Book 11_CBSE Notes_Straight Lines

Vidyamandir Classes

M_XI_Straight Line 1 VMC/School Level Notes

TOPIC: STRAIGHT LINE

Points to remember

Distance Formula

Distance between the point 1 1( , )P x y and 2 2( , )Q x y is

2 22 1 2 1 PQ x x y y

In particular, the distance of a point ( , )P x y from the origin (0,0)O is given by

2 2 . OP x y

Section Formula

The coordinates of a point dividing the line segment joining the points 1 1( , )x y and 2 2( , )x y internally, in ratio m : n

are 2 1 2 1, .

mx nx my nym n m n

The coordinates of the mid-point P of the join of the points 1 1( , )A x y and 2 2( , )B x y is

1 2 1 2,2 2

x x y y

Area of the triangle whose vertices are 1 1 2 2( , ),( , )x y x y and 3 3( , )x y is

1 2 3 2 3 1 3 1 21 .2

x y y x y y x y y

Slope of a line

If θ is the angle made by a line with positive direction of x-axis in anticlockwise direction, then the value of tanθ is called the slope of the line and is denoted by m.

The slope of a line passing through points 1 1( , )P x y and 2 2( , )Q x y is given by

2 1

2 1

tan θ

y ymx x

Angle between two lines

The angle θ between the two lines having slopes m1 and m2 is given by

1 2

1 2

( )tanθ1

m m

m m

If we take the acute angle between two lines, then 1 2

1 2

tanθ1

m m

m m

If the lines are parallel, then 1 2.m m

If the lines are perpendicular, then 1 2 1. m m

Vidyamandir Classes


Slope or gradient of a line is defined as tan θ, (θ 90 ), m where θ is angle which the line makes with

positive direction of x-axis measured in anticlockwise direction, 0 θ 180

Slop of x-axis is zero and slope of y-axis is not defined.

Slope of a line through given points 1 1( , )x y and 2 2( , )x y is given by 2 1

2 1

y yx x

Two lines are parallel to each other if and only if their slopes are equal.

Two lines are perpendicular to each other if and only if their slopes are negative reciprocal of each other.

Acute angle α between two line, whose slopes are m1 and m2 is given by 1 21 2

1 2

tan α ,1 01

m m m m

m m

x a is a line parallel to y-axis at a distance of a units from y-axis.

x a lies on right or left of y-axis according as a is positive or negative.

y b is a line parallel to x-axis at a distance of ‘b’ units from x-axis. y b lies above or below x-axis,

according as b is positive or negative.

Equation of a line passing through given point 1 1( , )x y and having slope m is given by 1 1( ) y y m x x

Equation of a line passing through given points 1 1( , )x y and 2 2( , )x y is given by 2 11 1

2 1

( )

y yy y x xx x

Equation of a line having slope m and y-intercept c is given by y mx c

Equation of line having intercepts a and b on x-axis and y-axis respectively is given by 1 x ya b

Equation of line in normal form is given by cosα sin α , x y p p = length of perpendicular segment from

origin to the line α Angle which the perpendicular segment makes with positive direction of x-axis

Equation of line in general form is given by 0, , Ax By C A B and C are real numbers and at least one

of A or B non zero.

Distance of a point 1 1( , )x y from line 0 Ax By C is given by 1 12 2

| |

Ax By CdA B

Distance between two parallel lines 1 0 Ax By C and 2 0 Ax By C is given by 1 22 2

| |

C CdA B

Shifting of origin to a new point without changing the direction of the axes is known as translation of axes.

Let OX, OY be the original axes and O’ be the new origin. Let coordinates of O’ referred to original axes be ( , ).h k

Let ( , )P x y be point in plane

Vidyamandir Classes


Let O’X’ and O’Y’ be drawn parallel to and in same direction as OX and OY respectively. Let coordinates of P

referred to new axes O’X’ and O’Y’ be ( ', ')x y then ' , ' x x h y y k or ' , ' x x h y y k

Thus

(i) The point whose coordinates were ( , )x y has now coordinates ( , ) x h y k when origin is shifted to

( , ).h k

(ii) Coordinates of old origin referred to new axes are ( , ). h k

Equation of family of lines parallel to 0 Ax By C is given by 0, Ax By k for different real values

of k

Equation of family of lines perpendicular to 0 Ax By C is given by 0, Bx Ay k for different real

values of k.

