STRAIGHT LINES EXERCISE - 3 Q. D C (3,4) E A(, ) Mid point of A, C is B 3 E, Point D rotation of point C(3, 4) by angle 90 o about E. 3 o 3 3 i4 cis90 i 5i 3 i i 5 i 5 D, point E mid point of B & D. So calculate B Q. A (3, 4) B (5, ) P (a,b) A (3,4) B (5, )
PA PB a 3 b 4 a 5 b 6a + 9 8b + 6 = 0a + 5 + 4b + 4 4a b = 4 a 3b = () a b 3 4 0 5 a (6) b ( ) + ( 6) = 0 6a + b = 46...() Solve () & () a = 7, b = So a + b = 9 Q.3 P(h, k) A (0,3) B (0, 3) PA + PB = 8 PA = 8 PB PA 64 PB 6PB h k 3 64 h k 3 6 h k 3 6k 6k 64 6 h k 3 Again take square & simplicity for (h, k) So locus of (h, k) is x y 6 7
Q.4 b 3, 4 a x y a b 3 4 a b given a + b = 4..().() After solving () & () we get a = b = 7 or a = 6, b = 8 So equation of line is x y 7 7 or x y 6 8 Q.5 Q point o o 3 r cos45 5 rsin 45 Q point lies on x + y = 6 o o So, 3 r cos45 5 rsin 45 6 r 4 So length PQ = r = 4 Q.6 b O 0,0 B(0, b) D p a A a,0 Equation of line AB is x y a b
OD p 0 0 a b Proved. p a b Q.7 C L L 3 A L B L x y 6 0 L x y 4 0 L3 x y 5 0 A(,8), B,, C7, So area of triangle 8 7 7 Q.8 3 m 4 3x 4y 0 Let slope of line is m, which is making m tan m tan So, m or m m tan m tan o 45 angle with given line. 3 m 4 7 3 4 or 3 4 m 3 7 4
x 3 7 So equation of line is (y ) = 7 (x + 3) or y = 7x y + 3 = 0 () or 7y + x = 0 () Q.9 Equation of line perpendicular to x y is x y k a b a b () k = So line is x y a b Q.0 A (,) H E B (3, ) D C (,6) 6 mac 5 mbc 4 Equation of BE is 5 y x 3 5y + x = 3..() Equation of AD is x 4y = 3 () 4 y x On solving () & () 37 6 x, y 9 9
37 6 So orthocenter H, 9 9 Q. A(0, ) B (3,5) O(h, k) C (5,8) For Circumcentre. OA = OB h k h 3 k 5 4 k + 4 = 6h + 9 + 5 0k 6h + 6k = 30 h + k = 5..() OA = OC h k h 5 k 8 4k + 4 = 0h + 5 + 64 6k 0 h + k = 85 By solving () & () 5 35 h,k So, 5 35 O, Q.
x y 3 (3() 4(3) 7) 3 4 3 4 5 x x 3 4 y 3 5 5 47 79 y 5 5 47 79 ( x, y), 5 5 Q.3 x xy 3y 0 a, b 3, h 3y xy x 0 y y 3 0 x x y 4 5 x 6 6, 3 y x or y x are the lines 3
tan 6 h ab 4 a b 3 5 tan 5 acute angle between lines Q.4 Equation of bisectors of 3x + 4y = 0 and 5x y + 6 = 0 3x 4y 5x y 6 3 4 5 (3x 4y )(3) 5 (5x y 6) 0 (39x 5y 3) (5x 60y 30) 0 So case I : (39x 5y 3) (5x 60y 30) 0 Positive sign 64x 8y + 7 = 0.(A) Case II : Negative sign 4x + y 43 = 0..(B) Lines are 64x 8y 7 0 and 4x y 43 0 Q.5 6x xy 0y x 3y 5 0
3 a 6, h, b 0, g, f, c 5 tan h ab a b 60 4 6 0 9 4 Acute angle between lines is 9 tan 4 Point of intersection is given by the formula. bg fh af gh, h ab h ab 3 3 ( 0) 6, 60 60 4 4 0 34 3, 36 36 = (, ) To find the lines 6x xy 0y x 3y 5 0 6 x x(y ) 0y 3y 5 0 x (y ) (y ) 4(6)( 0y 3y 5) x (y ) (y y 40y 4 3y 360)
x x (y ) 36y 7y 36 (y ) 9( y ) x ( y ) 9( y ) ( y ) 9( y ) or x 30y 8 8y 0 or 5y 3 y 5 x or x 3 x 5y 3 or 3x y 5 Q.6 Meeting of x + y = 0 and x y = 0 is x y x y 3 5 Equation of line with intercepts A & B is x y
Mid point of AB is (h, k) (h, k) 0 0,, So h and k 3, 5 5 x y lies on x y So 5 5 h And k 3 0h 0k Locus is 3 0x 0y i.e. 3 x y 0 Q.7 C lies on 9x + 7y + 4 = 0 9 7 4 0 Centroid is (h, k) (h, k) 4 5, 3 3
6 6, 3 3 6 6 3h k 3 3h 6 3k 6 9 7 4 0 9(3h 6) 7(3k 6) 4 0 7h 54 k 4 4 0 7h k 8 0 So locus of centroid is 7x y 8 0 Q.8 Slope PA Slope PB = k 3 k 5 h, h h (k 3) (k 5) + (h ) (h + ) = 0 locus of P is (x ) (x + ) + (y 3) (y 5) = 0 x x y y 8 5 0 x y x y x 8 3 0, Q.9
