Maths 1B 1st YEAR material notes jr inter PDF

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MINIMUM LEARNING MATERIAL (MLM) FOR (1st YEAR) INTERMEDIATE PUBLIC EXAMINATION 2020-21

MATHS 1B www.tmreis.telangana.gov.in

BLUE PRINT S.No

Name of the Chapter

VSAQ (2 Marks)

SAQ (4 Marks )

LAQ ( 7 Marks )

1

LOCUS

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2

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2

TRANSFORMATION OF AXEX

-

2

--

3

THE STRAIGHT LINE

2

1

2

4

PAIR OF STRAIGHT LINES

--

--

2

5

THREE DIMENSIONAL GEOMETRY

1

1

--

6

DIRECTION COSINES AND DIRECTION RATIOS

--

--

1

7

THE PLANE

1

---

---

8

LIMITS AND CONTINUITY

2

1

---

9

DIFFERENTIATION

2

1

2

10

APPLICATIONS OF DERIVATIVES

2

2

3

LOCUS 2

SHORT ANSWER QUESTIONS: 1. Find the locus of a third vertex of right angled triangle, the ends of whose Hypotenuse are (4,0) and ( 0,4 ).

B(0,4)

Ans: Given points A ( 4,0) B ( 0,4) Let P ( x,y ) By Pythagoras Theorem

AB2 = (4,0)

( 0,4)

AP2 + (4,0)

( x,y)

PB2 ( 0,4)

( x,y)

P(x,y)

A(4,0)

( 0-4)2 + ( 4-0)2 = (x-4)2 +(y - 0)2 +(x - 0)2+ (y - 4)2 16 + 16 = x2 + 16 -8x +y2 + x2 + y2 + 16 -8y 32 = 2x2 + 2y2 -8x -8y +32 2x2 + 2y2- 8x- 8y =0 2 [x2 +y2 -4x -4y] = 0 Equation of locus of P is X2 + y2 - 4x -4y =0

2. The ends of the hypotenuse of a right angled triangle are (0,6) and (6,0) . Find the equation of locus of its third vertex.

3. Find the equation locus of P, if the line joining (2,3) and (-1,5) subtends a right angle at P. 4. A(5,3) B ( 3,2) are two points. Find the locus of P so that the area of PAB is 9. Sol : Given points A( 5,3) B( 3,2) Let P (x,y) Area of triangle =

1

2

x1 ( y2–y3 ) + x2( y3 - y1) + ( y1 - y2 )

P ( x , y ) A ( 5, 3 ) B ( 3, - 2 ) (x1 y1) ( x2 y2 ) (x3 y3)

3

Given that PAB = 9 1 2

x ( 3+2) + 5( -2 – y ) + 3 ( y- 3) =9 5 x- 10 -5y +3y -9

=2X9

5x-2y-19 = 18 5x-2y-19 = ± 18 5x – 2y – 19 =± 18

5x – 2y -19 = 18 5x – 2y – 19 -18 = 0 5x – 2y – 37 = 0

5x - 2y - 19 = -18 5x – 2y – 19 + 18 = 0 5x – 2y – 1 = 0

Equation of locus P ( 5x – 2y – 37 ) ( 5x – 2y – 1 ) = 0 5. A ( 2,3 ) and B ( -3 , 4 ) are two given points. Find the equation of locus of P so that the area of the triangle PAB is 8.5 6. Find the equation of the locus of a point which forms a triangle of area 2 with the points A ( 1,1 ) B ( -2, 3 ).

7. If the distance from P to the point ( 2,3) and (2,-3) are in the ration 2:3 the find the equation of the locus of P. Sol : Given Points A ( 2, 3 ) B ( 2, -3 ) Let P ( x , y ) Given Condition AP: PB = 2 : 3 𝐴𝑃

𝑃𝐵

=

2 3

3 AP = 2 PB

Squaring on both side

9AP2 = (2,3)

( x,y)

4PB2 (2,-3)

( x,y)

