6y (d) y = 6

16. The equation of a line parallel to y -axis ata distance of 5 units to the left of y-axis, is:(a) y = –5 (b) x = –5(c) x + 5y = 0 (d) y + 5x = 0

17. The equation of a line parallel to x -axisand at a distance of 7 units below x -axisis :(a) y = –7 (b) x = 7(c) x = –7 (d) y = –7x

18. The area of the triangle whose vertices areP (4, 5), Q(–3, 8) and R (3, –4), (in squareunits) is :

(a) 66 (b) 21

16

(c) 33 (d) 3519. The points A(0, 0), B(0, 3) and C(4, 0) are

the vertices of a triangle which is :(a) Isosceles (b) Right angled(c) Equilateral (d) None of these

20. The co-ordinates of the centroid of PQRwith vertices P(–2, 0), Q(9, –3) and R (8, 3)is :

(a) (1, 0) (b)

0,

319

(c) (0, 5) (d) (5, 0)21. The equation of a line passing through the

points A (0, –3) and B(–5, 2) is :(a) x + y + 3 = 0 (b) x + y – 3 = 0(c) x – y + 3 = 0 (d) x – y – 3 = 0

22. The length of perpendicular from the ori-gin to the line 12x + 5y + 7 = 0 is :(a) 2 units (b) 1 unit

(c) 137

units (d)117

units

23. The angle which the line joining the points

,13 and 5,15 makes with x–axis is:(a) 30° (b) 45°(c) 60° (d) 90°

24. The lines whose equations are 2x – 5y + 7= 0 and 8x – 20y + 28 = 0 are :

(a) parallel (b) perpendicualr(c) coincident (d) intersecting

--------------- LEVEL - 2 ----------------

1. If the distance of the point P(x, y) fromA(a, 0) is a + x, then y2 = ?(a) 2 ax (b) 4ax(c) 6ax (d) 8ax

2. If the point (x, y) is equidistant from thepoints (a + b, b – a) and (a–b, a + b) then bx= ?(a) a2y (b) ay2

(c) ay (d) a2 y2

3. If the sum of the square of the distance ofthe point (x, y) from the point (a, 0) and (–a,0) is 2b2, then :(a) x2 + a2 = b2 + y2 (b) x2 + a2 = 2b2 – y2

(c) x2 – a2 = b2 + y2 (d) x2 + a2 = b2 – y2

4. P (– 4, a) and Q(2, a + 4) are two points andthe co-ordinates of the middle point of PQare (–1, 4). The value of a is :(a) 0 (b) 2(c) –2 (d) 3

5. If the points P(2, 3), Q(5, a) and R(6, 7) arecollinear, the value of a is :(a) 5/2 (b) – 4/3(c) 6 (d) 5

6. The equation of a line parallel to x-axis andpassing through (– 6 ,– 5 ) is :(a) y = – 5 (b) x = – 6(c) y = – 5x (d) y = – 6x – 5

7. The equation of a line parallel to y-axis andpassing through (2 ,– 5 ) is :(a)x = 2 (b) y = –5(c) y = 2x (d) x = – 5y

8. Two vertices of a triangle PQR are P(–1, 0)and Q(5, –2) and its centroid is (4, 0). Theco-ordinates of R are :(a) (8, –2) (b) (8, 2)(c) (–8, 2) (d) (–8, –2)

9. The co-ordinates of the point of intersec-tion of the medians of a triangle with ver-tices P(0, 6), Q(5, 3) and R(7, 3) are :(a) (4, 5) (b) (3, 4)(c) (4, 4) (d) (5, 4)

10. The ratio in which the line segment join-ing A(3, –5) and B(5, 4) is divided by x-axisis :(a) 4 : 5 (b) 5 : 4(c) 5 : 7 (d) 6 : 5

11. The ratio in which the line segment join-ing P(–3, 7) and Q (7, 5) is divided by y-axisis :(a) 3 : 7 (b) 4 : 7(c) 3 : 5 (d) 4 : 5

12. The ratio in which the point P

310,1 di-

vides the join of the point A(–3, 2) and B(3,

6

4) is :(a) 2 : 3 (b) 1 : 2(c) 2 : 1 (d) 3 : 1

13. The equation of a line with slope 5 andpassing through the point (–4, 1) is :(a) y = 5x + 21 (b) y = 5x – 21(c) 5y = x + 21 (d) 5y = x – 21

14. The value of a so that the lines x + 3y –8 =0 and ax + 12y + 5 = 0 are parallel is :(a) 0 (b) 1(c) 4 (d) – 4

15. The value of P for which the lines 3x + 8y +9 = 0 and 24x + py + 19 = 0 areperpendicualar is :(a) –12 (b) – 9(c) – 11 (d) 9