3x + y = x y 4 M is the mid point of AB so 4 0 0 M, M = (, 6) So Equation of line is y = 3x slope 3 ( m ) slope of L 3 m m 3 ( 3) 6 tan mm 3( 3) 8 So 3 3 tan tan 4 4 Q.0 Equation of BD is x + 7y = 9 D is intersection of 7x + y = 0 and B is intersection of 4x + 5y = 0 and
x + 7y = 9 x + 7y = 9 x y 0 7 0 7 9 7 9 7 x y 8 63 7 x y 0 5 4 0 4 5 9 7 9 7 x y ( 45) 36 7 7 x, y 3 3 x 5 4, y 3 3 D 7, 3 3 B 5 4, 3 3 Mid point of BD is 5 7 4 3 3, 3 3 P, Equation of OP and OC are the same Equation of OP is y = x Q. L : x + 3y = 6 L : 3x + y = 6 Intersection is x y 6 3 6 3 6 3 6 3 x y 6 6 5 x y 6 5
Equation of AB is x y AB passes through 6 6, 5 5. So it satisfies 6 6.(A) 5 5 Now mid point of AB is (h, k) So (h, k) 0 0,, h k h k Putting in (A) we get 6 6 0h 0k Locus is 3 3 3( x y) 5xy 5x 5y Q.
A intersection of 4x y = 4 and x y = 6 x y 4 4 4 4 6 6 So x y x, y 8 6 A (, 8) B intersection of x y = 6 & x +y = 6 is x = 4 & y = B (4, ) C intersection of 4x y = 4 & x + y = 6 is x = and y = 4 AH perpendicular to BC So
k 8 4 h 4 k 8 h k + 8 = h + (i) BH perpendicular to AC So k 48 h 4 k 4 h 4 4k 8 = 4 h (ii) k + 8 = h + 5k = 5 and k = h = k + 7 h = 8 Orthocenter is H (8, ) Q.3 AC perpendicular to BD Slope of BD = So slope of AC =
i )Equation of AC is (y ) = (x ) y = x y = x ii ) Perpendicular distance of A to BD is () 0 5 5 5 Units. In the figure AO (side) ( the diagonal) So (side) 0 units. Area 0 sq.units Q.4 Equation of AB is : y 3 = (x ) y x =..(i) Equation of BC is : y 5 (x 3) y + x = 3.(ii) Intersection B y x 3 3 y x 7 5
7 x, y 5 5 P is mid point of AC 3 3 5, (, 4) Equation of BD = Equation of BP 7 4 ( 4) 5 y ( x ) 5 7 ( y 4) ( x ) ( y 4) 7x 4 7x y0 0 Q.5 If A ( x, y ) Then x 4y 0 0..(i) E is the mid point of AC 3 x y E, E satisfies 6x + 0y = 59 3 x y 6 0 59
3(3 x ) 5( y ) 59 3x 5y 55..(ii) x 4y 0 (i) Solving we get x y 55 5 3 55 3 5 0 4 0 4 x y 70 85 7 x = 0, y = 5 A (0, 5) Image of (3, ) in angle bisector x 4y + 0 = 0 lies on AB So (, ) C 3 (3 4 0) 4 4 3 8, 7 C (, 7) Equation of AB = Equation AC AB:x 9y 65 Equation of AC :
5 y ( x 3) 0 3 7y 7 6x8 AC:6x 7y 5 Intersection of x + 9y = 65 and 6x + 0y = 59 is B B x y 65 9 65 9 59 0 6 59 6 0 x y (650 53) (8 390) 34 x y 9 7 34 x 7, y 8 B 7,8 C (3, ) Equation of BC is 8 ( y) ( x 3) 7 3 9 ( y) ( x 3) 3 8 ( y ) ( x 3) 3 3y + 8x = 4 Q.6
Centroid is G is given by intersection of x + y = 5 & x = 4 G x = 4 and y = 5 x y = G (4, ) x y 3 3 E, E lies on x + y = 5. So, x3 y3 5 x (i) 3 y3 7 B lies on x + y = 5. So, x y.(ii) 5 x x y y, (4, ) 3 3 3 3 G..(iii) C lies on x = 4 x3 4 x 3 4 Then y3 7 x3 from (ii) y3 3 C : (4, 4).(iv) From (iii) x x, y y 3 3 3
x 7 and y B (7, ).Equation of BC is B (7, ) C(4, 3) Equation of BC is 3 ( y 3) ( x 4) 4 7 ( y 3)( 3) 5( x 4) 5x3y 9 Q.7 Perpendicular of AB to line = perpendicular distance of B to line Hence by symmetry line must pass through mid point of AB i.e. 5 4, M (3, ) Line must be perpendicular to CM so that it is furthest. Slope of CM is 8 7 7 3 4 Equation of line is 4 ( y) ( x 3) 7
(7y 7) = 4 (x 3) 7y 7 = 4x + 7y4x 9 Q.8 A is the ex-centre of DFE A A 0(8) 0(7) 8(5) 0(6) 5(7) 9(5), 05 7 0 5 7 00 0, (50, 5) B is also ex-center of DEF B 40 60 0 75 30 35, 35 7 35 7 B (5,0) C 040 60 3575 30,