9 [ ( x-2)2 + ( y-3 )2 ] = 4 [( x-2)2 + ( y+3 )2] 4

9 [ x2 + 4 -4x +y2 + 9 -6y ] = 4 [x2 + 4 - 4x + y2 + 9 + 6y ] 9 [ x2 +y2 - 4x - 6y + 13 ] = 4 [x2 + y2 -4x + 6y+13 ] 9x2 + 9y2 -36x -54y + 117 = 4x2 + 4y2 - 16x + 24y + 52 9x2 + 9y2 -36x -54y + 117 - 4x2 - 4y2 + 16x - 24y – 52 = 0

5x2 + 5y2 -20x -78y + 65 =0 8. Find the equation of locus of P, if the ratio of the distances from P to A ( 5,-4) B (7,6) is 2 : 3 9. A (1,2 ) B ( 2,3) and C ( -2,3) are three points. If a point P moves such that PA2+ PB2 = 2PC2,then show that the equation to the locus of P is 7x – 7y + 4 = 0 Sol : Given points A (1,2 ), B ( 2,3) , C ( -2,3) Let P ( x, y )

PA2 + (1 , 2)

( x , y)

PB2 = (2,- 3)

2PC2 ( x , y)

( -2 , 3 )

( x , y)

( x-1)2 + ( y-2 )2 + ( x-2)2 + ( y+3 )2 = 2 [ ( x+2)2 + ( y-3 )2 ] x2 +1 – 2x+y2+ 4 – 4y + X2 + 4 – 4x + y2 + 9 + 6y = 2[x2+4+4x+ y2+9-6y] 2x2+2y2-6x+2y+18 = 2[x2+y2 +4x-6y+ 13]

2x2+2y2-6x+2y+18 = 2x2+2y2+8x-12y+26 -6x+ 2y+18 -8x + 12y – 26 = 0 -14x + 14 y -8 = 0 -2 [ 7x – 7y + 4 ] = 0 7x – 7y + 4 = 0

5

TRANSFORMATION OF AXES SHORT ANSWER QUESTIONS: 1. When the origin is shifted to the point (2,3) the transformed equation of a curve is x 2+3xy2y2+17x-7y-11=0. Find the original equation of the curve. Ans:Given Transformed equation X2+ 3xy-2y2+17x-7y-11=0 Given (h,k) = ( 2, 3) X = x+ h y = y+ k X=x+2 y=y+3

X = x-2

Y = y-3

( x-2 )

2+

3( x-2 ) (y-3) -2(y-3)2+ 17(x-2) -7 (y-3) – 11 =0

X2+4-4x+(3x-6) (y-3) – 2(y2+9-6y)+17x-34-7y+21-11=0 X2+4-4x+3xy-9x-6y+18-2y2-18+12y+17x-34-7y+10=0

X2+3xy-2y2+4x-y-20 =0 𝝅

2. When the axes are rotated through an angle

𝟒

Find the transformed equation of 3x2 + 10xy + 3y2 =9 ANS : Given Original equation 3x2 + 10xy + 3y2 =9 (1) 𝝅 Given  = 𝟒 x= Xcos  - Y sin  𝝅 𝝅 x = Xcos 𝟒 –Y sin 𝟒

y= xcos  - y sin 𝝅 𝝅 y = Xcos 𝟒 – Y sin 𝟒

y = X 1/√𝟐 - Y 1/√𝟐

x= X 1/√𝟐 - Y 1/√𝟐 x=

X-Y

y =

√𝟐

Sub in Equation ------(1)

3 X- Y 2 + 10

X-Y

X Y x cos  -Sin y Sin cos 

X-Y

√𝟐

X+Y

+3

X+Y

3(X +Y -2XY)+10(X -Y ) + 3 (X +Y +2XY) 2

= 9

√𝟐 √𝟐 √𝟐 3(x-y)2 + 10 (x-y) (x+y) + 3(x+y)2

2 2

2

2 2

2

√𝟐

2 2

2

3X2+3Y2-6XY+10X2-10Y2+3X2+3Y2+6XY = 2x9 16X2-4Y2 = 2x9 2[8X2-2Y2]=2x9 Transformed equation is 8x2-2y2 =9

6

2

= 9

3. When the axes are rotated through an angle 45° the transformed equation of a curve is 17x2-16xy+17y2=225. Find the original equation of the curve.

