CONCURRENT LINES- PROPERTIES RELATED TO A TRIANGLE THEOREM The medians of a triangle are concurrent. Proof: Let A(x 1, y 1 ), B(x, y ), C(x 3, y 3 ) be the vertices of the triangle A(x 1, y 1 ) F E B(x, y ) D C(x 3, y 3 ) Let D,E,F be the mid points of BC, CA, AB respectively x + x 3 3 3 1 3 1, y + y, x + D E x, y + y = = x 1+ x 1, y + F y = Slope of AD is y + y3 y1 y + y y = x + x x + x x 3 1 3 x 3 1 1 Equation of AD is y + y3 y1 y y1 = ( x x1) x + x3 x1 (y y 1 ) (x + x 3 x 1 ) = (x x 1 )(y + y 3 y 1 ) L 1 (x x 1 )(y + y 3 y 1 ) (y y 1 ) (x + x 3 x 1 ) = 0. Similarly, the equations to BE and CF respectively are L (x x )(y 3 + y 1 y ) (y y ) (x 3 + x 1 x ) = 0. L 3 (x x 3 )(y 1 + y y 3 ) (y y 3 ) (x 1 + x x 3 ) = 0. Now 1. L 1 + 1.L + 1. L 3 = 0 The medians L 1 = 0, L =0, L 3 = 0 are concurrent.

THEOREM The altitudes of a triangle are concurrent. Proof: Let A(x 1, y 1 ), B(x, y ), C(x 3, y 3 ) be the vertices of the triangle ABC. Let AD, BE,CF be the altitudes. Slope of y3 y BC is and AD BC x3 x x Slope of the altitude through A is 3 x y y 3 x Equation of the altitude through A is y y 1 = 3 x y3 y (y y 1 ) (y 3 y ) = (x x 1 ) (x 3 x ) L 1 = (x x 1 )(x x 3 ) + (y y 1 )(y y 3 ) = 0. Similarly equations of the altitudes through B,C are L = (x x ) (x 3 x 1 ) + (y y ) (y y 3 ) = 0, L 3 = (x x 3 ) (x 1 x ) + (y y 3 ) (y 1 y ) = 0. Now 1.L 1 + 1.L + 1.L 3 = 0 The altitudes L 1 = 0, L = 0, L 3 = 0 are concurrent. (x x 1 ) THEOREM The internal bisectors of the angles of a triangle are concurrent. THEOREM The perpendicular bisectors of the sides of a triangle are concurrent

EXERCISE 3 (e) I. 1. Find the in center of the triangle whose vertices are (1, 3)(, 0) and (0, 0) Sol. let A(0, 0), B (1, 3), C(, 0) be the vertices of ABC a = BC= b =CA= (1 ) + ( 3 0) = 1+ 3 = ( 0) (0 0) = 4 = C = AB= (0 1) + (0 3) = 4 = ABC is an equilateral triangle co-ordinates of the in centre are ax bx cx ay by cy + +, + + a+ b+ c a+ b+ c = 1 3 1 3 6 3 1 =, = 1, 6 6 3 =.0 +,1+..0 +. 3 +.0, + + + +. Find the orthocenter of the triangle are given by x + y + 10 = 0, x y = 0 and x + y 7 = 0 Sol. Let equation of AB be x + y + 10 = 0 ---(1) BC be x y = 0 ---() and AC be x + y 7 = 0 ---(3) Solving (1) and () B = (- 4, - 6 ) Solving (1) and (3) A =(17, -7) Equation of BC is x y = 0 Altitude AD is perpendicular to BC, therefore Equation of AD is x + y + k = 0 AD is passing through A (17, -7) 17 7 + k = 0 k = 10 Equation if AD is x + y + 10 = 0 ----(4)