16. The value of a so that line joining P(–2, 5)and Q (0, –7) and the line joining A (–4, –2)and B(8, a) are perpendicular to each otheris :(a) –1 (b) 5(c) 1 (d) 0

17. The angle between the lines representedby the equations 2y – 12 x – 9 = 0 and 3 y– x + 7 = 0, is :(a) 30° (b) 45°

(c) 60° (d)2

122

18. If P(3, 5), Q (4, 5) and R(4, 6) be any threepoints, the angle between PQ and PR is :(a) 30° (b) 45°(c) 60° (d) 90°

19. Given a PQR with vertices P (2, 3), Q (– 3,7) and R (– 1, –3). The equation of medianPM is :(a) x – y + 10 = 0 (b) x – 4y –10 = 0(c) x – 4y + 10 = 0 (d) None of these

20. The co-ordinates of the point P which di-

vides the join of A(3, –2) and B

221,

211

in

the ratio 2 : 3 are :(a) (4, 3) (b) (4, 5)

(c)54,2

(d)

27,

23

--------------- LEVEL - 3 ----------------

1. The length of the portion of the straight line8x + 15y = 120 intercepted between theaxes is :(a) 14 units (b) 15 units(c) 16 units (d) 17 units

2. The equation of the line passing throughthe point (1, 1) and perpendicular to the line3x + 4y – 5 = 0, is :(a) 3x + 4y – 7 = 0 (b) 3x + 4y + k = 0(c) 3x – 4y – 1 = 0 (d) 4x – 3y + 1 = 0

3. The equation of a line passing through thepoint (5, 3) and parallel to the line 2x – 5y +3 = 0, is :(a) 2x – 5y – 7 = 0 (b) 2x – 5y + 5 = 0(c) 2x – 2y + 5 = 0 (d) 2x – 5y = 0

4. The sides PQ, QR, RS and SP of a quadrilat-eral have the equations x + 2y = 3, x = 1,x – 3y = 4, 5x + y + 12 = 0 respectively, thenthe angle between the diagonals PR and QSis :(a) 30° (b) 45°(c) 60° (d) 90°

5. The equations of two equal sides of an isos-celes triangle are 7x – y + 3 = 0 and x + y –3 = 0 and its third side passes through thepoint (1, –10). The equation of the third sideis :(a) x – 3y – 31 = 0 but not x – 3y – 31 = 0(b) neither 3x + y + 7 = 0 nor x – 3y – 31 = 0(c) 3x + y + 7 = 0 or x – 3y – 31 = 0(d) 3x + y + 7 = 0 but not x – 3y – 31 = 0

6. If P1 and P2 be perpendicular from the ori-gin upon the straight lines x sec θ + ycosecθ = a and x cosθ – y sin θ = a cos 2θ

respectively, then the value of 4 22

21 PP is

:(a) a2 (b) 2a2

(c) 22 a (d) 3a2

7. Find the equation of the line passingthrough the point (2, 2) and cutting off in-tercepts on the axes whose sum is 9 ?(a) x + 2y – 6 = 0 but not 2x + y – 6 = 0(b) neither x + 2y – 6 = 0 nor 2x + y – 6 = 0(c) 2x + y – 6 = 0 but not x + 2y – 6 = 0(d) x + 2y – 6 = 0 or 2x + y – 6 = 0

LEVEL - 11. (b) 2. (d) 3. (a) 4. (b)5. (d) 6. (b) 7. (a) 8. (c)9. (d) 10. (a) 11. (c) 12. (b)

13. (c) 14. (a) 15. (d) 16. (b)17. (a) 18. (c) 19. (b) 20. (d)21. (a) 22. (c) 23. (a) 24. (c)

LEVEL - 21. (b) 2. (c) 3. (d) 4. (b)5. (c) 6. (a) 7. (a) 8. (b)9. (c) 10. (b) 11. (a) 12. (c)

13. (a) 14. (c) 15. (b) 16. (d)17. (a) 18. (b) 19. (c) 20. (a)

LEVEL - 31. (d) 2. (c) 3. (b) 4. (d)5. (c) 6. (a) 7. (d)

7

Hints and Solutions :LEVEL - 11. (b)

The point (–5, 7) lies in the second quad-rant.

2. (d)The point (7, –5) lies in the fourth quad-rant.