ANS : Given transformed equation 17x2-16xy+17y2=225

X (1)

x

Given =45°

y

X= xcos  + y sin  X =xsin45°+ycos45o

cos  sin 

Y -sin 

cos 

Y= -x sin  + y cos  Y= -x sin 45o + y cos 45o

X= x+y √𝟐

Sub In Equation (1)

Y = -x+y √𝟐

17 x+y 2- 16 x+y –x+y + 17 –x+y 2 = 225 √𝟐 √𝟐 √𝟐 √𝟐 17 (x+y)2 - 16 (y+x) ( y-x) + 17 ( y-x)2 = 225 2

2

2

17(x+y)2 -16(y+x)( y-x) + 17 (y-x)2 =225 2 17 ( x2+ y2 + 2xy) - 16 ( y2-x2)+17(y2+ x2 - 2xy ) = 2x225 17x2+ 17y2+34xy-16y2+16x2+17y2+17x2-34xy = 2x225 50x2+18y2 = 2x225 2 [25x2+9y2] = 2x 225 25x2 + 9y2 = 225 2

25x 225

(or) + 9y2 225

1

x2 + y2 = 1 9 25

4. When the axes are rotated through an angle of x2+2 √𝟑 xy – y2 = 2a2 .

7

𝝅 𝟔

find the transformed equation

ANS : Given original equation x2+2 √𝟑 xy – y2 = 2a2

1

𝝅

Given  = 𝟔 X

Y

x cos 

-Sin

y Sin

cos 

x = Xcos  - YSin 𝝅 𝝅 x = Xcos 𝟔 -Y sin 𝟔

y = Xsin - X cos 𝝅 𝝅 y = Xsin𝟔 - Y cos 𝟔

x = x √𝟑 - y 1 2

y=x1 +y 1

2

2

x = √𝟑 𝑿 − 𝒀 2

2

y= X + √𝟑𝒀 2

= √𝟑 𝑿 − 𝒀2 2 + 2 √𝟑√𝟑 𝑿 − 𝒀 2

X + √𝟑𝒀 2

- x + √𝟑𝒚 2

(√𝟑 𝑿 − 𝒀)2 + 2 √𝟑 (√𝟑𝑿 − 𝒀) ( 𝑿 + √𝟑𝒀) 4 4

- (X+ √𝟑𝒀)2 4

2

= 2a2 = 2a2

3X2+Y2-2√𝟑𝑿𝒀 + ( 6X-2𝒀)(𝑿 + √𝟑𝒀)-( X2+3Y2+2√𝟑𝑿𝒀) =2a2 √𝟑𝑿𝒀 + = 8a2

3X2+Y2-2 8X2-8Y2

4

√𝟑𝑿𝒀-6Y2-X2-3Y2-2√𝟑𝑿𝒀 = 4x2a2

6X2+6√𝟑𝒀)-2

8( X2-Y2) = 8a2

x2-y2=a2

5. When the axes are rotated through an angle  find the transformed equation of 𝒙 𝒄𝒐𝒔 + 𝐲 𝐬𝐢𝐧  = 𝐏 ANS: Given original equation 8

𝒙 𝒄𝒐𝒔  + 𝐲 𝐬𝐢𝐧  = 𝐏

1

Given  =  𝒙 = 𝑿cos  - Y Sin

𝒙=Xcos  - Y sin 

y = 𝑿sin +Y cos

y= 𝑿 sin  + Y cos 

(𝑋cos - Ysin) cos + (𝑋 sin + Ycos  ) sin = P 𝑿 cos2 - Ysin  cos + 𝑿 sin2 + Ysin cos  =P

𝑿cos2 + Xsin2 =P

𝑿 ( sin2 +cos2 )=P 𝑿 (1) = P 𝑿 =P

Transformed equation is

x=P

THE STRAIGHT LINE VERY SHORT ANSWER QUESTIONS(VSAQ) 1. Find the value of P, if the lines 𝒙 +P=0 𝒚+2=0, 3𝒙+2𝒚+5=0 are concurrent. ANS : Given straight lines 𝑥 +P=0 1 𝑦+2=0, 2 3𝑥+2𝑦+5=0 3 From (1) 𝑥=-P From (2) 𝑦 = -2 (𝑥 , 𝑦) = (-P,-2) Given lines area concurrent Substitute (-P,-2) in equation (3) 3(-P)+2(-2)+5=0 -3P-4+5=0 -3P=-1 p=1/3

