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1 LENGTH OF THE PERPENDICULAR FROM A POINT TO A STRAIGHT LINE AND DISTANCE BETWEEN TWO PAPALLEL LINES THEOREM The perpendicular distance from a point P(x 1, y 1 ) to the line ax + by + c 0 is ax1+ by1+ c Proof: Let the axes be translated to the point P(x 1, y 1 ). Let (X,Y) be the new coordinates of (x, y). Then x X + x 1, y Y + y 1 The transformed equation of the given line is a(x + x 1 ) + b(y + y 1 ) + c 0 ax + by + (ax 1 + by 1 + c) 0 The perpendicular distance from the new origin P to the line is (from normal form) The perpendicular distance from a point ax1+ by1+ c P(x 1, y 1 ) to the line ax+ by + c 0 is. ax + by + c 1 1 a + b. DISTANCE BETWEEN PARALLEL LINES THEOREM The distance between the two parallel lines ax + by + c 1 0 and ax + by + c 0 is c1 c. Proof: Given lines are ax + by + c (1) ax + by + c () Let P(x 1, y 1 ) be a point on the line ().

2 Then L 0 ax 1 + by 1 + c 0 P(x 1, y 1 ) ax 1 + by 1 c. L 10 Distance between the parallel lines Perpendicular distance from P to line (1) ax1+ by1+ c1 c1 c FOOT OF THE PERPENDICULAR THEOREM If (h, k) is the foot of the perpendicular from (x 1, y 1 ) to the line ax + by + c 0 h x1 k y1 ( ax1+ by1+ c) (a0, b0) then. a b Proof : Let A (x 1, y 1 ) P (h, k) P lies on ax + by + c 0 ah + bk + c 0 ah + bk c k y1 Slope of AP is h x1 a Slope of given line is b AP is perpendicular to the given line k y1 a 1 h x1 b k y1 h x1 b a By the law of multipliers in ratio and proportion h x1 k y1 a( h x1) + b( k y1) a b

3 ah + bk ax1 by1 ax1 by1 c ( ax1+ by1+ c) h x1 k y1 ( ax1+ by1+ c) Hence a b IMAGE OF A POINT THEOREM If (h, k) is the image of (x 1, y 1 ) w.r.t the line ax + by + c 0 (a 0, b 0), h x1 k y1 ( x1+ by1+ c) then. a b Proof: Let A(x 1, y 1 ), B(h, k) 1 1 Mid point of is P x + h, y + k Since B is the image of A,therefore mid point P lies on ax + by + c 0. x 1+ h y 1+ a b k + +c 0 ax 1 + by 1 + ah + bk + c 0 ah + bk ax 1 + by 1 c. k y1 Slope of AB is h x1 a And Slope of given line is b AB is perpendicular to the given line k y1 a 1 h x1 b k y1 h x1 b a By the law of multipliers in ratio and proportion

4 ( ) + ( ) h x1 k y1 a h x1 b k y1 a b ah + bk ax1 by1 ax1 by1 c ax1 by 1 ( ax1+ by1+ c) h x1 k y1 ( x1+ by1+ c) Hence a b Note 1: The image of (x 1, y 1 ) w.r.t the line x y is (y 1, x 1 ) Note : The image of (x 1, y 1 ) w.r.t the line x + y 0 is ( y 1, x 1 ) THEOREM If the four straight lines ax + by + p 0, ax + by + q 0, cx + dy + r 0 and cx + dy + s 0 form a parallelogram. Then the area of the parallelogram so formed is ( p q)( r s) bc ad Proof: Let L 1 ax + by + p 0 L ax + by + q 0 L 3 cx + dy + r 0 L 4 cx + dy + s 0 Clearly L 1 L and L 3 L 4. So L 1 and L 3 are nonparallel. Let be the angle between L 1 and L 3. Let d 1 distance between L 1 and L Let d distance between L 3 and L 4 Now cos θ bc ad ( )( c + d ) ac + bd ( )( c + d ) and Now area of the parallelogram is 1 sin sin θ p q a + b r s c + d ( )( c + d ) ( ac+ bd) ( )( c + d ) ddθ ( p q))( r s) bc ad

