POINT & STRAIGHT LINE

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1 J-Mathematics POINT & STRAIGHT LIN NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 BASIC THORMS & RSULTS OF TRIANGLS : ( a ) Two polgons are similar if (i) their corresponding angles are equal, (ii) the length of their corresponding sides are proportional. (Both condition are independent & necessar) In case of a triangle, an one of the condition is sufficient, other satisfies automaticall. ( b ) Thales Theorem (Basic Proportionalit Theorem) : In a triangle, a line drawn parallel to one side, to intersect the other sides in distinct points, divides the two sides in the same ratio. Converse : If a line divides an two sides of a triangle in the same ratio then the line must be parallel to the third side. ( c ) Similarit Theorem : (i) AAA similarit : If in two triangles, corresponding angles are equal i.e. two triangles are equiangular, then the triangles are similar. (ii) SSS similarit : If the corresponding sides of two triangles are proportional, then the are similar. (iii) SAS similarit : If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar. (iv) If two triangles are similar then () The are equiangular. () The ratio of the corresponding (I) Sides (all), (II) Perimeters, (III) Medians, (IV) Angle bisector segments, (V) Altitudes are same (converse also true) () The ratio of the areas is equal to the ratio of the squares of corresponding (I) Sides (all), (II) Perimeters, (III) Medians, (IV) Angle bisector segments, (V) Altitudes (converse also true) ( d ) Congruenc theorem : Congruent triangles : Two triangles are conguent, iff one of them can be made to superpose on the other, so as to cover it eactl. Sufficient- conditions (criteria) for congruence of triangles : (i) Side-Angle-Side (SAS ) : Two triangles are congruent, if two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of other triangle. (ii) Angle-Side-Angle (ASA) : Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle. (iii) Angle-Angle-Side(AAS) : If an two angles and a non-included side of one triangle are equal to the corresponding to angles & the non-included side of the other triangle then the two triangles are congurent. (iv) Side-Side-Side (SSS) : Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle. (v) Right angle-hpotenuse-side(rhs) : Two right-triangles are congruent, if the hpotenuse and one side of one triangle are respectivel equal to the hpotenuse and one side of other triangle. ( d ) Pthagoras theorem : (i) In a right triangle the square of hpotenuse is equal to the sum of squares of the other two sides. Converse : In a triangle if square of one side is equal to sum of the squares of the other two side. then the angle opposite to the side is a right angle. (e) (ii) In obtuse : (iii) In Acute : A D B A C AC = AB + BC + BC. BD & AC > AB + BC AC = AB + BC BC. BD & AC < AB + BC B D C In an triangle ABC, AB + AC = (AD + DC ), where D is the mid point of BC

2 J-Mathematics ( f ) The internal/eternal bisector of an angle of a triangle divides the opposite side internall/eternall in the ratio of sides containing the angle (converse is also true). BP BQ AB PC CQ AC c A b c b B P C ( g ) The line joining the mid points of two sides of a is parallel & half of the third side. (It's converse also true) ( h ) (i) The diagonals of a trapezium divide each other proportionall. (converse is also true) (ii) An line parallel to the parallel sides of a trapezium divides the non parallel sides proportionall. (iii) If three or more parallel lines are intersected b two transversals, then intercepts made b them on the transversals are proportional. (i) In an triangle three times the sum of squares of the sides of a triangle is equal to four times the sum of the squares of its medians. ( j ) The altitudes, medians and angle bisectors of a triangle are concurrent among themselves.. INTRODUCTION OF COORDINAT GOMTRY : Coordinate geometr is the combination of algebra and geometr. A sstematic stud of geometr b the use of algebra was first carried out b celebrated French philosopher and mathematician René Descartes. The resulting combination of analsis and geometr is referred as analtical geometr.. CARTSIAN CO- ORDINATS SYSTM : P(,) In two dimensional coordinate sstem, two lines are used; the lines are at right angles, forming a rectangular coordinate sstem. The horizontal ais is the ' O -ais and the vertical ais is -ais. The point of intersection O is the origin of the coordinate sstem. Distances along the -ais to the right of the origin are taken as positive, distances to the left as negative. Distances along the -ais ' above the origin are positive; distances below are negative. The position of a point anwhere in the plane can be specified b two numbers, the coordinates of the point, written as (, ). The -coordinate (or abscissa) is the distance of the point from the -ais in a direction parallel to the -ais (i.e. horizontall). The -coordinate (or ordinate) is the distance from the -ais in a direction parallel to the -ais (verticall). The origin O is the point (0, 0).. POLAR CO-ORDINATS SYSTM : A coordinate sstem in which the position of a point is determined b the length of a line segment from a fied origin together with the angle that the line segment makes with a fied line. The origin is called the pole and the line segment is the radius vector (r). The angle between the polar ais and the radius vector is called the vectorial angle. B convention, positive values of are measured in an anticlockwise sense, negative values in clockwise sense. The coordinates of the point are then specified as (r, ). If (,) are cartesian co-ordinates of a point P, then : =r cos, = r sin and r, tan 4. DISTANC FORMULA AND ITS APPLICATIONS : If A(, ) and B(, ) are two points, then AB ( ) ( ) O r Q (,) P (r, ) Note : (i) Three given points A,B and C are collinear, when sum of an two distances out of AB,BC, CA is equal to the remaining third otherwise the points will be the vertices of a triangle. (ii) Let A,B,C & D be the four given points in a plane. Then the quadrilateral will be : (a) Square if AB = BC = CD = DA & AC = BD ; AC BD (b) Rhombus if AB = BC = CD = DA and AC BD ; AC BD (c) Parallelogram if AB = DC, BC = AD; AC BD ; AC BD (d) Rectangle if AB = CD, BC = DA, AC = BD ; AC BD NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