2.Find the value of 𝒙 if the slope of the line passing through (2,5)and (𝒙, 𝟑) is 2 ANS : Given points A (2,5) B (𝑥, 3) Given that, slope of AB = 2 𝑦2- 𝑦1 = 2 9

𝑥2- 𝑥1 3-5 =2 𝑥-2 -2=2(𝑥-2) 𝑥 =-1+2 𝑥-2 =-1 x=1 3) Find the value of y if the line joining the points (3,y)and (2,7) is parallel to the line joining the points (-1,4) and (0,6) ANS : Let the given points A (3,y) B( 2,7) C ( -1,4) D( 0,6) Given that ,AB CD Slope of AB = Slope of CD 7-y = 6-4 2-3 0+1 7-y = 2 -1 7-y = -2 y=7+2 y=9 4) Find the equation of the straight line passing through the point (2,3) and making non zero intercepts on the axes of coordinate whose sum is zero. Sol : Let the equation of straight line 𝑥

𝑎

+ =1

𝑥

𝑦

𝑦

𝑏

( 1 ) , Given that a+b=0 b = -a Substitute in (1)

+ =1 𝑎 −𝑎 𝑥−𝑦 =a ( 2 ) it passes through (2,3) 2-3=a a = -1 Substitute in (2) x-y = -1 𝑥−𝑦+1 =0

10

5.Find the equation of straight line passing through the point (-2,4) and making non zero intercepts whose sum is zero. 6.Find the equation of straight line passing through (-4,5) and cutting off equal and non zero intercepts on the co ordinate axes 7.Transform the equation 4 𝒙 − 𝟑𝒚 + 𝟏𝟐 = 𝟎 into ( i ) slope intercept form ( ii ) Intercept form ANS : (i) Slope intercept form (𝑦 = 𝑚𝑥 + 𝑐) 4 𝑥 − 3𝑦 + 12 = 0 3 𝑦 = 4𝑥 + 12 12 4 𝑦= 3𝑥+ 3

4

𝑦 = 𝑥 +4 3

𝑥

𝑦

(ii) Intercept form [ + = 1] 𝑏 𝑎 4 𝑥 − 3𝑦 + 12 = 0 4𝑥 − 3𝑦 = −12 Divided on both sides by -12 4𝑥 −12 3𝑦 = -

12 −12 𝑥 𝑦 + =1 −3 4

−12

8.Transform the equation 3 𝒙 + 𝒚 = 𝟒 in slope intercept form (ii) Intercept form 9.Transform the equation 3 𝒙 + 𝟒𝒚 + 𝟏𝟐 = 𝟎 into normal form. Sol : Given straight line 3 𝑥 + 4𝑦 + 12 = 0 , a=3, b=4 -3 𝑥 − 4𝑦 = 12 (1) √𝑎2+b2= √32+42 = √9 + 16 = √25 =5 Divide eq (1 ) on bothsides by 5 −3𝑥 4 𝑦 12 = 5 −3

5

5

4

12

𝑥( )+ 𝑦 (− ) = 5 5 5 (𝑥𝑐𝑜𝑠  + sin  = 𝑃) 11

3

Where, cos 



4

12 = − , sin = − , 𝑃= 5 5 5 𝟏 = 𝟎into normal form. 10. Transform the equation 𝒙 + 𝒚 + ANS : Given Stright line 𝑥 + 𝑦 + 1 = 0 a=1, b= 1 −𝑥 − 𝑦 = 1 2 2 2 2 √𝑎 +b =√1 +1 =√2 Divided eq (1) on both side by -12 −1

−1

𝑥 ( √2 ) + 𝑦(√2 )=

1

√2 −1

−1

1

Where cos  = sin  = P= √2 √2 √2 cos  = cos(180 + 45) sin = sin( 180 + 45) sin  = sin 225 cos  = cos 225 equation of straight line 1 𝑥𝑐𝑜𝑠 225 + 𝑦𝑠𝑖𝑛 225 = √2 11 . Find the length of the perpendicular from the point (3,4) to the line 3𝒙 − 𝟒𝒚 + 𝟏𝟎 = 𝟎