50 360 C, (5,30) Circumcenter be S (x, y) ( x 5) (4 0) ( x 5) ( y 30) ( x 50) ( y 5) (x 0)(0) (y 40)(0) 0 x 0 + y 40 = 0 x + y = 50.(i) and (x 65)(35) + (y 5) ( 35) = 0 x 65 y + 5 = 0 x y = 40 x y = 0 (ii) y = 0 & x = 30 S (30, 0) co ordinates of circumcenter. Q.9 slope of BC so slope of AB is 3 3 Equation of AB is y = 3 (x 8)
y + 3x = 6 : AC y 3x 6 (equation of AC)..(i) x 3y 0 (equation of BC)..(ii) 3 3 x x3 y y3 7, 5 (iii) (mid point) x x 4 & y y 30 3 3 3 x 4 x & y 30 y putting in (ii) we get 3 3 (4 x ) 3(30 y ) 0 4 x 90 3y 0 4 x 3y 4 0 3y x 78..(A) y 3x 6 from(i) y x 78 3 78 3 6 3 6 3 y 6, x 0 A (0, 6) C (4, 4) Equation of AC is 6 4 y 6 ( x 0) 0 4 ( x ) 7 7 yx 6 7 7yx 8 : equation of AC
Q.30 Using parametric locus equation of line is x rcos y 7 rsin For point A : Let r r and A ( rcos, 7 rsin ) For point B : Let r r ( r cos, 7 r sin ) B 4( r cos ) 3( 7 r sin ) r (4cos 3sin ) 3 8 3 4( r cos ) ( 7 r sin ) 3 r (4cos 3sin ) 4 ( r r)(4cos 3sin ) 9 r r 9 squaring we get 9(6cos 9sin 4sin cos ) 8 6 cos 9sin 4sin cos 9
7 cos 4sin cos 0 7 cos 0 or tan 4 Case I : cos 0 Case II : 90 or 7 tan 4 Equation is x 7 ( y 7) ( x ) 4 i.e. 4y7x 8 4y 7x8 0 Q.3 L& L are inclined at 60 w.r.t. x + y = m ( ) tan 60 m( ) m m 3 3 m m m 3( m ) m( 3) ( 3) ( 3) m ( 3)
( 3) m ( 3) m 3 Line trough the centroid A (, ) then 0 0 (0 0 ) ( ) A (, ) y ( 3)( x ) Q.3 M is intersection of y + 8x 7 = 0 & y 8x + 7 = 0 i.e. y = 5 and x 3 M 3,5 Slope of BM = 8 = tan 8 sin, cos 65 65 3 r 8r B,5 65 65
3 r 8r D,5 65 65 slope of y -8 x + 7 = 0 is 8 tan 8 8 sin, cos 65 65 A 3 r,5 8 r 65 65 C 3 r 8r,5 65 65 r = length of diagonal Area 8 AB BC AB BC r 65 6r 65 3r 8 65 65 65 r r (taking positive sign) 8 A (, 9) B (, 9) C (, ) D (, )
Equation of Altitude : 7x 0y + = 0 Intersection of 3x y + 5 = 0 & 7x 0y + = 0 gives us B B x y 5 3 5 3 0 7 7 0 x y (50 ) (3) 6 x 3, y B ( 3, ) Reflection of A in 3x y + 5 = 0 lies of BC Reflection A is given by x y (6 5) 3 9 4 x y 3 A ( 4,3) A lies on BC Equation of BC is 3 y ( x 3) 43 (y + )+ 5 (x + 3) = 0 BC: y 5x 7 0 Q.34
Let inclination of line be parametric locus of line is x rcos y 5 rsin B r r ( cos,5 sin ) 3( r cos ) (5 r sin ) 4 r (3cos sin ) 4..(A) A r r ( cos,5 sin ) 5( r cos ) (5 r sin ) 4 r sin 5r cos 4 r (sin 5cos ) 4...(B) ( A) ( B) we get 3cos sin sin 5cos 3 tan tan 5 tan 0 tan
equation of line is y 5 = x i.e. yx 4 Q.35 Point (, ) C 4x y = 3 & 3x 7y = 8 x y 3 4 4 8 7 3 8 3 7 x y 09 9 5 09 9 x, y 5 5 B x, y 5 5 B, 5 5 For point A sign of 3x 7y 8 = 0 is 99 6 8 5 = 4 8 = 34 (positive) 3 48 0
3..(A) For point B and line 4x y 3 = 0 Sign of 4x y 3 is 44 3 = 9 3 = (negative) 5 5 4 3 0 4 33 33 4 (B) For point C and line x + y = 0 Sign of x + y is 09 9 5 5 = 7. = 5. (positive) 0 0..(C) Intersection of (A), (B), (C) gives 33 3 4 Q.36 Image of A(, ) in x y + 5 = 0 is B
point B x y ( 5) 8 x 7 y = 6 B ( 7,6) Image of A (, ) in x + y = 0 is C point C x y 4 6 5 x, y 5 5 C, 5 5 Equation of BC is 6 6 5 y ( x 7) 7 5 8 4 y 6 ( x 7) ( x 7) 46 3 3( y 6) 4( x 7) 0 3y4x 38 98 3y4x 40 Q.37
From angle bisector theorem AD : DC = : D D () ( ) 8 7, 3 3, 3 3 Equation of AD 3 y ( x 5) 5 3 ( x 5) ( x 5) 4 7 7y 7 = x 5 7y x Q.38