Altitude BE is perpendicular to AC. Let the equation of DE be x y = k BE is passing through D (-4, -6) -4 + 1 = k k = 8 Equation of BE is x y = 8-----(5) Solving (4) and (5), the point of intersection is (-4, -6). Therefore the orthocenter of the triangle is (-4, -6). 3. Find the orthocentre of the triangle whose sides are given by 4x 7y +10 = 0, x + y = 5 and 7x + 4y = 15 Sol. Ans: O (1, ) 4. Find the circumcentre of the triangle whose sides are x = 1, y = 1 and x + y = 1 Sol. Let equation of AB be x = 1----(1) BC be y = 1 ----() and AC be x + y = 1 ----(3) lines (1) and () are perpendicular to each other. Therefore, given triangle is a right triangle and B=90º. Therefore, circumcentre is the mid point of hypotenuse AC. Solving (1) and (3), vertex A =(1, 0) Solving () and (3), vertex c =(0, 1) Circumcentre = mid point of AC= 1 1, 5. Find the incentre of the triangle formed by the lines x = 1, y = 1 and x + y = 1 Sol. ANS: 1, 1

6. Find the circumcentre of the triangle whose vertices are (1, 0), (-1, ) and (3, ) Sol. vertices of the triangle are A (1, 0), B (-1, ), (3, ) Let S (x, y) be the circumcentre of ABC. Then SA = SB = SC Let SA = SB SA = SB (x 1) + y = (x + 1) + (y ) x x+ 1+ y = x + x+ 1+ y 4y+ 4 4x 4y = -4 x y = -1 ---(1) SB = SC SB = SC (x + 1) + (y ) = (x 3) + (y ) x + x+ 1= x 6x+ 9 8x = 8 x = 1 From (1), 1 y = - 1 y = Circum centre is (1, ) 7. Find the value of k, if the angle between the straight lines kx + y + 9 = 0 and 3x y + 4 = 0 is π /4 Sol. Given lines are kx + y + 9 = 0 3x y + 4 = 0 and angle between the lines is π /4. π 3k 1 1 3k 1 cos = = 4 k + 1 9+ 1 10 k + 1 Squaring 5k + 5 = (3k 1) = 9k 6k + 1 4k 6k 4 = 0 k 3k = 0 (k - ) (k + 1) = 0 k= or -1/ 8. Find the equation of the straight line passing through the origin and also the point of intersection of the lines. x y + 5 = 0 and x + y + 1 = 0 Sol. Given lines are L 1 = x y + 5 = 0 L = x + y + 1 = 0

Equation of any line passing through the point of intersection of the lines L 1 =0 and L =0 is L1+ KL = 0 (x y + 5) + k (x + y + 1) = 0 -----(1) This line is passing through O (0, 0) 5 + k = 0 k = - 5 Substituting in (1), equation of OA is (x y + 5) 5 ( x + y + 1 ) = 0 x y + 5 5y 5 = 0-3x 6y = 0 x + y = 0 9. Find the equation of the straight line parallel to the lines 3x + 4y = 7 and passing through the point of intersection of the lines x y 3 = 0 and x + 3y 6 = 0 Sol. Given lines are L1 = x y 3= 0 and L = x_3y 6= 0 Equation of any line passing through the point of intersection of the lines L 1 =0 and L =0 is L1+ KL = 0 (x y 3 ) + k( x + 3y 6 ) = 0 (1 + k)x + (- + 3k)y + (-3-6k) = 0----(1) This line is parallel to 3x + 4y = 7 a1 b1 3 4 = = a b (1+ k) ( + 3k) 3( + 3k) = (1 + k)4 6 + 9k = 4 + 4k 5k = 10 k = Equation of the required line is 3x + 4y 15 = 0 10. Find the equation of the straight line perpendicular to the line x + 3y = 0 and passing through the point of intersection of the lines x + 3y 1 = 0 and x y + 4 = 0 Sol. L 1 =x + 3y 1 = 0 L =x y + 4 = 0 Equation of any line passing through the point of intersection of the lines L 1 =0 and L =0 is L1+ KL = 0 (x + 3y 1 ) + k ( x y + 4 ) = 0 (1 + k)x + (3 k )y + (4k 1) = 0---(1) This line is perpendicular to x + 3y = 0, a1a + b1b = 0 (1+ k) + 3(3 k) = 0 11 + k+ 9 6k = 0 4k = 11 k = 4 Substituting in (1), equation of the required line is