5 ANGLE BETWEEN TWO LINES THEOREM If θ is an angle between the lines a 1 x + b 1 y + c 1 0, a x + b y + c 0 then cosθ ± aa + bb Proof: The lines passing through the origin and parallel to the given lines are a 1 x + b 1 y 0, -- (1) a x + b y () Let θ1, θ be the inclinations of (1) and () respectively ( θ>θ 1 ) Now θ is an angle between (1) and () θθ1 θ P( b 1,a 1 ) satisfies eq(1), the point lies on (1) Similarly, Q( b, a ) lies on () Let L, M be the projection of P,Q respectively on the x - axis. OL b1 PL a1 cos θ 1, sin θ 1 OP OP OM b OM a cos θ, sin θ OQ a b OQ θθ1 θ cos cos ( θ θ) 1 cos θ 1 cos θ + sin θ 1 sin θ ( b 1) ( b ) a + 1 a aa + bb a1 + b1 1 b1 a a 1 + b1

6 Note 1: If is the acute angle between the lines then aa 1 1+ bb 1 cos θ 1 1 Note : If is an angle between two lines, then is another angle between the lines. Note 3: If is an angle between two lines are not a right angle then the angle between the lines means the acute angle between the lines. Note 4: If θ is an angle between the lines a 1 x + b 1 y + c 1 0, a x + b y + c 0 a b a b 1 1 then tan θ a1b 1 Note 5: If is the acute angle between the lines a 1 x + b 1 y + c 1 0, a x + b y + c 0 then ab 1 ab1 a1/ b1 a/ b tan θ ab aa / bb + 1 THEOREM ( ) ( ) ( a ) ( ) 1/ b1 a / b where m 1 + ( a / b )( a / b ) 1, m are the slopes of the lines. 1 1 The equation of the line parallel to ax + by + c 0 and passing through (x 1, y 1 ) is a(x x 1 ) + b(y y 1 ) 0. Proof: Slope of the given line is a/b. Slope of the required line is a/b.(lines are parallel) Equation of the required line is a y y 1 (x x b 1 ) b(y y 1 ) a(x x 1 ) a(x x 1 ) + b(y y 1 ) 0. Note 1: The equation of a line parallel to ax + by + c 0 may be taken as ax + by + k 0. Note : The equation of a line parallel to ax + by + c 0 and passing through the origin is ax + by 0.

7 THEOREM The equation of the line perpendicular to ax + by + c 0 and passing through (x 1, y 1 ) is b(x x 1 ) a(y y 1 ) 0. Proof: Slope of the given line is a/b. Slope of the required line is b/a. (since product of slopes -1) Equation of the required line is y y 1 b a (x x 1 ) a(y y 1 ) b(x x 1 ) b(x x 1 ) a(y y 1 ) 0. Note 1: The equation of a line perpendicular to ax + by + c 0 may be taken as bx ay + k 0 Note : The equation of a line perpendicular to ax + by + c 0 and passing through the origin is bx ay 0. EXERCISE -3 (d) I. Find the angle between the following straight lines. 1. y 4 x, y 3x + 7 Sol: given lines are y 4 x x + y 4 0and 3x y Let θ be the angle between the lines, then a1a+ b1b cos θ ( 1) π θ 4. 3x + 5y 7, x y Sol. ans : θ cos

8 3. y 3x + 5, y 1 x 3 3 Sol. slope of 1 st line is m1 3 Slope of nd line is m mm 1 ( 3) 1. 3 The lines are perpendicular,hence angle between the lines is 4. ax + by, a(x y) + b(x + y) b Sol. given lines ax + by, () x + (- ) y b let θ be the angle between the lines, then a() + b( ) cos θ + (a+ b) + ( a+ b) π θ a + ab ab + b 1 ( ) ( ) π θ 4 Find the length of the perpendicular drawn form the point given against the following straight lines. 5. 5x y + 4 0, (-, -3) ax Sol. Length of the perpendicular 1+ by1+ c 5( ) ( 3) x 4y , (3, 4) Sol. Length of the perpendicular x 3y 4 0, (0, 0) 4 Sol. Ans; 10 Find the distance between the following parallel lines. 8. 3x 4y 1, 3x 4y 7 Sol. Given lines are 3x 4y 1, 3x 4y 7