Reflection of a in 3x -y + 8 = 0 gives C x y (3 8) 3 9 4 3 3 x3 y3 (3) 3 (3) x 6 y 4 x C 3 3 5 y 3 3 3 ( 5, 3) Equation of BC 3 ( y) ( x 3) 53 ( y) ( x 3) 8 4(y ) + (x 3) = 0 4yx 7 Q.39 A line through the intersection of x y & x y a b b a is of the form x y x y 0 a b b a x ( b a) a b y ab ab x y ab( ) ab ( ) ab a b A is the x intercept
ab( ) A,0 a b B is the y intercept B 0, ab ( ) a b Mid point of AB ab( ) ab( ), ( ab) ( a b) ( hk, ) h k ( ab) ( a b) ab( ) ab( ) ( ab)( ) ab( ) ( a b) h k ab Locus is ( a b) x y ab i.e. ( x y) ab xy( a b) Q.40 Assume angle = 60 o
Proof using vectors BA ( x x ) i ( y y ) j BC ( x x ) i ( y y ) j 3 3 BA BC Area of triangle BA BC sin 60 Area of the triangle x x 3 which is rational. y y x y x y 3 If xx x3... 3 y y y are rational BA BC BA BC sin 60 BABC ( x x )( x x ) ( y y )( y y ) 3 3 BA BC is also rational. BA BC BA BC cos 60 BA BC BA BC must be rational. But ratio of BA BC BA BC is tan 60 which is irrational. not possible that angle is 60 Q.4 A (ae, 0), P (h, k) B ( ae, 0) ( ) ( ) h ae k h ae k a
( h ae) k a ( h ae) k Squaring we get ( h ae) k 4 a ( h ae ) k 4 a ( h ae) k 4 a ( h ae) k 4a 4aeh ( ) h ae k a eh h a e aeh k a aeh e h h ( e ) k a ( e ) h k a a ( e ) Locus is x a y b b a ( e ) Q.4 Centroid C cost sin t sin t cos t, 3 3 ( h, k) 3h cos t sin t (A) 3k sin t cos t.(b) Squaring and adding we get (3 ) (3 ) cos sin cos sin h k t t t t
= Locus is (3x) (3y ) Q.43 SA = SB = SC (circumcentre) x x tan x x tan x x tan 3 3 3 x sec x sec x sec 3 3 x sec x sec x sec k 3 3 Centroid G x tan x, 3 3 x x x x cos cos cos cos 3 k G 3 x sin cos cos, 3 3 k cos sin G k, 3 3 k G kcos sin, k 3 3
S, G, H are collinear Equation of SG is k sin y 0 3 ( x 0) k cos 3 y sin cos x H lies on SG. H ( x, y) satisfies equation of SG sin y x cos y( cos ) x ( sin ) Q.44 L : x y 0 L :3x y 6 0 Assume origin to lie in Area. For (0, 0) L 0 L 0 Region I : L 0 L 0
Region II : L 0 L 0 Region III : L 0 L 0 Region IV : L 0 L 0 Point L L Region (, 3) 6 3 + 6 + 6 I (3, ) 3 + 4 9 4 + 6 II (, 4) + 8 3 8 + 6 III ( 4, ) 4 + + 6 IV So they all lie in different regions. Q.45 AB C Equation of AB is x Ccos y Csin xsin y cos C sin cos xsin y cos C sin cos 0 Foot of perpendicular is Q(h, k) given by h C cos k C sin C cos sin C sin cos csin cos sin cos sin cos h C cos k C sin C sincos sin cos h C cos C cos sin
3 h Ccos k C sin C sin cos 3 k Csin h C 3 cos k C 3 sin sin cos 3 3 h k 3 3 C C 3 3 3 x y C Q.46 h ( u cos ) t k ( usin ) t py t u h cos h h k ( u cos ) p ucos ucos k htan ph u sec locus is y xtan px sec u Q.47
AB = units. Parameter Equation of line L : x rcos y3 rsin PA r PB r AB A ( r cos, 3 r sin ) B ( ( r ) cos, 3 ( n )sin ) A satisfies t + x = 3 B satisfies y + x = 5 3 r sin ( r cos ) 3 3 ( r sin ( ( n ) cos )) 5 r (sin cos ) 4.(A) r (sin cos ) 3 sin 4 4cos 5 form (A), (B) we get 4 sin 4cos sin cos sin cos t t t t t t t 3t t 0 r (sin cos ) sin 4cos..(B)
t, 3 Case I : If tan 90 So Equation of line x...ans. : i Case II : tan 3 3 tan 3 8 9 9 6 tan 8 3 tan 4 So equation of L is 3 y 3 ( x ) 4 4y3x 8...Ans. : ii Q.48 Let equation of line parallel to BC passing through A be called L.