11 11 1+ x+ 3 y + (11 1) = 0 4 15 5 x y+ 10= 0 4 15x 10y = 40 = 0 3x y + 8 = 0 11. Find the equation of the straight line making non zero equal intercepts on the axes and passing through the point of intersection of the lines x 5y + 1 = 0 and x 3y - 4 = 0 Sol. Let L 1 = x+ 5y+ 1= 0,L = x 3y 4= 0 Equation of any line passing through the point of intersection of the lines L 1 =0 and L =0 is L1+ KL = 0 (x 5y + 1) + k(x 3y 4 ) = 0 ( + k)x (5 + 3k)y + (1 4k) = 0 (1) Intercepts on co-ordinates axes are equal, coefficient of x = coefficient of y + k = -5 3k 4k = - 7 k = - 7/4 Substituting in (1) Equation of the required line is 7 x 5 1 y + (1+ 7) = 0 4 4 1 x+ 1 y+ 8= 0 x + y + 3 = 0 4 4 1. Find the length of the perpendicular drawn from the point of intersection of the lines 3x + y + 4 = 0 and x+5y-1= to the straight line 7x + 4y 15 = 0 Sol. Given lines are 3x + y + 4 = 0 -----(1) x + 5y 1 = 0----() Solving (1) and (), point of intersection is P (-, 1). Length of the perpendicular from P (-, 1) to the line 7x + 4y 15 = 0 is 14 + 4 15 5 1 = = =. 49 + 576 5 5 13. Find the value of a if the distance of the points (, 3) and (-4, a) from the straight line 3x + 4y 8 = 0 are equal. Sol. Equation of the line is 3x + 4y 8 = 0 ---(1)

Given pointsp (, 3), (-4, a) Perpendicular from P(,3) to (1) = perpendicular from Q(-4,a) to (1) 3. + 4.3 8 3.( 4) + 4a 8 = 9+ 16 9+ 16 10 = 4a 0 4a 0 =± 10 4a = 0 ± 10 = 30or10 30 10 a = or 4 4 a = 15 or5 / 14. Fund the circumcentre of the triangle formed by the straight lines x + y = 0, x + y + 5 = 0 and x y = Sol. let the equation of AB be x + y = 0 ---(1) BC be x + y + 5 = 0 ---() And AC be x y = ---(3) A x + y = 0 x y = B x + y + 5 = 0 Solving (1) and (), vertex B = (-5, 5) Solving () and (3),vertex C= (-1, -3) Solving (1) and (3), vertex A = (1, -1) Let S (x, y) be the circumcentre of ABC. Then SA = SB = SC SA = SB SA = SB (x + 5) + (y 5) = (x + 1) + (y + 3) x + 10x + 5 + y 10y + 5 = x + x + 1+ y + 6y + 9 8x 16y = - 40 x y = -5 ---(4) SB = SC SB = SC (x + 1) + (y+ 3) = (x 1) + (y+ 1) x + x+ 1+ y + 6y+ 9= x x+ 1+ y + y+ 1 4x + 4y = -8 x + y = - ---(5) C

Solving (4) & (5), point of intersection is (-3, 1) circumcentre is S(-3, 1) 15. If θ is the angle between the lines x + y = 1and x + x = 1, find the value of sin θ, a b b a when a > b. Sol. Given equations are x + y = 1 bx+ ay= ab a b And x + y = 1 ax+ by= ab b a Let θ be angle between the lines, then aa 1 + bb 1 cos θ= a + b a + b 1 1 ab + ab ab = = b + a b + a a + b sin θ= 1 cos θ= 1 4a b ( a + b ) a sin θ= a b + b II. 1. Find the equation of the straight lines passing through the point ( 10, 4) and making an angle θ with the line x y = 10 such that tan θ =. Sol: Given line is x y = 10---- (1) and point ( 10, 4). 1 tan θ= cos θ= 5 Let m be the slope of the require line. This line is passing through (-10, 4), therefore equation of the line is y 4 = m(x + 10) = mx + 10m mx y + (10m + 4) = 0 ------() aa 1 + bb 1 Given θ is the angle between (1) and (), therefore, cos θ= a + b a + b 1 m+ = + + Squaring 5 1 4 m 1 ( ) m + 1= m+ 4m + 3 = 0 m = = m + 4m+ 4 3 4 1 1