Equation of family of lines through the intersection of lines 1 1 1 0 A x B y C and 2 2 2 0 A x B y C is

given by 1 1 1 2 2 2( ) ( ) 0 A x B y C k A x B y C for different real values of k.

PRACTICE QUESTIONS

1. If the angle between two lines is π4

and slope of one of the lines is 1 ,2

find the slope of the other line.

[Ans. 13or 3

m m ]

2. Three point 1 1( , ), ( , )P h k Q x y and 2 2( , )R x y lie on a line. Show that

1 2 1 1 2 1( )( ) ( )( ). h x y y k y x x

3. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at

the origin. Find vertices of the triangle.

[Ans. (0, ),(0, )a a and 3 ,0 a or 0, , 0, ,a a and 3 ,0a ]

Vidyamandir Classes


4. Find the distance between 1 1( , )P x y and 2 2( , )Q x y when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel

to the x-axis. [Ans. (i) 2 1| |,y y (ii) 2 1| |x x ]

5. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured

anticlockwise. [Ans. 3 ]

6. If three points ( ,0),( , )h a b and (0, )k lie on a line, show that 1. a bh k

7. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1 ,3

find the

slopes of the lines. [Ans. 1 and 2, or 12

and 1, or –1 and –2, or 12

and –1]

8. The vertices Δ PQR are (2,1), ( 2,3)P Q and (4, 5).R Find equation of the median through the vertex R.

[Ans. 3 4 8 0 x y ]

9. A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find

the equation of the line. [Ans. (1 ) 3(1 ) 11 n x n y n ]

10. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is

9. [Ans. 2 6 0,2 6 0 x y x y ]

11. Point ( , )R h k divides a line segment between the axes in the ratio 1 : 2. Find equation of the line.

[Ans. 2 3 kx hy kh ]

12. Reduce the equation 3 8 0 x y into normal form. Find the values of p and ω .

[Ans. 4p and ω 30 ]

13. Find the points on the x-axis, whose distances from the line 13 4

x y are 4 units.

[Ans. (–2, 0) and (8, 0)l]

14. Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2,

find equation of the other line.

[Ans. 3 2 2 3 1 8 3 1 x y or 3 2 1 2 3 1 8 3 x y ]

15. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3 4 16 0. x y

[Ans. 68 49,25 25

]

16. The perpendicular from the origin to the line y mx c meets it at the point (–1, 2). Find the values of m

and c. [Ans. 1 5,2 2

m c ]

Vidyamandir Classes


17. If p and q are the lengths of perpendiculars from the origin to the lines cosθ sin θ cos2θ x y k and

secθ cosecθ , x y k respectively, prove that 2 2 24 . p q k

Sol. The equations of given lines are

cosθ sin θ cos2θ x y k … (1)

secθ cosecθ x y k … (2)

The perpendicular distance (d) of a line 0 Ax By C from a point 1 1( , )x y is given by

1 12 2

| | .

Ax By CdA B

On comparing equation (1) to the general equation of line i.e., 0, Ax By C we obtain

cosθ, sin θ, A B and cos2θ. C k

It is given that p is the length of the perpendicular from (0, 0) to line (1).

2 2 2 2 2 2

| (0) (0) | | | | cos2θ | | cos2θ |cos θ sin θ

A B C C kp kA B A B

… (3)

On comparing equation (2) to the general equation of line i.e., 0, Ax By C we obtain

secθ, cosecθ, A B and . C k

It is given that q is the length of the perpendicular from (0, 0) to line (2).