ANS: Given ight line 3𝑥 − 4𝑦 + 10 = 0 Given point (𝑥 1, 𝑦1) = ( 3,4) Length of the perpendicular d= a 𝑥1+ b 𝑦1+c1 √𝑎2+ b2 d= 3(3)-4(4)+10 √32+ (-4)2 = 9-16+10 √9 + 16 = 3 √25 3 d= 5

12

12. Find the length of the perpendicular from the point (-2,-3) to the line

5 𝒙 − 𝟐𝒚 +4=0. 13. find the distance between the parallel lines 5 𝒙 − 𝟑𝒚 −4=0 , 10 𝒙 − 𝟔𝒚 −9=0 Sol : Given straight lines 5 𝑥 − 3𝑦 −4=0 (1) 10 𝑥 − 6𝑦 −9=0 (2) 2x(1) 10 𝑥 -6 𝑦-8=0 [ a 𝑥 + 𝑏𝑦 + 𝑐1=0] 10 𝑥-6 𝑦-9=0 [ a 𝑥 + 𝑏𝑦 + 𝑐2=0] a = 10, b =-6, c1 = -8, c2 = -9 Distance between the parallel lines d= c1−𝑐2 √𝑎2+ b2 d=

-8+9 √102+ (-6)2 d= 1 √100+36 d= 1 √100+36 d= 1 √136 14. Find the distance between the parallel straight lines 3 x +4 𝒚 – 3 = 0 and 6𝒙+8 𝒚-1=0 15. Find the value of P if the straight lines 3 𝒙+7 𝒚-1=0 And 7𝒙-p𝒚+3=0 are mutually perpendicular ANS : Given straight lines 3𝑥+7 𝑦-1=0 [ a1x+b1y+c1=0] 7𝑥-p 𝑦+3=0 [ a2x+b2y+c2=0] a1= 3 , b1 =7 a2=7 , b2 =-p given that, the lines are mutually perpendicular i..e a1a2 +b1b2 =0 3(7) + 7(-P) =0 7[3-P]=0 3-P=0 13

P=3 16. Find the value of K, if the straight lines 6x-10y+3=0 Kx-5y+8=0 are parallel. Sol: Given straight lines 6x-10y+3=0 [ a1x+b1y+c1=0] K𝑥-5 𝑦+8=0 [ a2x+b2y+c2=0] a1= 6 , b1 =-10 a2=K , b2 =-5 given that the lines are parallel

i..e a1

=

a2

6

=

K

6

=

b1 b2

-10 -5

2

k K=3 17. Find the equation of straight line parallel to the line 2x+3y+7=0 and passing through the point (5,4). ANS: Given straight line 2x+3y+7=0 (ax+by+c=0) Given point (x1,y1) = (5,4) Required equation is a(𝑥 − 𝑥 1) + b (𝑦 − 𝑦1)=0 2(x-5) + 3(y-4) =0 2 𝑥 − 10 + 6𝑦 − 12=0 2 𝑥 + 6𝑦 − 22 =0

18. Find the equation of the straight line perpendicular to the line 5𝒙-3 𝒚+1=0 and

passing through (4,-3)

ANS : Given straight line 5𝑥-3 𝑦+1=0 (a𝑥+b 𝑦+C=0) Given point (𝒙1, 𝒚1) = (4,-3) Required equation b(𝑥 − 𝑥 1) -a(𝑦 − 𝑦 1)=0 -3(𝑥 − 4) − 5 (𝑦 + 3)=0 -3𝑥 +12-5 𝑦-15 =0 -3𝑥 -5 𝑦-3 =0, 3𝑥+5 𝑦+3 =0 14

19. Find the sum of the squares of the intercepts of the line 4x-3y=12 on the axes of co ordinates ANS : Given straight line 4x-3y=12