9 c Distance between parallel lines 1 c x - 3y 4 0, 10x 6y 9 0 Sol. Equations of the lines can be taken as 10x 6y 8 0,10x 6y Distance between parallel lines Find the equation of the straight line parallel to the line x + 3y and passing through the point (5, 4). Sol. Given line is x + 3y Equation of the parallel to x + 3y is x + 3y k. This line is passing through P (5, 4) k k Equation of the required line is x + 3y Find the equation of the straight line perpendicular to the line 5x 3y and passing through the point (4, -3). Sol. Equation of the given line is 5x 3 y Equation of the perpendicular to 5x 3 y is 3x + 5y + k 0 This line is passing through P (4, -3) k 0 k 3 Equation of the required line is 3x + 5y Find the value of k, if the straight lines 6x 10y and k x 5y are parallel. Sol. Given lines are 6x 10y and k x 5y a1 b1 Since the lines are parallel a b k k Fund the value of P, if the straight lines 3x + 7y 1 0 and 7x p y are mutually perpendicular. Sol. Given lines are 3x + 7y 1 0, 7x p y Since the lines are perpendicular ab p 0 7p 1 p ( ) 14. Find the value of k, the straight lines y 3kx and (k 1) x (8k 1) y 6 are perpendicular. Sol. Given lines are -3kx + y + 4 0

10 (k 1)x (8k 1)y 6 0 These lines are perpendicular aa 1 + bb 1 0 3k(k 1) 1(8k 1) 0 6k + 3k 8k k + 5k 1 0 (k + 1)(6k 1) 0 k 1or 1/6 15. (-4, 5) is a vertex of a square and one of its diagonal is 7x y Find the equation of the other diagonal. ( 4, 5) D C A B Sol. let ABCD be the square. Let the equation of the diagonal AC be 7x y The point (-4, 5) is not satisfying the equation. Let D(-4, 5) The other diagonal BD is perpendicular to AC. Equation of BD is can be taken as x + 7y + k 0 D (-4, 5) is a point on this line k 0 k Equation of BD is x + 7y 31 0 II. 1. Find the equation of the straight lines passing through (1, 3) and i)parallel to ii) perpendicular to the line passing through the points (3, -5) and (-6, 1). Sol. Given points A (3, -5), B (-6, 1) Slope of AB i) slope of the line parallel to AB is 3 equation of the line parallel to AB and passing through (1, 3) is y 3 (x 1) 3y 9 x+ x+ 3y ii) slope of the line perpendicular to AB is 3/.

11 Equation of the line passing through (1, 3) and having slope 3/ is x+3y-50. y 3 (x 1) 3. The line x y 1 meets the X axis at P. Find the equation of the line a b perpendicular to this line at P. Sol. given line is x y (1) a b On x- axis y 0 x 0 1 x a a b Point P(a,0) Equation of the line perpendicular to (1) is x y + k b a This line is passing through P (a,0) a 0 k k a/b b + Equation of the line is x + y a. b a b 3. Find the equation of the line perpendicular to the line 3x + 4y and making intercept -4 on X axis. Sol. Given line is 3x + 4y Equation of the perpendicular to 3x + 4y is 4x 3y k 4x 3y x y k k k k 4 3 k x int ercept 4 k 16 4 Equation of the required line is 4x 3y 16 4x 3y A (-1, 1), B (5,3) are opposite vertices of a square in the XY plane. Find the equation of the other diagonal (not passing through A, B) of a square. Sol. A(-1,1), B ( 5, 3) are opposite vertices of the square. Slope of AB

12 D B(5, 3) O A( 1, 1) C The other diagonal is perpendicular to AB 1 Slope of CD -3 m Diagonal CD is passing through O (,), Equation of CD is y 3 (x ) 3x + 6 3x + y 8 0. Let O is the point of intersection of the diagonals then O, (,) 5. Find the foot of the perpendicular drawn from (4,1) upon the straight line 3x 4y Sol. Equation of the line is 3x - 4y If (h,k)is the foot of the perpendicular from (x 1,y 1) on the line ax + by + c 0, then h x1 k y1 ( ax1+ by1+ c) a b h 4 k 1 ( ) h 4 k h 4 h k 1 k Foot of the perpendicular, Find the foot of the perpendicular drawn from (3,0) upon the straight line 5x + 1y Sol. Ans:, 13 13

13 7. x - y is the perpendicular bisector of the line segment joining the points A,B. If A (-1,-3), find the co-ordinates of B. Sol. If PQ is the perpendicular bisector of AB, then B is the image of A in the line PQ. Equation of the line is x 3y -5 0 Given point A (-1,-3). Here B is the image of A w.r.t x 3y Let B (h,k) be the image of A (-1,-3), then h x1 k y1 ax ( 1+ by1+ c) a b h+ 1 k+ 3 ( ) h+ 1 y h h y h k+ 3 k 3 6/ Co-ordinates of B are, Find the image of the point straight line 3x + 4y 1 0. Sol. 7 6 Ans:, Show that the distance of the point (6,-) from the line 4x + 3y 1 is half of the distance of the point (3,4) from the line 4x 3y Find the locus of foot of the perpendicular from the origin to a variable straight line which always passes through a fixed point (a,b). Sol. Let P(x,y) be the foot of the perpendicular from O(0,0) to the line.