2 2 2 2 2 2

| (0) (0) | | | | |sec θ cosec θ

A B C C kpA B A B

… (4)

From (3) and (4), we have

2

22 2

2 2

| |4 | cos 2θ | 4sec θ cosec θ

kp q k

22 2

2 2

4cos 2θsec θ cosec θ

kk

2

2 2

2 2

4cos 2θ1 1

cos θ sin θ

kk

2

2 22 2

2 2

4cos 2θsin θ cos θsin θcos θ

kk

2

2 2

2 2

4cos 2θ1

sin θcos θ

kk

2 2 2 2 2cos 2θ 4 sin θcos θ k k

Vidyamandir Classes


22 2 2cos 2θ 2sinθcosθ k k

2 2 2 2cos 2θ sin 2θ k k

2 2 2cos 2θ sin 2θ k

2 k

Hence, we proved that 2 2 24 . p q k

18. If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then

show that 2 2 21 1 1 . p a b

19. Find the distance of the line 4 0 x y from the point (4,1)P measured along the line making an angle of

135° with the positive x-axis. [Ans. 3 2 units]

20. Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the

line 3 4 0. x y [Ans. 6 7,5 5

]

21. Show that the area of the triangle formed by the lines

1 1 2 2, y m x c y m x c and 0x is 21 2

1 22 | |

c cm m

22. A line is such that its segment between the lines 5 4 0 x y and 3 4 4 0 x y is bisected at the point

(1, 5). Obtain its equation. [Ans. 107 3 92 0 x y ]

23. Show that the path of a moving point such that its distances from twolines 3 2 5 x y and 3 2 5 x y are

equal is a straight line.

24. Find the value of k for which the line 2 2( 3) (4 ) 7 6 0 k x k y k k is

(a) Parallel to the x-axis, (b) Parallel to the y-axis,

(c) Passing through the origin. [Ans. (a) 3, (b) 2, (c) 6 or 1]

25. Find the values of θ and ,p if the equation cosθ sin θ x y p is the normal form of the line

3 2 0. x y [Ans. 7π ,16

]

26. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6,

respectively. [Ans. 2 3 6, 3 2 6 x y x y ]

27. What are the points on the y-axis whose distance from the line 13 4

x y is 4 units.

[Ans. 8 320, , 0,3 3

]

Vidyamandir Classes


28. Find perpendicular distance from the origin of the line joining the points (cosθ,sin θ) and (cos ,sin ).

Sol. The equation of the line joining the points cosθ,sin θ and cos ,sin is given by

sin sinθsinθ ( cosθ)cos cosθ

y x

cos cosθ sinθ(cos cosθ) (sin sin θ) cosθ(sin sin θ) y x

cosθ sin (cos cosθ) cosθsin cosθsin θ sin θcos sin θcosθ 0 y y

(sin θ sin ) (cos cosθ) sin( θ) 0 x y

0, Ax By C where sin θ sin , cos cosθ, A B and sin( θ) C

It is know that the perpendicular distance (d) of a line 0 Ax By C from a point

1 1( , )x y is given by 1 12 2

| |

Ax By CdA B

Therefore, the perpendicular distance (d) of the given line from point 1 1( , ) (0,0)x y is

2 2

| (sin θ sin )(0) (cos cosθ)(0) sin( θ) |

sin θ sin cos cosθ

d

2 2 2 2

| sin( θ)sin θ sin 2sinθsin cos cos θ 2cos cosθ

2 2 2 2

| sin( θ)

sin θ cos θ sin cos 2 sinθsin cosθcos

| sin( θ)1 1 2 cos θ

| sin( θ)2 1 cos θ

2

| sin( θ)θ2 2sin

2

sin( θ)θ2sin

2

29. Find the equation of a line drawn perpendicular to the line 14 6

x y through the point, where it meets the

y-axis. [Ans. 2 3 18 0 x y ]

Vidyamandir Classes


30. Find the equation of a line drawn perpendicular to the line 14 6

x y through the point, where it meets the

y-axis. [Ans. 2 3 18 0 x y ]

31. If three lines whose equations are 1 1 2 2, y m x c y m x c and 3 3 y m x c are concurrent, then show that

1 2 3 2 3 1 3 1 2( ) ( ) ( ) 0. m c c m c c m c c

32. Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line 2 3. x y

Sol. Let the slope of the required line be m1.

The given line can be written as 1 3 ,2 2

y x which is of the form y mx c

Slope of the given line 212

m

It is given that the angle between the required line and line 2 3 x y is 45°.

We know that if θ is the acute angle between line 1l and 2l with slopes 1m and 2m respectively, then

2 1

1 2

tanθ1

m m

m m.