Divide on both sides by 12 4𝑥 3𝑦 12 = 12 𝑥12 𝑦 −12

-

=1

3 4 𝑦 𝑥 + =1 3 −4 𝑦 𝑥 + =1 𝑎 𝑏 x-intercept (a) =3 y-intercept (b) =-4 a2+b2 = 32+ (-4)2 =9+16 = 25 20. Find the area of the triangle formed by the straight line 𝒙-4 𝒚+12=0 with the co ordinate axes . ANS : Given straight line 𝑥-4 𝑦+12=0 (a𝒙+b 𝒚+C=0) a = 1 , b = -4 , c = 12 𝟏 Area of the triangle = 𝟐 c2/ab 𝟏

=𝟐 =

𝟏

𝟐

122/1(-4) 144/-4

= 18 sq units

STRAIGHT LINE : ( SAQ’s)

1. Transform the equation 3𝒙+4 𝒚+12=0 into

(i) Slope intercept form ( ii) Intercept form and , (iii) Normal form

ANS : (i) Slop intercept form : 3𝑥+4 𝑦+12=0 4 𝑦=-3x-12 −𝑥 𝑦=( )𝑥 − 3 4

[ 𝑦=mx+c ]

−3

𝑦=( 4 )𝑥 + (−3) 15

𝑥

(ii) Intercept form : ( + 3𝑥+4 𝑦+12=0 𝑎 3𝑥+4 𝑦=-12 3𝑥

−12

−12 −12

4𝑦

+ −12 =

a=-4

𝑦 =1) 𝑏

𝑥 −4

b=-3

+

𝑦

−3

=1

iii) Normal form; ( 𝑥𝑐𝑜𝑠  + 𝑦 sin  = 𝑃) 3𝑥 + 4𝑦 + 12 = 0 a=3 b=4 12=-3 𝑥 − 4𝑦 −3𝑥 − 4𝑦 = 12 (1) √𝑎2 + 𝑏 2 = √32 + 42 = √9 + 16 = √25 =5 Divide eq. (1) on bothsides by 5 −3𝑥 −4𝑦 12 - 5 = 5 5 −3

−4

12

) +y( 5 ) = 5 5 ( 𝑥𝑐𝑜𝑠  + 𝑦 sin  = 𝑝) x(

𝑐𝑜𝑠 =

−3 5

, sin  =

−4 5

,𝑝 =

12 5

2. Find the value of K , if the lines 𝟐𝒙 − 𝟑𝒚 + 𝒌 = 𝟎, 𝟑𝒙 − 𝟒𝒚 − 𝟏𝟑 = 𝟎 and 𝟖𝒙 − 𝟏𝟏𝒚 − 𝟑𝟑 = 𝟎 ate concurrent

ANS :Given straight lines

2𝑥 − 3𝑦 + 𝑘 = 0 3𝑥 − 4𝑦 − 13 = 0 8𝑥 − 11𝑦 − 33 = 0 are concurrent then

= 0 2

-3

k

3

-4

-13

8

-11

=0

-33 16

2[132 -143] + 3[-99+ 104] + k [ -33+ 32]=0 2(-11) + 3(5) + k (-1) =0 -22+15-k=0 K=-7

3. Find th value of P if the lines 3 𝒙+4 𝒚 =5 , 2 𝒙+3 𝒚 =4, P 𝒙+4 𝒚 =6, are concurrent.

4. If the straight lines a𝒙+b 𝒚+C=0, b𝒙+c 𝒚+a=0, ,c𝒙+a 𝒚+b=0 are concurrent then prove that a3+b3+c3= 3abc.

Sol :

Given straight lines ax+by+c=0 b𝒙+cy+a=0 cx+ay+b=0

are concurrent

=0 a

b

c

b

c

a

c

a

b

=0

a[bc-a2] – b[b2-ac]+c[ab-c2]=0

abc-a3-b3+abc+abc-c3=0 3abc = a3+b3+c3 4. A Straight line with slope 1 passes through Q (-3,5) and meets the straight line 𝒙+ 𝒚-6=0, at P find the distance PQ

Ans : Given straight line 𝑥- 𝑦+6=0 Given that m=1 , Q ( -3,5) Equation of the straight line y-y1 = m (𝑥 -x1) 𝑦-5 =1(𝑥 + 3) 𝑦-5 =𝑥 + 3 x+3-y+5=0 x-y+8=0 It meets eq (1 ) at P

(1)