3. (a)Distance between two points

212 yyxx 212

here (x1, y1) = (–6, 2) and (x2, y2) (2, 4) Required distance =

22 4226–

unit17268464 4. (b)

2 2 2 2AB b – 0 0 – a b a

22 ba 5. (d)

AB2 = (7 – 3)2 + (4 – 1)2 = 42 + 32 = 16 + 9 = 25

AB = units525 6. (b)

Clearly, the point of x -axis has ordinate 0and abscissa 5.So, the point is (5, 0)

7. (a)Clearly, the point on y-axis has abscissa 0.So, the point is (0, 7)

8. (c)Clearly, the point is (0, 6)

9. (d)

Required distance = 2 26 – 0 8 – 0

units101006436 10. (a)

2x + 7y = 1 ........ (i)4x + 5y = 11 ......(ii)on solving (i) and (ii), we get x = 4 and y =–1 Required point of intersection = (4, –1)

11. (c)Equation of x-axis is y = 0put y = 0 in 4x + 7y = 12 we get x = 3 Required point = (3, 0)

12. (b)Equation of y-axis is x = 0

put x = 0 in 4x – 9y = 11 we get y = 911

Required point =110, –9

13. (c)3x + 7y + 8 = 0 7y = – 3x – 8

78x

73y

73islinetheofSlope

14. (a)

32

64

4273yy

PQofSlope12

12

xx

15. (d)Clearly; the equation of the line is, y = 6

16. (b)Clearly, the equation of the line is, x = – 5

17. (a)Clearly, the equation of the line is y = –7

18. (c)

1 2 3 2 3 1 3 1 21 y y y y y y2

x x x

85354384421

unitssq.336621

19. (b)

2 2AB 0 0 0 3 3

2 2AC 4 0 0 0 4

2 2and BC 4 0 0 3 5

AB2 + AC2 = BC2

ABC is a right angled triangle.

A

B

C

(0, 3)

(4,0)(0, 0)

20. (d)The co-ordinates of the centroid of PQRare -

05,3

330,3

892

21. (a)The required equation is

2 3y 3 05 0

x

y + 3 = – x x + y + 3 = o

22. (c)Length of perpendicular =

2 2

12 0 5 0 7 7 units1312 5

8

23. (a)

The slope of the line is31

31515

30θor31θtan

24. (c)Here,

41

287

ccand

41

205

bb,

41

82

aa

2

1

2

1

2

1

2

1

2

1

2

1

cc

bb

aa

, So the given lines are co-

incident.

LEVEL - 21. (b)

2 2a (y 0) ax x

2 22x a y a x

222 axaxy

axy 42 2. (c)

Let (x, y), Q(a + b, b – a) and R(a – b, a + b)are given points. PQ = PR.

22 abba yx

22 baba yx x 2 – 2x (a + b) + (a + b)2 + y2 – 2y(b –a) +(b+ a)2 = x 2 + (a – b)2 – 2x (a – b) + y2 + (a + b)2– 2y (a + b) ax + bx + by – ay = ax – bx + ay + by 2bx = 2ay bx = ay.

3. (d)Let A (x, y), P(a, 0) and Q(–a, 0), Then, AP2 + AQ2 = 2b2

[(x – a)2 + (y – 0)2] + [(x + a)2 + (y – 0)2] = 2b2

x 2 + a2 – 2ax + y2 + x 2 + a2 + 2ax + y2 = 2b2

2(x 2 + a2 + y2) = 2b2

x 2 + a2 + y2 = b2

x 2 + a2 = b2 – y2

4. (d)co-ordinates of middle point (–1, 4)

42a842a42

4aa

a = 25. (c)

Since, P,Q and R collinearslope of PQ = slope of PR

44

33–a

2–63–7

2–53–a

a – 3 = 3 a = 66. (a)

The equation of a line parallel to x-axis is y= b.Since, it passes through (–6, –5), so b = –5 The required equation is, y = –5

7. (a)The equation of a line parallel to y-axis is,x = a.Since, it passes through (2, –5), so a = 2 The required equation is, x = 2

8. (b)Let the co-ordinates of R be (x, y). Then,

–1 5 0 – 2 y4 and 03 3

x

or 4 + x = 12 and –2 + y = 0or x = 8 and y = 2 R = (x, y) = (8, 2)

9. (c)Since, point of intersection of median is“centroid”. co-ordinates of centroid

3336,

3750

312,

312

44,10. (b)

Let the ratio be k : 1The ordinate of a point lying on x-axis mustbe zero

45k54k0

1k154k

4:51:45isratioRequired

11. (a)Let the ratio be k : 1The abcissa of a point lying on y-axis mustbe zero

73k037k0

1k137k

7:31:73isratioRequired

12. (c)Let the ratio be k : 1

11k

133k

3k – 3 = k + 1 2k = 4 k = 2 Required ratio is 2 : 1

13. (a)Let the equation be y = 5x + cSince it passes through (–4, 1), we have 1= 5(–4) + c c = 21 so, its equation is, y = 5x + 21

9

14. (c)