Case (i): Co-efficient of m = 0 One of the root is Hence the line is vertical. Equation of the vertical line passing through ( 10, 4) is x + 10 = 0 3 Case (ii): m = 4 Substituting in (1) 3 30 Equation of the line is x y+ + 4 = 0 4 4 3x 4y 14 = 0 3x + 4y + 14 = 0 4. Find the equation of the straight lines passing through the point (1, ) and making an angle of 60º with the line 3x + y = 0. Sol: equation of the given line is 3x + y = 0.-----(1) Let P(1, ). let m be the slope of the required line. Equation of the line passing through P(1, ) and having slope m is y = m(x 1)= mx m mx y + ( m) = 0 ---() This line is making an angle of 60º with (1), therefore, cos θ= aa + bb 1 1 1 + 1 + a b a b 1 3m 1 = m + 1 cos 60º = Squaring on both sides, m + 1= ( 3m 1) m 3m = 0 m( m 3) = 0 3m 1 3+ 1 m + 1 = 3m + 1 3m m= 0 or 3 Case (i): m = 0, P(1, )

Equation of the line is y+ = 0 or y = 0 Case (ii): m= 3, P(1, ) Equation is 3x y + ( 3) = 0 3. The base of an equilateral triangle is x y 0 Find the equation of the remaining sides. y+ 1= 3 x ANS: y 1 ( 3)( x ) + = +, ( )( ). + = and the opposite vertex is (, 1) 4. Find the orthocentre of the triangle whose sides are given below. 5,, 1, and 1, 4 i) (, 1 ),( 6, 1) and (,5) ii) ( ) ( ) ( ) Sol: i) A(, 1 ),B( 6, 1 ),C(,5) are the vertices of ABC. A (-, -1) O E Slope of 5+ 1 6 3 BC = = = 6 4 B(6, -1) D C(, 5) AD is perpendicular to BC Slope of Equation of AD is y+ 1= ( x+ ) 3 x 3y + 1= 0 ---(1) 5+ 1 6 3 Slope of AC = = = + 4 r BE is l to AC y+ 1= x 6 3 x 3y 9 = 0 ---() solving (1), () x y 1 Equation of BE is ( ) 3-9 3-3 1-3 x y 1 = = 3 7 18 6 6 AD = 3

x y 1 = = 4 0 1 4 0 5 x = =, y= = 1 1 3 Co-ordinates of the orthocenter 5 Oare=, 3 ii) A5, (,B ) ( 1,,C1,4 ) ( ) are the vertices of ABC. 1 14 ANS:, 5 5 5. Find the circumcentre of the triangle whose vertices are given below. i) (,3)(, 1) and ( 4,0) ii) ( 1, 3 ), ( 0, ) and ( 3,1) 3 5 Sol: i), Ans ii) ( 1, 3 ), ( 0, ) and ( 3,1) 1 ANS:, 3 3 6. Let PS be the median of the triangle with vertices P(,) Q( 6, 1) and R( 7,3). Find the equation of the straight line passing through (1, 1) and parallel to the median PS. Sol: P,,Q6, ( ) ( 1,R7,3 ) ( ) are the vertices of ABC. Let A(1, 1) Given S is the midpoint of QR 6+ 7 1+ 3 13 Co-ordinates of S are, =,1 1 1 Slope of PS = = = 13 9 9