1 2

1 2

| |tan 451

m m

m m

1

1

1211

2

m

m

1

1

1 221 22

m

m

1

1

1 212

mm

1

1

1 212

mm

1 1

1 1

1 2 1 21 or 12 2

m mm m

1 1 1 12 1 2 or 2 1 2 m m m m

1 11 or 33

m m

Vidyamandir Classes


Case I: 1 3m

The equation of the line passing through (3, 2) and having a slope of 3 is:

2 3( 3) y x

2 3 9 y x

3 7 x y

Case II: 113

m

The equation of the line passing through (3, 2) and having a slope of 13

is:

12 33

y x

3 6 3 y x

3 9 x y

Thus, the equations of the lines are 3 7 x y and 3 9. x y

3 7 x y , 3 9. x y

32. Show that the equation of the line passing through the origin an making an angle θ with the line y mx c

is tanθ .1 tanθ

y mx m

Sol. Let the equation of the line passing through the origin be 1 .y m x

If this line makes an angle of θ with line , y mx c then angle θ is given by

1

1

tanθ1

m m

m m

tan θ1

y mx

y mx

tan θ1

y mx

y mx

tan θ or tan θ1 1

y ym mx x

y ym mx x

Vidyamandir Classes


Case I: tan θ1

y mx

y mx

tan θ1

y mx

y mx

tanθ tanθ y ym mx x

tanθ 1 tanθ ym mx

tanθ1 tanθ

y mx m

Case II: tan θ1

y mx

y mx

tan θ1

y mx

y mx

tanθ tanθ y ym mx x

(1 tanθ) tanθ y m mx

tanθ1 tanθ

y mx m

Therefore, the required line is given by tanθ1 tanθ

y mx m

33. Find the direction in which a straight line must be drawn through the point ( 1,2) so that its point of

intersection with the line 4 x y may be at a distance of 3 units from this point.

Sol. Let y mx c be the line through point ( 1,2).

Accordingly, 2 ( 1) . m c

2 m c

2 c m

2 y mx m … (1)

The given line is

Vidyamandir Classes


4 x y … (2)

On solving equations (1) and (2), we obtain

2 5 2and 1 1

m mx y

m m

2 5 2,1 1

m m

m m is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (–1, 2), according to distance formula,

2 22 5 21 2 3

1 1

m mm m

2 2

22 1 5 2 2 2 31 1

m m m mm m

2

2 2

9 9 9( 1) ( 1)

mm m

2

21 1

1

m

m

2 21 1 2 m m m

2 0 m

0 m

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.

34. Find the distance of the line 4 7 5 0 x x from the point (1, 2) along the line 2 0. x y

[Ans. 23 518

units]

35. Find the image of the point (3, 8) with respect to the line 3 7 x y assuming the line to be a plane mirror.

Sol. The equation of the given line is

3 7 x y … (1)

Let point ( , )B a b be the image of point (3, 8).A

Accordingly, line (1) is the perpendicular bisector of AB.

Slope of 8 ,3

bABa

while the slope of line 1(1)3

Since line (1) is perpendicular to AB,

8 1 13 3

ba

Vidyamandir Classes


8 13 9

ba

8 3 9 b a

3 1 a b … (2)

Mid-point of 3 8,2 2

a bAB

The mid-point of line segment AB will also satisfy line (1).

Hence, from equation (1), we have

3 83 72 2

a b

3 3 24 14 a b

3 13 a b … (3)

On solving equations (2) and (3), we obtained 1 a and 4. b

Thus, the image of the given point with respect to the given line is (–1, –4).

36. If the lines 3 1 y x and 2 3 y x are equally inclined to the line 4, y mx find the value of m.

[Ans. 1 5 27

]

37. Find equation of the line which is equidistant from parallel lines 9 6 7 0 x y and 3 2 6 0. x y

[Ans. 18 12 11 0 x y ]

38. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes

through the point (5, 3). Find the coordinates of A. [Ans. 13, 05

]

39. Prove that the product of the lengths of the perpendiculars drawn from the point 2 2 ,0a b and

2 2 ,0a b to the line cosθ sinθ 1x ya b

is 2.b

40. A person standing at the junction (crossing) of two straight paths represented by the equations

2 3 4 0x y and 3 4 5 0x y wants to reach the path whose equation is 6 7 8 0x y in the least

time. Find equation of the path that he should follow. [Ans.119 102 125x y ]

41. What are the possible slopes of a line which makes equal angle with both axes? [Ans. 1 ]

42. If the line joining two points A(2, 0) and B(3, 1) is rotated about A in anticlockwise direction through an

angle of 15°. Find the equation of the line in newposition. [Ans. 3 2 3 0y x ]

Vidyamandir Classes


43 If the slope of a line passing through the point (3,2)A is 3 ,4

then find points on the line which are 5 units

away from the point A.