L 8x 3y 3 (7x 5y ) 0 x(8 7 ) (3 5 ) y (3 ()) 0 slope condition we get 3 5 8 3 5 64 56 8 7 L (8x 3y 3) (7x 5y ) Let equation of line parallel to AC passing through B be called L. L 8x 3y 3 ( x 8y 9) 0 (8 ) x (3 8) y (3 9) 0 L parallel to AC 8 7 40 5 56 38 5 L : 8 x 3 y 3 ( x 8 y 9) 0 Let equation of L3 be line parallel to AB passing through C. L3 x 8y 9 (7x 5y ) 0 ( 7 ) x (8 5 ) y (9 ) L3 parallel to 8x + 3y + 3 = 0. So, ( 7 ) 8 (8 5 ) 3 Substituting co ordinates of origin (0, 0)
L, L : 3 ( ) = 46 positive 0 lies BC : 9 negative bet wee n L : x 8 y 9 7 x 5 y 0 3 L & BC All 3 conditions are satisfied L, L : 3 ( 9) = 50 positive 0 lies AC : negative L & AC bet wee n 0 lies in the triangle Q.49 L3, L3 : 9 ) = 0 negative 0 lies AB : 3 positive bet wee n L3 & AB 5x + 3y = 4 3x + 8y = 3 H intersection orthocenter x y 4 3 5 4 5 3 3 8 3 3 3 8 x y 7 77 3 7 77 x, y 3 3 slope of BC 3 5 Equation of BC 3 y 5 ( x 4) 5
BC:5y 3x 3 0 5y 5 3x Equation of AB =? slope of AB is 8 3 Equation of AB 8 y 5 ( x 4) 3 AB:3y 8x 7 7 77 H, 3 3 Equation of BH is 77 5 y 5 3 ( x 4) 7 4 3 78 y 5 ( x 4) 95 ( 4) 5 x 5y + 5 = x + 8 So slope of AC 5 (perpendicular to BH) (A) Let C (h, k) slope CH 77 k 3 3 8 7 h 3 38k 3h3 778 7(3) 0 3(8k 3 h) 403
8k 3h 3..(B) Also BC equation is satisfied by C 5k 3h + 3 = 0..(C) h =, k = C (, ) Equation of AC is from (A) 5 y ( x ) y + 4 + 5x 5 = 0 AC:y 5x Q.50 Slope of AH b a perpendicular to BC b Equation of AH or HD is y (passing through origin) ax Now a3x b3y 0 ax by 0 b x are concurrent ay 0 So
a a b b 3 3 b a 0 0 a 3 3 3 3 a a b b 0 0 b b a 0 ( a a )( a ) ( b )( b b ) 0 3 3 a a a a ( b b b b ) 3 3 aa bb aa3 bb3 Using the equation of BH we can prove the other conditions. Q.5 AH perpendicular to BC y 5 3 x 4 y 3 x 4 4( y ) 3( x ).. (i) 4y 8 3x 3 3x 4y 5..(i) BH perpendicular to AC y 5 3 x 4
y 5 3 x 5 5( y 5) 3( x ) (ii) 5y 5 3x 6 3 3x 5y (ii) solving (i) and (ii) we get y = 6, x = 3 A (33, 6) Q.5 slope of BC slope of AH 3 3 Equation of AH = y = slope of AC slope of BH 4 5 5 4 (x ) Equation of BH = (y ) 3 5 4 (x ) 3(y )+(x ) = 0 4(y ) = 5(x ) x + 3y = 5.(i) A is intersection of (AH) x + 3y = 5 & (AC) 4x + 5y = 0 x y 6 3 5 3 0 5 4 0 4 5 = 5x 4y (ii) B is intersection of (BH) = 5x 4y & (BC) 6 = 3x y x y 4 5 5 4 6 3 6 3
x y 35 0 x y 6 33 x 35, y = 0 x = 3, y 33 A 35, 0 B 3, 33 Equation of slope AB 33 0 3 3 35 3 6 6 3 35 y0 x 6 ( y0) 6x 35 3 6xy 0 35 3 Equation of third side is 6x y 675 Q.53 Let inclination of line L be A, A..., A n be points of intersection. Using parametric form x x r cos For A ( x, y ). Take length as r
x x r cos y y r sin Now A ( x, y ) satisfies ax b y c 0 a ( x r cos ) b ( y r sin ) c 0 r ( a cos b sin ) ( a x b y c ) r ( a x b y c ) OA ( acos bsin ) OA similarly ( anx bn y cn ) OAn ( a cos b sin ) n A be any point of the line L A(h, k) h x r cos k y r sin n OA OR r a x b y c acos bsin n ( ancos bnsin ) r ( anx bn y cn ) ( anr cos bnr sin ) n ( anx bn y cn ) an( h x ) ( k y ) b n n anx bn y cn Locus of A is n
an( x x ) ( y y ) b n n anx bn y cn a... ' an s constant b... ' bn s constant c... ' cn s constant x, y is given Q.54 Equation of PB : k 0 y 0 ( x a) h a yh ay kx ak yh kx ak ay yh kx ak ay y( h a) kx ak..(i) Equation of PA will be y( h a) kx ak (ii) S is intersection of PA & y = x Co ordinates of S : y( h a) kx ak yx 0
y x ak k h a ak h a k 0 0 y x ( ak) ( ak) k ( h a) ak x y k a h ak ak S, k a h k a h Co ordinates of R will be ak ak R, h a k h a k (replace a by a) Length SR ak ak h a k k a h ak ( k a h) ( h a k) a ( h k) Length SR = K k h K ak a ( k h) Replacing k by y, h by x K 8 ay ( x y) a ( y x) is the locus of point P. Q.55 y at t x at ( ) Let t & t be lines in the family such that l & l are perpendicular.