Required line is parallel to PS and passing through A1, ( 1), 9 9y + 9 = x + x + 9y + 7 = 0 Equation of the line is y+ 1= ( x 1) 7. Find the orthocentre of the triangle formed by the lines. x+ y = 0,4x+ 3y 5= 0 and 3x + y = 0. Sol: Given equations are x + y = 0 ---(1) 4x + 3y 5= 0 ---() 3x + y = 0 ---(3) Solving (1) and (), vertex A = (0, 0) Solving (1) and (3), Vertex B (, 1) Equation of BC is 4x + 3y 5 = 0 AB is perpendicular to BC and passes through A(0, 0) Equation of AB is 3x 4y = 0 ---(4) BE is perpendicular to AC Equation of BE is x 3y = k BE passes through B, ( 1) + 3= k k= 5 Equation of BE is x 3y 5= 0 ---(5) Solving (4) and (5), Orthocentre is O( 4, 3) 8. Find the circumference of the triangle whose sides are given by x+ y+ = 0, 5x y = 0 and x y+ 5= 0. Sol: Given lines are x+ y+ = 0 ---(1) 5x y = 0 ---() x y+ 5= 0 ---(3)

Point of intersection of (1) and () is A= ( 0, ) Point of intersection of () and (3) is B= ( 1,3) Point of intersection of (1) and (3) is C= ( 3,1) Let S = ( α, β ) the orthocentre of ABC then SA = SB = SC SA = SB = SC ( α 0) + ( β+ ) = ( α 1) + ( β 3) = ( α+ 3) + ( β 1) α +β + 4β + 4=α +β α 6β + 10=α +β + 6α β + 10 SA = SB α +β + 4β+ 4 =α +β α β+ 6 10 α+ 10β 6= 0 α+ 5β 3= 0 ---(4) SA = SC α +β + 4β+ 4 =α +β +α β+ 6 10 6α 6β + 6= 0 α β + 1= 0 ---(5) From (4) and (5) α β 1 5-3 1 5-1 1 1-1 α β 1 α β 1 = = = = 5 3 3 1 1 5 4 6 1 α= = 6 3 4 β = = 6 3 1 Circumcentre S =, 3 3 9. Find the equation of the straight lines passing through (1, 1) and which are at a distance of 3 units from (-, 3). Sol: let A(1, 1). Let m be the slope of the line. Equation of the line is y - 1 = m(x 1) mx y + ( 1 m) = 0 ---(1)

Give distance from (-, 3) to (1) = 3 m 3+ 1 m = 3 m + 1 ( 3m + ) = 9( m + 1) 9m + 4 + 1m = 9m + 9 5 1m = 5 m = 1 Co-efficient of m = 0 m= Case i) m = line is a vertical line Equation of the vertical line passing through A(1, 1) is x = 1 5 Case ii) m =, point (1,1) 1 5 y 1= x 1 = 0 1 Equation of the line is ( ) 5x 1y + 7 = 0 10. If p and q are lengths of the perpendiculars from the origin to the straight lines x secα+ ycosecα= a and xcosα ysinα= acosα, prove that Sol: Equation of AB is xsecα+ ycosecα= a x y a cosα + sin α = x sin α+ y cos α= a sin αcos α x sin α+ y cos α asin αcos α= 0 p = length of the perpendicular from O to AB = sin α = asin α.cosα = a. p = a sin α ---(1) Equation of CD is xcosα ysinα= a cosα xcosα ysinα acosα= 0 q = Length of the perpendicular from O to CD Squaring and adding (1) and () 4p + q = a sin α+ a cos α 0+ 0 asinαcosα sin α+ cos α 4p + q = a. 0+ 0 acosα = acos α cos α+ sin α ---()

( ) = a sin α+ cos α = a.1= a 11. Two adjacent sides of a parallelogram are given by 4x + 5y = 0 and 7x + y = 0 and one diagonal is 11x + 7y = 9. Find the equations of the remaining sides and the other diagonal. Sol: Let 4x + 5y = 0 ---(1) and 7x + y = 0 ---() respectively denote the side OA and OB of the parallelogram OABC. Equation of the diagonal AB is 11x + 7y 9 = 0 ---(3) Solving (1) and () vertex O = (0, 0) 5 4 Solving (1) and (3), A =, 3 3 7 Solving () and (3), B =, 3 3 11 Midpoint of AB is P,. Slope of OP is 1 Equation to OC is y = (1) x x y = 0 x = y. 5 4 Equation of AC is 4 x + 3 y+ = 0 4x+ 5y = 9 3 3 7 Equation of BC is 7 x+ + y = 0 7x+ y = 9 3 3 1. Find the in centre of the triangle whose sides are given below. i) x + 1 = 0, 3x 4y = 5 and 5x + 1y = 7 ii) x+ y 7 = 0, x y+ 1= 0 and x 3y+ 5= 0 Sol: i) Sides are