17

Solve eq (1) and (2) 𝑥+𝑦−6 =0 𝑥−𝑦+8 =0 2x+2=0 2x=-2 𝑥 = −1

sub in (1)

-1+y-6=0 y-7=0 y=7 P (𝑥, 𝑦) = (-1,7) P ( -1,7) Q (-3,5)

PQ =√(x2-x1)2+(y2-y1)2

= √(-3-+1)2+(5-7)2 = √(-2)2+(-2)2

= √4+4

= √8 PQ = 2√2

5. A Straight line through Q (√3,2) make an angle

𝝅

𝟔

with the positive

direction of the x-axis. If the straight line intercects the line √𝟑𝒙 4 𝒚 + 𝟖 = 𝟎, at P. find the distance PQ 6. Find the value of k if the angle between the straight lines 4x-y+7=0, and kx-5y-9=0 is 45 𝑆𝑜𝑙: 4𝑥 − 𝑦+7 =0 (1) 𝑘𝑥 − 5𝑦-9 =0 (2) −4 Slop of eq (1) m1 = , m1=4 slope of eq (2)

−𝑘

m2=

−5

−1

, m2 =

𝑘

5

angle between 1 and 2 is 45° Tan 

=

m1-m2 1+m1m2

18

𝑘

Tan 45° =

4-5 𝑘 1+4. 5

20-k 1 =

5

5+4k 5

1

=

20-k 5+4k 20 – k = ± 1 5+ 4k 20 –k = 5+ 4k 20-5=4k+k 15=5k K=3

7. Find the equation of straight lines passing through (1,3) and (i) parallel to( ii) perpendicular to the line passing through the points (3,-5) (-6,1) ANS : Given two points (3,-5) (-6,1) Equation of straight line (y-y1) (x2-x1) = (x-x1) (y2-y1) (y+5) (-6-3) = (x-3) (1+5) (y+5) (-9) = (x-3) (6) (y+5) (-3) = (x-3) (2) -3y-15 = 2x-6 2x+3y-6+15=0 2x+3y-9=0 (i) Equation of straight line parallel to 2x+3y-9=0 and passing through (1,3) 2(x-1)+3(y-3)=0 [a(𝑥 − 𝒙1)+b(𝑦 − 𝒚1)=0] 2x-2+3y-9=0 2 𝑥 + 3𝑦 − 11 = 0 19

(ii) Equation of straight line perpendicular to 2x+3y-9=0 and passing through (1,3) 3(x-1)-2(y-3)=0 [b(𝑥 − 𝒙 1) - a(𝑦 − 𝒚1) = 0] 3 𝑥 − 3 − 2𝑦 + 6 = 0 3 𝑥 − 2𝑦 + 3 = 0

9) Find the equation of straight line perpendicular to the line 𝟑𝒙 + 𝟒𝒚 + 𝟔 = 𝟎 and making an intercept -4 on the x-axis? 10 ) Find the image of the point (1,2) w.r.t the straight line 3x+4y-1=0. Sol: Given P(1,2) Let Q (h,k) be the image of P w.r.t 3x+4y-1=0 [ ax+by+c=0] h-x1 = k-y1 = -2(ax1+by1+c) a b a2+b2 h-1 = k-2 = -2(3(1)+4(2)-1) 3 4 32+42 h-1 = k-2 = -2(3+8-1) 3 4 9+16 h-1 = k-2 = -2(10) 3 4 25 h-1 = k-2 = -4 3 4 5 h-1 = -4 k-2 = -4 3 5 4 5 h-1 = -12 k-2 = -16 5 5 h = -12 +1 5 h = -12 +5 5 h = -7 k 5 Q (h,k) = - 7 5

k

=

k = =

-16 +2 5 -16 +10 5

-6 5

-6 5

20

11) x-3y-5=0 is the perpendicular bisector of the line segment joining the points A,B if A= (-1,-3) find the co ordinate of B. Sol : Given straight line, x-3y-5=0 Given A(-1,-3) x-3y-5=0 Let B ( h,k) B Point B be the image of A w.r.t straight line h-x1 = k-y1 = -2(ax1+by1+c) a b a2+b2 h+1 = k+3 = -2(-1-3(-3)-5) -3 12+(-3)2 h+1 = k+3 = -2(3) ...