Sol. Equation of the line passing through (3, 2) having slope 34

is given by

32 ( 3)4

y x

4 3 1 0y x … (1)

Let ( , )h k be the points on the line such that

2 2( 3) ( 2) 25h k … (2)

Also, we have

4 3 1 0k h … (3)

or 3 14

hk

Putting the value of k in (2) and on simplifying, we get

225 150 175 0h h

or 2 6 7 0h h

or ( 1)( 7) 0 1, 7h h h h

Putting these values of k in (4), we get 1k and 5.k Therefore, the coordinates of the required points

are either ( 1, 1) or (7,5).

( 1, 1) or (7,5).

44. Find the equation to the straight line passing through the point of intersection of the lines 5 6 1 0x y

and 3 2 5 0x y and perpendicular to the line 3 5 11 0.x y [Ans. 5 3 8 0x y ]

45. If one diagonal of a square is along the line 8 15 0x y and one of its vertex is at (1, 2), then find the

equation of sides of the square passing through thisvertex.

Sol. Let ABCD be the given square and the coordinates of the vertex D be (1, 2).

We are required to find the equations of its sides DC and AD.

Given that BD is along the line 8 15 0,x y so its slope is

8 .15

The angles made by BD with sides AD and DC is 45°.

Let the slope of DC be m. Then

Vidyamandir Classes


815tan 45 8115

m

m

or 15 8 15 8m m

or 7 23,m which gives 237

m

Therefore, the equation of the side DC is given by

232 ( 1)7

y x or 23 7 9 0.x y

Similarly, the equation of another side AD is given by

72 ( 1)23

y x or 7 23 53 0.x y

23 7 9 0, 7 23 53 0x y x y

46. Prove that medians drawn from two vertices to equal sides of an isosceles triangle areequal.

Sol. Let OAB be an equilateral triangle with vertices (0, 0), (a, 0) and ,2a a

Let C and D be the mid-points of AB and OB respectively.

C has coordinates 3 ,4 2a a

and D has coordinates ,4 2a a

Length of 2 2 2 23 9 13

4 2 16 4 4a a a a aOC

and Length of 2 2

0 134 2 4a a aAD a

Length of OC = length of AD

Hence the proof.

47. Prove that the line segment joining the mid-points of two sides of a triangle is parallelto the third side and is

half of it.

Vidyamandir Classes


Sol. Let the coordinates of the vertices be (0,0), ( ,0)O A a and ( ,. ).B b c Let C and D be the mid-points of OB

and AB respectively

The coordinates of C are ,2 2b c

The coordinates of D are ,2 2

a b c

Slope of 2 2 0

2 2

c c

CD a b b

Slope of OA, i.e.,x-axis = 0

CD is parallel to OA

Length of 2 2

2 2 2 2 2a b b c c aCD

Length of OA a

12

CD OA Which proves the above result

48. Prove that if the diagonals of a quadrilateral are perpendicular and bisect each other,then the quadrilateral is

a rhombus.

Sol. Let the diagonals be represented along OX and OY as shown in

Figure and suppose that the lengths of their diagonals be 2a and 2b

respectively.

OA = OC = a and OD = OB = b

2 2 2 2AD OA OD a b

2 2 2 2AD OB OA a b

Similarly 2 2BC a b CD

As AB BC CD DA

ABCD is a rhombus.

49. Find the points on the x-axis whose perpendicular distance from the straight line 1x ya b is a.

Sol. Let 1( ,0)x be any point on x-axis.