x + 1 = 0 ---(1) 3x 4y 5 = 0 ---() 5x + 1y 7 = 0 ---(3) The point of intersection of (1), () is A= ( 1,, ) The point of intersection of (), (3), B= ( 3,1) The point of intersection of (3), (1) is a BC ( 3 1) = = + + 1+ 3 8 b CA ( 1 1) = = + + 3 ( ) ( ) 8 c= AB= 1 3 + 1 = 16+ 9= 5 Incentre = I = 8 C = 1, 3 5 169 13 = 16 + = = 9 9 3 14 14 14 = 0 + = = 3 3 3 13 14 13 14 1 3 5 1 1 5 8 + + + + 3 3 3 5 5 + + + + 3 3 3 3 ax1+ bx + cx3 ay1+ by + cy, 3 ( ) ( ) ( ) ( ) () a+ b+ c a+ b+ c = 3 3, 13 14 13 14 14 8 1 =, =, 4 4 3 3 1 Incentre =, 3 3 ii)ans:( 3,1+ 5 ) e 13. A l is formed by the lines ax + by + c = 0, l x+ my+ n = 0 and px + qy + r = 0. Given ax + by + c lx + my + n that the straight line = passes through the orthocentre of the ap + bq lp + mq e l.

Sol: (1) () (3) Sides of the triangle are ax + by + c = 0 ---(1) l x+ my+ n = 0 ---() px + qy + r = 0 ---(3) Equation of the line passing through intersecting points of (1), () is ax + by + c + k l x + my + n = 0 ---(4) ( ) ( l ) ( ) ( ) a + k x + b + km y + c + nk = 0 r If (4) is the altitude of the triangle then it is l to (3), ap + bq pa ( + kl) + qb ( + km) = 0 k = lp+ mq From (4) ap + bq ( ax + by + c) ( x + my + n) = 0 lp+ mq l ax + by + c lx + my + n = ap + bq lp + mq is the required straight line equation which is passing through orthocenter. (it is altitude) 14. The Cartesian equations of the sides BC, CA, AB of a are respectively u1 = a1x+ b1y+ c1 = 0, u = ax+ by+ c = 0. and u3 = a3x+ b3y+ c3 = 0. Show that the equation of the straight line through A Parallel to the side BC is u3 u =. ab ab a b ab 3 1 1 3 1 1 e l

Sol: A is the point of intersecting of the lines u = 0 and u3 = 0 Equation to a line passing through A is u u 0 a x b y c λ ax+ by+ c ---(1) +λ = ( + + ) + ( 3 3 3) ( a a ) x ( b b ) y ( c c ) 0 3 +λ + +λ + +λ = 3 3 3 If this is parallel to ax 1 + by 1 + c1 = 0, ( a +λ a3 ) ( b +λb3 = ) a1 b1 a +λ a b = b +λ b a ( ) ( ) 3 1 1 3 1 ab1+λ ab 3 1= ab 1 +λ ab 1 3 λ( ab 3 1 ab 1 3) = ( ab 1 ab 1 ) ( ab 1 ab 1 ) λ= ab 3 1 ab 1 3 Substituting this value of λ in (1), the required equation is ( ab1 a1b) ( ax+ by+ c) ( ax 3 + by 3 + c3) = 0 ( ab 3 1 ab 1 3) ab ab ax by c ab ab ax+ by+ c = 0 ( 31 13)( + + ) ( 1 1) ( 3 3 3) ( ab 3 1 ab 1 3) u ( ab 1 ab 1 ) u3= 0 ( ab ab) u = ( ab ab) u 3 1 1 3 1 1 3 u = 3 ( ab ab ) ( a b ab ) 3 1 1 3 1 1 u. PROBLEMS FOR PRACTICE 1. Find the equation of the straight line passing through the point (, 3) and making non-zero intercepts on the axes of co-ordinates whose sum is zero. at, at and. Find the equation of the straight line passing through the points ( 1 1) ( at, at ).