3 J-Mathematics Illustration : The number of points on -ais which are at a distance c(c < ) from the point (, ) is (A) (B) (C) infinite (D) no point Solution : Let a point on -ais is (, 0) then its distance from the point (, ) 9 c or c 9 since c 9 will be imaginar. c c 9 0 Illustration : The distance between the point P a cos, a sin and Q a cos, a sin is - Ans. (D) (A) 4a sin (B) a sin (C) a sin Solution : d a cos a cos a sin a sin a cos cos a sin sin (D) acos a sin sin a cos sin 4a sin sin cos 4a sin d a sin Ans. (C) Do ourself - : (i) Find the distance between the points P(, ) and Q(, ). (ii) If the distance between the points P(, 5) and Q(, ) is 58, then find the value(s) of. (iii) A line segment is of the length 5 units and one end is at the point (, ), if the abscissa of the other end is 5, then find possible ordinates. 5. SCTION FORMULA : The co-ordinates of a point dividing a line joining the points P(, ) and Q(, ) in the ratio m:n is given b : ( a ) For internal division : P - R - Q R divides line segment PQ, internall. NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 m n m n (, ), m n m n m R(,) P(, ) Q(, ) ( b ) For eternal division : R - P - Q or P - Q - R R divides line segment PQ, eternall. m n m n (, ), m n m n ( P R ) ( Q R ) < R lies on the left of P & P(, ) Q(, ) m ( P R ) > R lies on the right of Q ( Q R ) n n R(,) ( c ) Harmonic conjugate : If P divides AB internall in the ratio m : n & Q divides AB eternall in the ratio m : n then P & Q are said to be harmonic conjugate of each other w.r.t. AB. Mathematicall ; i.e. AP, AB & AQ are in H.P. AB AP AQ

4 J-Mathematics Illustration : Determine the ratio in which + = 0 divides the line joining (, ) and (8, 9). Solution : Suppose the line + = 0 divides the line segment joining A(, ) and B(8, 9) in the ratio : at a point P, then the co-ordinates of the point P are 8 9, But P lies on + = 0 therefore = = 0 or = Do ourself - : So, the required ratio is :, i.e., : (internall) since here is positive. (i) Find the co-ordinates of the point dividing the join of A(, ) and B(4, 7) : (ii) (a) Internall in the ratio : (b) ternall in the ratio of : In what ratio is the line joining A(8, 9) and B( 7, 4) is divided b (a) the point (, 7) (b) the -ais (c) the -ais. 6. CO-ORDINATS OF SOM PARTICUL AR POINTS : Let A,, B, and ( a ) Centroid : C, are vertices of an triangle ABC, then The centroid is the point of intersection of the medians (line joining the mid point of sides and opposite vertices). Centroid divides each median in the ratio of :. Co-ordinates of centroid ( b ) Incenter : G, The incenter is the point of intersection of internal bisectors of the angles of a triangle. Also it is a centre of the circle touching all the sides of a triangle. a b c a b c Co-ordinates of incenter I, a b c a b c where a, b, c are the sides of triangle ABC. Note : 4 (i) Angle bisector divides the opposite sides in the ratio of remaining sides. e.g. (ii) Incenter divides the angle bisectors in the ratio b c : a, c a : b, a b : c. ( c ) Circumcenter : It is the point of intersection of perpendicular bisectors of the sides of a triangle. If O is the circumcenter of an triangle ABC, then OA OB OC. Also it is a centre of a circle touching all the vertices of a triangle. F A(, ) G B(, ) D C(, ) F A(, ) I B(, ) D C(, ) D BD AB c DC AC b A (, ) O (, ) (, ) B C F NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