L y at t x at 3 L y at t x at 3 L perpendicular L slope L L = t t tt L : y t x at at 3 L : y t x at at 3 Intersection is y x at at t at at 3 3 3 3 at at t at at t t 3 3 ( t( at at )) t( at at) y ( t t ) 3 3 ( at at ) ( at at) x ( t t ) att att att ( t t ) y ( t t ) y at t ( t t ) [ t t ] y a( t t ) x a a( t t t t ) y ( t t ) t t t t a y x a t t tt a a
y a ax a y ax 3a y a x a ( 3 ) Q.56 ABCD 8 7 4 8 6 3 5 6 [ 4 35 4 4 ( 4 6 0 6)] [79 ( 7)] (5) PDA 4 8 h 4 6 k [(4 8 k h ) ( 6 6 h 4 k )] [40 4 h k ] [0 h 3 k] Now the line x 3y0 0 passes through D does not pass through ABCD. So for any point inside the Quadrilateral 0 h 3k will be positive.
PDA [0 h 3 k] PBC 7 h 5 3 k 5 [6 7 k 5 h (35 3 h k )] [ 9 h 5 k ] x5y9 0 is the equation of BC. for all points inside the Quadrilateral 9 5 PBC [9 h 5 k ] PBC PDA 9 5 (0 h 6 k) h k 69 7k h 5 4 ABCD 38 4h4k 5 4h4k 3 4x4y 3 Q.57 3x + 4y = 9 mx y = h k will be negative.
x y 9 4 3 9 9 4 m m x y 5 3 9m 3 4m 5 3 9m x, y 34m 34m x and y must be integers. 5 x 3 4m 3 + 4m must be a factor of 5 or 5 3 + 4m can be :,, 5, 5 For 3 4m m For 3 4m m For 3 4m 5 m For 3 4m 5 m Put m & in y and see if y is also integer. m = then m = then 3 9 6 y 6 3 4 3 8 5 y 3 38 5 so m is, Q.58 dx cy cd passes through (a, b) parametric from of the line. x a r cos d tan y b r sin c
When x 0, a r cos 0 a r cos y b a tan d (y intercept) When y = 0 b r sin b x a cos sin x a bcot c (x intercept) c ( a bcot ) d ( b a tan ) c d a bbcot atan c 0 so a asin bcos bcot 0..() sin d 0 so bcos asin b atan 0..() cos Multiplying () & () sincos 0 tan 0 Now c d a b b( cot ) a( tan ) AB makes 45 with 3x 4y 49 0 AP also makes 45 with 3x 4y 49 0 slope 3 4
tan 45 3 m 4 3m 4 4m 3 4 3m (4m 3) (4 3 m) 0 7m 0 or m 7 m 7 m 7 or 7 So equation of AP or AB is ( y ) ( x ) and ( y ) 7( x ) 7 7y 7 x 0 y 7x 6.(ii) 7yx 8..(i) 7x y 6.(iii) Intersection with 3x4y 49 will be give P & B x7y 8 3x4y 49 x y 8 7 8 7 49 4 3 49 3 4 x y 375 5 5 x5, y (5, ) (A) A(, )
Intersection of 7x y 6 & 3x 4y 49 x y 6 7 6 7 49 4 3 49 3 4 x y 5 35 5 x, y 3 (, 3) (B) Case : I A(, ) P (5, ), B (, 3) Then Q ( x, y) Mid point PQ = mid- point of AB 5 x x 3 and y 3 y So Q ( 3, ) One possibility A (, ), B (, 3), P (5, ), Q ( 3, ) Case : II A (,), P (, 3), B (5, ) Let Q ( x, y) Mid point PQ = Mid point AB x 5 x 5
and y 3 y 3 Q (5,3) A (,), P (, 3), Q (5,3), B (5, ). There are parallelograms possible. Represented by the dotted lines Q.59 Q.60 Let line be x y Q (, 0) and S (, ) R(0, ) Now P( a, b ) satisfies the line so
a b a b x y Q.6 C ( a ib) (0 i) 0 ib C ( ai b) ib C b i( a b) C ( b, a b) a b a b AC Mid point of, Mid point of AC lies on the line y x Q.6 Condition for co cyclic points mm x y 3 0 m y kx m k mm k
k Q.63 Reflection of A (h, k) in lx + my + n = 0 gives A x h y k ( lh my n) l m l m lh mk n Let l m p x h pl y k pm A lies on ax by c 0 a( h pl) b( k pm) c 0 ah bk c p(al bm) lh mk n ah bk c ( al bm) l m locus is ( l m )( ax by c) ( al bm)( lx my n) Q.64 L : x y 5 L :3x y 8 Point P intersection given by
x y 5 5 8 3 8 3 x y 3 x 3 y P (3, ) L rotated 4 clockwise so m 3 New slope 3 tan m tan...( 45 ) 3 4 3() Equation of L : y ( x 3) : y x c 0 : yx 7 slope of sin L 5 is cos 5 r 0