3. Find the equation of the straight line passing through the point A( 1,3) and i) parallel ii) perpendicular to the straight line passing through B, ( 5) and C4,6 ( ). 4. Prove that the points ( 1,11 ), (,15 ) and ( 3, 5) the line containing them. 5. A straight line passing through A1, ( ) makes an angle Sol: are collinear and find the equation of tan 1 4 with the positive 3 direction of the X-axis in the anti clock-wise access. Find the points on the straight line whose distance from A is ± 5units. -5 5 Given C 1 4 4 α= tan tan α= 3 3 A( 1, ) B 5 4 3 4 cos α=, sin α= 5 5 ( x,y 1 1) = ( 1, ) = x1= 1,y1= Case i): r = 5 4 x = x1 + rcosα = 1+ 5. = 1+ 4= 5 3 3 y = y1 + rsinα = + 5. = + 3= 1 5 Co-ordinate of B are (5, 1) α 3 Case ii): 4 x = x1 + rcosα = 1 5. = 1 4= 3 5 3 y= y1 + rsinα = 5. = 3= 5 4 Co-ordinate of C are ( 3, 5)

6. A straight line parallel to the line y= 3x passes through Q,3 ( ) and cuts the line x + 4y 7 = 0 at P. Find the length of PQ. Sol: PQ is parallel to the straight line y= tan α= 3 = tan 60º α= 60º Q,3 ( ) is a given point 3x Q(, 3) P x + 4y - 7 =0 y = Co-ordinates of any point P are x rcos y rsin + rcos60º, 3+ rsin60º ( + α + α ) = ( ) 1 1 r 3 = P +, 3+ r P is a point on the line x + 4y 7 = 0 r 3 + + 4 3+ r 7= 0 4+ r+ 1+ 3r 7= 0 r ( 3 + 1) = 7 16 = 11 ( ) 11 3 1 11 3 1 r =. = 3+ 1 3 1 11 7. Transform the equation 3x + 4y + 1 = 0 into i) slope intercept form ii) intercept form and iii) normal form 8. If the area of the triangle formed by the straight line x = 0, y = 0 and 3x + 4y = a ( a > 0), find the value of a. 9. Find the value of k, if the lines x 3y + k = 0, 3x 4y 13 = 0 and 8x 11y 33 = 0 are concurrent. 10. If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent, 3 3 3 then prove that a + b + c = 3abc. Sol: The equations of the given lines are ax + by + c = 0 ---(1) 3x

bx + cy + a = 0 ---() cx + ay + b = 0 ---(3) Solving (1) and () points of intersection is got by x y 1 b c a b c a b c x = y = 1 ab c bc a ca b ab c bc a Point of intersection is, ca b ca b ab c bc a c + a + b= 0 ca b ca b cab c + abc a + bca b = 0 ( ) ( ) ( ) 3 3 3 abc c + abc a + abc b = 0 3 3 3 a + b + c = 3abc. 11. A variable straight line drawn through the point of intersection of the straight lines x y x y + = 1and + = 1 meets the co-ordinate axes at A and B. Show that the locus the a b b a mid point of AB is ( a+ b) xy= ab( x+ y). Sol: Equations of the given lines are x + y = 1 a b and x + y = 1 b a Solving the point of intersection Q( x 0,y 0) is any point on the locus ab ab P, a+ b a+ b The line with x-intercept x 0, y-intercept y 0, passes through P x y P lies on the straight line + = 1 x y ab 1 1 i.e., + = 1 a+ b x0 y0 ab x0 + y. 0 = 0 a+ b x y 0 0 0 0