5 J-Mathematics Note : (i) If the triangle is right angled, then its circumcenter is the mid point of hpotenuse. (ii) sin A sin B sin C sin A sin B sin C Co-ordinates of circumcenter, sin A sin B sin C sin A sin B sin C ( d ) Orthocenter : A (, ) It is the point of intersection of perpendiculars drawn from vertices on the opposite sides of a triangle and it can be obtained b solving the equation of an two altitudes. (, ) (, ) B C Note : (i) If a triangle is right angled, then orthocenter is the point where right angle is formed. tan A tan B tan C tan A tan B tan C (ii) Co-ordinates of circumcenter, tan A tan B tan C tan A tan B tan C Remarks : (i) If the triangle is equilateral, then centroid, incentre, orthocenter, circumcenter, coincide. (ii) (iii) (e) -centers : Orthocentre, centroid and circumcentre are alwas collinear and centroid divides the line joining orthocentre and circumcentre in the ratio : In an isosceles triangle centroid, orthocentre, incentre & circumcentre lie on the same line. The centre of a circle which touches side BC and the etended portions of sides AB and AC is called the e-centre of ABC with respect to the verte A. It is denoted b I and its coordinates are a b c a b c I, a b c a b c Similarl e-centers of ABC with respect to vertices B and C are denoted b I and I respectivel, and D I B O A I I C a b c a b c I, a b c a b c I, a b c a b c, a b c a b c NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Illustration 4 : Solution : 5 If, is the centroid of a triangle and its two vertices are (0, ) and (, ), then find its third verte, circumcentre, circumradius & orthocentre. Let the third verte of triangle be (, ), then 5 0 and 5. So third verte is (, 5). Now three vertices are A(0, ), B(, ) and C(, 5) Let circumcentre be P(h, k), then AP = BP = CP = R (circumradius) AP = BP = CP = R h + (k ) = (h ) + (k ) = (h ) + (k 5) = R... (i) from the first two equations, we have h + k =... (ii) from the first and third equation, we obtain 6h + 6k =... (iii) On solving, (ii) & (iii), we get h = 9, k = 5 5

6 J-Mathematics Substituting these values in (i), we have R = 5 0 O(, ) 5 G C HG, K J 9 5, HG K J Illustration 5 : Let O(, ) be the orthocentre, then 9 5 4, = 6. Hence orthocentre of the triangle is (4, 6). 5 The vertices of a triangle are A(0, 6), B( 6, 0) and C(,) respectivel, then coordinates of the ecentre opposite to verte A is : (A), Solution : (B) 4, a BC b CA c AB coordinates of e-centre opposite to verte A will be : (C), a b c 4 4 a b c (D) ( 4, 6) a b c 6 6 a b c Hence coordinates of e-centre is ( 4, 6) Ans. (D) Do ourself - : (i) The coordinates of the vertices of a triangle are (0, ), (, ) and (, 5) : (a) Find centroid of the triangle. (b) Find circumcentre & the circumradius. (c) Find Orthocentre of the triangle. 7. ARA OF TRIANGL : Let A(, ), B(, ) and C(, ) are vertices of a triangle, then Area of ABC = = [ ( ) + ( ) + ( )] To remember the above formula, take the help of the following method : = = [( ) + ( )+( )] Remarks : (i) If the area of triangle joining three points is zero, then the points are collinear. (ii) Area of quilateral triangle : If altitude of an equilateral triangle is P, then its area = a side of equilateral triangle, then its area = 4-6 P. If 'a' be the NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