x 3 0 5 y 0 5 x 3 y A (3, ) or B (3, ) Q.65 L xcos ysin p 0 k k k k k 0,,,3 L cos( ) L cos( ) L cos( ) 3 3 3 Family of lines through A is L L 0 slope of AH : x(cos cos ) (sin sin ) p p (cos cos ) sin sin slope of AH slope AH slope BC (cos cos ) cot 3 (sin sin 3) (cos cot sin ) (sin cos cot ) 3 3 3 sin sin3 cos cos 3 ( ) cos cos3 sin sin3
cos ( 3 ) cos ( ) 3 Equation of AH is L i.e. L 0 cos( ) L L 3 cos( 3) 0 L cos( ) L cos( ) and H satisfies this 3 3 Similarly L cos( ) L cos( ) 3 3 Can be proves using slope BH slope AC Q.66 y xtan meets the circle A a a at ( acos, asin ) at A (solving simultaneously) x y a ( cos, sin ) similarly we get ( acos, asin ) B C a 3 a 3 ( cos, sin ) Circumcentre is the origin S(0, 0) Centroid co ordinates G a cosi asini, 3 3 Equation of SG is asin i 0 y 0 3 ( x 0) a cos i 0 3
Equation of SG is Q.67 y sin i x cos i and H lies on this line Acute angle between L & L will be less than the acute angle between L 3 & L. To get the slope of L 3 rotate L clockwise by 5 m 3 (slope of L 3 ) ( 3) m tan 0 m tan ( 3) 0 0 0 3 0 3 0 3 8 3 (0 3 ) (8 3) m3 6
30 68 3 80 3 m3 6 m 3 0 3 98 6 Point N is given by intersection of x0y 34 and x 3y 5 x y 34 0 34 0 5 3 5 3 x y 5 39 3 x = 4, y = 3 N(4, 3) Equation of L 3 is 0 3 98 ( y3) ( x 4) 6 Q.68 Parametric form of Line x rcos y 3 rsin For point Q Let PQ r x r cos
y 3 r sin Also x y k r cos 3 r sin k r (sin cos ) k 5 k 5 r (sin cos ) For point R let PR r x r cos y 3 r sin x 3y 9 r cos 3( 3 r sin ) 9 r (cos 3sin ) 0 0 r (cos 3sin ) rr 0 k 5 (sin cos )(cos 3sin ) k 5 (sin cos )(cos 3sin ) If k We get 3 3sin cos 4sin cos sin 4sin cos sin 4sin cos
cos 4sin cos cos 0 i.e. 90 OR cos 4sin tan Equation is x OR ( y 3) ( x ) y 6 x x OR y x 4 0 Now, k 5 (sin cos )(cos 3sin ) If cos 0 then divide by cos ( k 5)(sec ) ( tan )( 3tan ) ( 5)( tan ) 4 tan 3tan k For unique line tan must be unique so roots must be equal ( k 5 3) tan 4 tan ( k 4) 0 ( k ) tan 4 tan ( k 4) 0 4 6 4( k )( k 4) tan ( k ) If D = 0 value is unique 4( k 6k8) 6 k 6k 4 0
6 36 6 k k 3 5 Q.69 Slope of AC is 3 5 AC and BD are perpendicular ABCD is a square. Rotate AC clockwise 45 scale to get D (say) i C (4 i) 3i C ( i)( i) ( 3 i) C i i 3i C 3i 3i C is (, 0) Rotate AC anticlockwise 45 scale i B (4 i) 3i B ( i)( i) ( 3 i) to get B (say)
B i i 3i B 44i B (4,4) Q.70 XM MY line parallel to x + y = 3 passing through (3, 4) is x + y = 7 Let equation of CD : x + y = C length C7 C7 MY ( C 7) 4 4 MX MX MY 4 C 7 Equation of x + y = 5 Co ordinates of A A : intersection of x + y = 3 x = x y = 3 y = AC CM 3
9 4 C, 3 3 C (5,0) D is intersection of x y = 3 and x + y = 5 x = 6, y = 9 D (6,9) DB : BM = 3 : ( ) 9 6 9 B, 3 3 B 3 3, Equation of BC is 3 0 0 y ( x 0) 3 5 7 y0 ( x 0) 7 BC : 7y0 7x
AB units DC units MX 6 units Area MX AB DC 3 6 9 A 9 sq units Q.7 Let t ; represent the sides. ( ) x y ( ) ( L ) ( ) x y ( ) ( L ) B and C are the x - intercepts of L& L slope of L is slope of L is ( ) ( ) L CH perpendicular k 0 h