ab( x0+ y0) = ( a + b) x0y0 Q( x 0,y 0) lies on the curve ( a+ b) xy= ab( x+ y) Locks the midpoint of AB is ( a+ b) xy= ab( x+ y). 1. If a, b, c are in arithmetic progression, then show that the equation ax + by + c = 0 represents a family of concurrent lines and find the point of concurrency. 13. Find the value of k, if the angle between the straight lines 4x y + 7 and kx 5y + 9 = 0 is 45º. 14. Find the equation of the straight line passing through ( x,y 0 0) and i) parallel ii) perpendicular to the straight line ax + by + c = 0. 15. Find the equation of the straight line perpendicular to the line 5x y = 7 and passing through the point of intersection of the lines x + 3y = 1 and 3x + 4y = 6. 16. If x 3y 5 = 0 is the perpendicular bisectors of the line segment joining (3-4) and ( α, β ) find α+β. 17. 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, show that the area of the parallelogram bc formed is. ( p q)( r s) bc ad 18. Find the orthocentre of the triangle whose vertices are ( 5, 7)( 13, ) and ( 5,6). 19. If the equations of the sides of a triangle are 7x + y 10 = 0, x y+ 5= 0 and x+ y+ = 0, find the orthocentre of the triangle. 0. Find the circumcentre of the triangle whose vertices are ( 1, 3 ), ( 3, 5) and ( 5, 1). 1. Find the circumcentre of t\he triangle whose sides are 3x y 5 = 0, x+ y 4= 0 and 5x + 3y + 1 = 0.

Sol: Let the given equations 3x y 5 = 0, x+ y 4= 0 and 5x + 3y + 1 = 0 represents the sides BC, CA and AB respectively of ABC. Solving the above equations two by two, we obtain the vertices A(,3 ),B( 1, ) and (,1) of the given triangle. The midpoints of the sides BCand CA are respectively 3 1 D =, =. and E ( 0,). Let O be any point in the plane of ABC such that O does not lie on any side of the triangle. If the line joining O to the vertices A, B, C meet the opposite sides in D, E, F respectively, then prove that BD CE AF = 1 (Ceva s Theorem) DC EA FB Sol: Without loss of generality take the point P as origin O. Let A( x 1,y1) B( x,y) C( x 3,y 3) y be the vertices. Slope of AP is 1 0 y = 1 x 0 x 1 1 A E F y x Equation of AP is y 0= 1 ( x 0) yx1 = xy1 xy1 yx1 = 0 BD ( xy1 x1y) x1y xy = = 1 DC xy 3 1 xy 1 3 xy 3 1 xy 1 3 Slope of BP y is 0 y = x 0 x Equation of BP y is y 0= ( x 0 ) x xy= yx xy xy= 0 CE ( xy xy) x y x y = = EA xy x y xy x y y Slope of 3 0 y CP = = 3 x3 0 x3 y Equation of CP is y 0 = 3 ( x 0) x3 x3y = xy3 xy3 x3y = 0 3 3 3 3 1 1 1 1 1 B D C

( ) AF xy xy xy xy = = FB x y x y x y x y 1 3 3 1 3 1 1 3 3 3 3 3 BD CE AF.. DC EA FB xy 1 xy 1 xy3 x3y x3y1 x1y.. 3 = 1 xy xy xy x y x y xy 3 1 1 3 1 1 3 3 3. If a transversal cuts the side BC,CA and AB of ABC in D, E and F respectively. Sol: Then prove that BD CE AF = 1. (Meneclau s Theorem) DC EA FB A F E Let A( x 1,y 1), B( x,y ),C( x 3,y 3) Let the transversal be ax + by + c = 0 BD = The ratio in which ax by c 0 DC + + = divides. ( ax + by + c) BC = ax3+ by3+ c CE = The ratio in which ax by c 0 EA + + = divides. ( ax3+ by3+ c) CA = ax1+ by1+ c AF = The ratio in which ax by c 0 FB + + = divides. ( ax1+ by1+ c) AB = ax + by + c BD CE AF.. = 1 DC EA FB B C D

4. Find the incentre of the triangle formed by straight lines y= 3x, y= 3x and y = 3. Sol: The straight lines y= 3x and y= 3x respectively make angles 60º and 10º with the positive directions of X-axis. Since y = 3 is a horizontal line, the triangle formed by the three given lines is equilateral. So in-centre is same and centriod. D 3,3 Vertices of the triangle and ( ) o+ 3 3 0+ 3+ 3 Incentre is, 3 3 = (0,). 0,0, A( 3,3 ) and ( )