Condition of parallelism2

1

2

1

bb

aa

4a123

a1

Alternatively,

31m

38

31y083y 1

xx

12ay005y1

12

52

xax

2am12

for parallelism, m1 = m2

412

aa31

15. (b)Condition of perpendicularism, a1a2 + b1b2= 0 3 24 + 8 p = 0 8p = –3 24 p = –9Alternatively-3x + 8y + 9 = 0

83m

89

83y 1

x

p19

p24y019py24

xx

224mp

for perpendicularism, m1.m2 = –1

3 24 18 p

9–P 16. (d)

m1 = Slope of PQ 6212

2057

m2 = Slope of AB 1248a 22

a

m1m2 = –1a 26 112

a + 2 = 2 a = 017. (a)

29

212y09.122y xx

3212m1

37

31y07y3

xx

31m2

1 2

1 2

13m – m 13 tan θ11 m .m 31 3.3

30θSo,18. (b)

Slope of PQ, 03455m1

Slope of PR, 13456m2

10110

mm1mm

θtan21

21

45θSo,19. (c)

Clearly, M is the mid-point of QR.

Co-ordinates of M are

237,

213

i.e. (–2, 2)Now, find the equation of the line joiningP(2, 3) and M(–2, 2)

Required equation is , (y – 3) = 2–2–3–2

(x –

2)

2–12–4y2413y xx

0104y– x20. (a)

Required point is :

23221223

,23

211233

34,5

15,520

LEVEL - 31. (d)

Point of intersection at x -axis = (x, 0) 8x + 15 y = 120 8x + 15 0 = 120 x = 15 point ot intersection = (15, 0)Point of intersection at y-axis = (0, y) 8x + 15 y = 120 0 + 15y = 120 y = 8 Point of intersection = (0, 8) Required length = AB

10

O

BA

(0, 8)

(15,O)

22 80015

28964225

units172. (c)

Given line - 3x + 4y – 5 = 045

43y

x

its slope,43m1

Let m2 be the slope of required line.

Then, m1m2 = –1 or

43

m2 = –1

m2 = 34

Let the required equation be, y = m2x + c

cx 34y

Since, it passes through (1, 1)

31

341cc1

341

31

34y,isequationrequiredthe x

or 4x – 3y – 1 = 03. (b)

2x – 5y + 3 = 0

53

52y x

52mslopeits 1

Let the slope of line which is parall;el tothe given line is m2

52mm 12

Let the required equation be,2y5

x c

Since, it passes through (5, 3)

1cc5523

152yis,equationRequired x

055y2or x4. (d)

x + 2y = 3 .......... (i)5x + y = –12 ........(ii)On solving (i) and (ii), we get x = –3, y = 3 co-ordinates of P(– 3, 3)

R

P

QS5+ y +12=0

x

x = 1x – 3y = 4

x + 2y= 3

Similarly, Q(1, 1), R(1, –1) and S(–2, 2)

Now, m1 = slope of PR = 13131

m2 = slope of QS = 11–2–12

1mm 21

90isanglerequiredthe5. (c)

Third side passes through (1, –10)so its equation y + 10 = m(x – 1) .......(i)This side makes equal angle with thegiven two sides.let this angle be θ .Now, slope of line 7x – y + 3 = 0 is m1, m1 = 1

and slope of line x + y – 3 = 0 is m2,m2 = –1

angle between (i) and 7x – y + 3 = 0 = anglebetween (i) and x + y – 3 = 0

1–m1

1––m7m1

7–mθtan

m –3 or 1/3 Hence possible equations of third side arey + 10 = –3 (x – 1)

1and y 10 –13

x

or 3x + y + 7 = 0 and x – 3y – 31 = 06. (a)

P1 = lenght of perpendicular from (0, 0) on xsecθ + y cosec θ = a

1 2 2

2 2

a aP1 1sec θ cosec θ

cos θ sin θ

.cosθsinθaor 2P1 = a(2sinθ . cosθ ) 2P1 = a sin 2θ

11

Similarly, 2θcosaθsinθcos

2θcosaP222

2 2 2 2 2 21 2 4P P a sin 2θ. cos 2θ a

7. (d)Let a and b are the intercepts on x and y-axes respectively. a + b = 9 b = 9 – a .........(i)and the equation of the line is

1by

a

x ..........(ii)

From (i) and (ii)

iii......1a–9

ya

x

this line also passes through the point (2,2)

1a9

2a2(iii)from

On solving we get a = 6 or a = 3If a = 6 then b = 9 – 6 = 3

equation of the line is 1y6

3

x

or x + 2y – 6 = 0If a = 3 then b = 9 – 3 = 6

equation of the line is 1y3

6

x

or 2x + y – 6 = 0Hence, required equation is

x + 2y – 6 = 0 or 2x + y – 6 = 0Note : Solve this type of question with thehelp of given options.