14 Slope of OP y x III. Line is passing through A(a,b). Slope of AP y a x b Line AP is perpendicular to OP, product of slopes -1 y y b. 1 y by ( x ax) x x a x + y ax by 0 1. Show that the lines x 7y 0, 3x + 4y and 7x + y 54 0 form a right angled isosceles triangle. Sol. Given lines x 7y (1) 3x + 4y () 7x + y (3) C (1) (3) A () B Let A be the angle between (1),() a cos A 1a+ b1b a1 + b A Let B be the angle between (),(3) then cos B B 45 Let C be the angle between (3), (1) 7 7 cosc 0 C Since A B 45 andc 90 Given lines form a right angled isosceles triangle.. Find the equation of the straight lines passing through the point (-3,) and making an angle of 45 with the straight line 3x y Sol. Given point P (-3,) Given line 3x y (1)

15 Slope m 1 - a b 3 Let m be the slope of the required line. m 3 Then tan m m 3 1 m m 1+ 3m m m m -4 or m - m 3 - m 3-1 3m 1+ 3m 4m m ½ case (1) m - and point (-3,) Equation of the line is y - -(x + 3) -x 6 x + y case () m 1,point (-3,) Equation of the line is y 1 (x + 3) y 4 x + 3 x -y Find the angle of the triangle whose sides are x + y 4 0, x + y 6 0, 5x + 3y Sol. Ans: cos, cos,cos Prove that the foot of the perpendiculars from the origin on the lines x + y 4, x + 5y 6 and 15x 7y 44 are collinear. 5. Find the equation of the lines passing through the point of intersection of the lines 3x + y + 4 0, x + 5y 1 and whose distance form (, -1) is. Sol. Equation of the lines passing through the point of intersection of the line

16 L1 3x+ y+ 4 0,L x+ 5y 1 0 is L1+λ L1 0 (3x + y + 4) + λ (x + 5y 1) 0 (3+ λ )x + ( + 5 λ )y + (4 λ ) (1) Given distance from (, -1) to (1) 3+ λ ) + (+ 5 λ)( 1) + (4 λ ) (3 + λ ) + ( + 5 λ) λ+ 8 (3 + λ ) + ( + 5 λ) ( λ + 4) 9 + 4λ + 1λ λ + 0λ 8λ + 40λ 3 0 8λ λ+ 4λ (λ+ 3)(14λ 1) 0 λ, λ 14 From (1) 1 If λ, then equation of the line is 4x + 3y If λ, then equation of the line is y Each side of a square is of length 4 units. The center of the square is (3, 7) and one of its diagonals is parallel to y x. Find the Co-ordinates of its vertices. Sol. Let ABCD be the square. Side AB 4 Point of intersection of the diagonals is the center P(3, 7) D C P A M B From P drawn PM AB. Then M is midpoint of AB AM MB PM Since a diagonal is parallel to y x, its sides are parallel to the co-ordinate axes. M(3, 5) A (3-, 5), B(3+, 5), C(3+5, 7+),D(3-,7+)

17 A (1, 5), B(5, 5), C(5, 9),D(1,9) 7. If ab > 0, find the area of the rhombus enclosed by the four straight lines ax ± by ± c 0. Sol. let the Equation of AB be ax + by + c 0 ---(1) Equation of CD be ax + by c 0 ---() Equation of BC be ax by + c 0 ---(3) Equation of AD be ax by c 0 ---(4) c Solving (1) and (3), B,0 a c Solving (1) and (4), A 0, b c Solving () and (3), C 0, b c Solving () and (4), D,0 a Area of rhombus ABCD 1 x 1 (y y 4 ) 1 c c c c c c 0(0 0) + + 0(0 0) + a b b a b b 1 4c c. sq. units ab ab 8. Find the area of the parallelogram whose sides are 3x + 4y + 5 0, 3x + 4y 0, x + 3y and x + 3y 7 0 Sol. Given sides are 3x + 4y (1) 3x + 4y () x + 3y (3) x + 3y (4)

18 Area of parallelogram formed by (1), (), (3), (4) (c 1 c )(d1 d ) (5 + )(1 + 7) ab a b 3(3) (4) sq. units 9 8 1

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