Equation of the given line is 0.bx ay ab The perpendicular distance of the point 1( ,0)x from the given

line is

Vidyamandir Classes


1

2 2

.0bx a abaa b

2 21

ax b a bb

Thus, the point on x-axis is 2 2 ,0a b a bb

50. Find the equation of line parallel to the y-axis and drawn through thepoint of intersection of 7 5 0x y

and 3 7 0x y

Sol. The equation of any line through the point of intersection of the given lines is of the form

7 5 (3 7) 0x y k x y

i.e., (1 3 ) ( 7) 5 7 0k x k y k … (1)

If this line is parallel to y-axis, then the coefficient of y should be zero, i.e.,

7 0k which gives 7.k

Substituting this value of k in the equation (1), we get

22 44 0,x i.e., 2 0,x which is the required equation

51. Find the equation of the line through the intersection of lines 3 4 7x y and 2 0x y and whose slope

is 5. [Ans. 35 7 18 0x y ]

52. Find the equation of the line through the intersection of lines 2 3 0x y and 4 7 0x y and which is

parallel to 5 4 20 0x y [Ans. 15 12 7 0x y ]

53. Find the equation of the line through the intersection of 5 3 1x y and 2 3 23 0x y and perpendicular

to the line 5 3 1 0.x y [Ans. 63 105 781 0x y ]

54. Find the new coordinates of point (3, –4) if the origin is shifted to(1, 2) by a translation.

Sol. The coordinates of the new origin are 1, 2,h k and the original coordinates are given to be 3, 4.x y

The transformation relation between the old coordinates ( , )x y and the new coordinates ( ', ')x y are given

by

'x x h i.e., 'x x h

and 'y y k i.e., 'y y k

Substituting the values, we have

' 3 1 2x and ' 4 2 6y

Hence, the coordinates of the point (3, –4) in the new system are (2, –6).

55. Find the transformed equation of the straight line 2 3 5 0,x y when the origin is shifted to the point (3, –

1) after translation of axes.

Vidyamandir Classes


Sol. Let coordinates of a point P changes from ( , )x y to ( ', ')x y in new coordinate axes whose origin has the

coordinates 3, 1.h k Therefore, we can write the transformation formulae as ' 3x x and ' 1.y y

Substituting, these values in the given equation of the straight line, we get

2( ' 3) 3( ' 1) 5 0x y

or 2 ' 3 ' 14 0x y

Therefore, the equation of the straight line in new system is 2 3 14 0x y

56. Find what the following equations become when the origin is shifted to the point (1, 1)

(i) 2 23 2 0x xy y y [Ans. 2 23 3 6 1 0x y xy x y ]

(ii) 2 0xy y x y [Ans. 2 0xy y ]

(iii) 1 0xy x y [Ans. 0xy ]

57. Find the point to which the origin should be shifted after shifting of origin so that the equation 2 12 4 0x x will have no first degree term.

Sol. Let origin be shifted to (h, k) and P(x, y) becomes

( , ).P X h Y k Substituting in the given equation we get

2( ) 12( ) 4 0X h X h

2 22 12 12 4 0X hX h X h

Since there is no first degree term : 2 12 0h or h = 6

Hence origin should be shifted to (6, k) for any real value k.

58. Verify that the area of the triangle with vertices (4, 6), (7, 10) and (1, –2) remains invariant under the

translation of axes when origin is shifted to the point (–2, 1)

Sol. Let (4,6), (7,10)P Q and ( 1,2)R be the given points

Area of 1Δ [4(10 2) 7(2 6) 1(6 10)]2

PQR

1 [32 28 4] 42

sq. U.

Now shifting ( , )x y to ( 2, 1)X Y

New coordinates are 2, 1X x Y y

(4,6) (6,5)P

(7,10) (9,9)Q

( 1,2) (1,1)R

Area of 1Δ [6(9 1) 9(1 5) 1(5 9)]2

Vidyamandir Classes


1 [48 36 4] 42

square units

Hence the area remains invarient.

59. If the origin is shifted to the point (1, –2), what do the following equations become?

(i) 2 22 4 4 0x y x y (ii) 2 4 4 8 0y x y

[Ans. (i) 2 22 6x y , (ii) 2 4y x ]

60. Find what the following equations becomes when origin is shifted to the point (2, 3)

(i) 2 22 3 0x xy y y (ii) 23 0xy x y x

[Ans. 2 2

2

(i) 2 10 5 0(ii) 3 6 5 13 0

x y xy x yxy x x y

]

Documents

Book 11_CBSE Notes_Straight Lines