k( ) h..(i) Similarly using CH perpendicular to k( ) h Solving we get hk the orthocenter lies on the line y x Q.7 L we get AC = BC 5 units. And AB = 0 units. ABC is a right angled isosceles triangle Equation of AC is 0 y 0 ( x8) 8 7 AC : y + x = 6 Equation of BC is (0 ) y 0 ( x8) 8 6 BC : x y = 8 Equation of AB is y ( x 6) 7 6 y 3( x 6) AB : y 3x 9 ABC DEF
distance BC and EF is 8 5 3 or 8 5 5 we take as (decreasing y intercept of EF)
distance AB & DE is 9 0 0 9 0 9 or 9 We take 9 DE is shifted upward so y intercept is decreasing 9 distance of DF to AC is 6 5 5 or DF y intercept in increasing D (6, 9) F (, ) E (, 6) Q.73 slope of PR = 0 4 6 3 3 8 4
3 tan 4 3 sin 5 4 cos 5 x 3 rcos 4r x 3 5 3r y 4 5 parametric Equation of PR PQ.5 r 5 (.5) x 3 4 5 x 3 x 5 or (.5) y 4 3 5 y 5.5 or.5 5 Q 5, or, Q.74 Let this be equation of the line y mx + ma = 0
a ma ma m a a ma m a Squaring we get ( a ma) a ( m ) 4a m a 4ama a a m 4a 4a m m equation of line is y x a Q.75 9a b 6ab c 0 (3 a b) c 0 (3 a b c)(3 a b c) 0 Either 3a + b c = 0 OR 3a + b + c = 0 Case I : 3a + b c = 0 c 3a b The line ax + by + c = 0 ax by 3a b 0 a( x 3) b( y ) 0 Always passes through ( 3, ) Case II : 3a b c 0 c 3a b The line ax by c 0 ax by 3a b 0
a( x 3) b( y ) 0 Always passes through (3, ) Either line passes through ( 3, ) or (3, ) or both. Q.76 Equation of AB : 3 0 y 0 ( x 5) 5 4y = 3 (x 5) 4y = 3x 5 5= 3x + 4y c) y x for a is 3 = y x for B is 0 5 = 5 y x for C is + = 4 A & C positive B negative so this line passes through the interior of the triangle (y x = 0). So not always true. c) is false. For A : x + 3y = + 9 = For B : x + 3y = 0 + 0 = 0 For C : x + 3y = + c = 4 So positive for all points A, B, C. it will be positive for all points inside a) is true b) For A : x 3y is 9 = 7
for B : x 3y is 0 C : x 3y is 6 = 8 So for A & C it is negative for B it is positive. So A & C lie on one side and B lies on the other side. So the line passé as shown. not always true b) is false. Q.77 x y 5 7 Rotate AC by 45 anticlockwise scale by Co ordinates of B i (5 0 0 7 i) (0 7 i) to get B.
( i) (5 7 i) 7i 5 7i 5i 7 7i 6i7i 66i B (6, 6) Co ordinates of D : rotate by 45 clockwise scale by to get D i (5 0 0 7 i) (0 7 i) i (5 7 i) 7i (5 7 i) ( i ) 7 i 5 7i 5i 7 7i 6i 7i i D (,) B(6, 6) D(,) Q.78 ( cos 3sin ) x (3cos sin ) y5cos sin divide by cos we get cos 0 ( 3tan ) x (3 5tan ) y 5 tan (x 3y 5) (3x 5y ) tan 0
Represents family of straights passing through the intersection of (x 3y 5)0 & (3x 5y ) 0 x 3y 5 0 3x 5y 0 It of intersection. x y 5 3 5 3 5 3 3 5 x y 9 9 9 x, y point of intersection is (,) Q.79 Area = 5 units sq. Side = 5 unit Distance between AB & CD is 5 unit Equation of CD is of the form 3x 4y + c = 0 ( CD ++ AB) Distance AB and CD is C 0 3 4 5
C 5 C 5 Equation of CD is 3x 495 0 similarly Equation of BC is of the form 4x + 3y + C = 0 distance between AD & BC is C ' 0 4 3 5 C ' 5 C ' 5 Equation of BC is 4x 3y 5 0 Q.80 A (4, 3) B(, 5) AH perpendicular BC So 3 k 5 4 h 5 k 5 3 h 5( k5) ( h )3 5k3h 3.(i)
CH perpendicular to AB 3 5 k 4 h 4 k 3 h 4( k ) 3( h ) 4k3h 5..(ii) k h 3 3 5 3 5 3 5 3 4 5 4 3 h 33 k h 78 99 3 k 6 h 33 C (33, 6) centroid 4 33 5 3 6 G, 3 3 G 35 8, 3 3