7 J-Mathematics (iii) Area of quadrilateral with given vertices A(, ), B(, ), C(, ), D( 4, 4 ) Area of quad. ABCD = 4 4 Note : Area of a polgon can be obtained b dividing the polgon into disjoined triangles and then adding their areas. Illustration 6 : If the vertices of a triangle are (, ), (4, 6) and (, 5) then its area is (A) 5 sq. units (B) sq. units (C) 5 sq. units (D) 5 sq. units square units Ans. (A) Solution : Illustration 7 : The point A divides the join of the points ( 5, ) and (, 5) in the ratio k: and coordinates of points B and C are (, 5) and (7, ) respectivel. If the area of ABC be units, then k equals - (A) 7, 9 (B) 6, 7 (C) 7, 9 (D) 9, 9 Solution : k 5 5k A, k k NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Illustration 8 : Area of ABC units 4k 66 4 k 7 k 5 5k 5k k k k k 7 or Ans. (C) 9 Prove that the co-ordinates of the vertices of an equilateral triangle can not all be rational. Solution : Let A( ), B(, ) and C(, ) be the vertices of a triangle ABC. If possible let,,,,, be all rational. Now area of ABC = ( ) + ( ) + ( ) = Rational... (i) Since ABC is equilateral Area of ABC = From (i) and (ii), Rational = Irrational which is contradiction. Hence,,,,, cannot all be rational. 8. CONDITIONS FOR COLLINARITY OF THR GIVN POINTS : (side) (AB) {( ) ( ) } = Irrational... (ii) Three given points A (, ), B (, ), C (, ) are collinear if an one of the following conditions are satisfied. ( a ) Area of triangle ABC is zero i.e. ( b ) Slope of AB = slope of BC = slope of AC. i.e. = 0 ( c ) Find the equation of line passing through given points, if the third point satisfies the given equation of the line, then three points are collinear.

8 J-Mathematics Do ourself - 4 : (i) Find the area of the triangle whose vertices are A(,), B(7, ) and C(, ) (ii) Find the area of the quadrilateral whose vertices are A(,) B(7, ), C(,) and D(7, ) (iii) (iv) Prove that the points A(a, b + c), B(b, c + a) and C(c, a + b) are collinear (B determinant method) Prove that the points (, ), (, ) and (8, ) are collinear. ( v ) Find the value of so that the points (, ), (, ) and (4, 5) are collinear. 9. LOCUS : The locus of a moving point is the path traced out b that point under one or more geometrical conditions. ( a ) quation of Locus : The equation to a locus is the relation which eists between the coordinates of an point on the path, and which holds for no other point ecept those ling on the path. ( b ) Procedure for finding the equation of the locus of a point : (i) If we are finding the equation of the locus of a point P, assign coordinates (h, k) to P. (ii) (iii) (iv) press the given condition as equations in terms of the known quantities to facilitate calculations. We sometimes include some unknown quantities known as parameters. liminate the parameters, so that the eliminant contains onl h, k and known quantities. Replace h b, and k b, in the eliminant. The resulting equation would be the equation of the locus of P. Illustration 9 : The ends of the rod of length moves on two mutuall perpendicular lines, find the locus of the point on the rod which divides it in the ratio m : m (A) m m m m m m (B) m m m m m m (C) m m m m 8 (D) none of these Solution : Let, be the point that divide the rod AB, in the ratio m : m, and OA a, OB b sa a b... (i) m a m m Now a m m m m b m m b m m m putting these values in (i) Locus of, is m m m m m m m m m m m m B b O (0,b) m a (, ) m A(a,0) Ans. (C) Illustration 0 : A(a, 0) and B( a, 0) are two fied points of ABC. If its verte C moves in such a wa that cota + cotb =, where is a constant, then the locus of the point C is - (A) = a (B) = a (C) a = (D) none of these NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

9 J-Mathematics Solution : Given that coordinates of two fied points A and B are (a, 0) and ( a, 0) respectivel. Let variable point C is (h, k). From the adjoining figure DA a h cot A CD k BD a h cot B CD k But cota + cotb =, so we have C(h,k) a h a h k k = a k Hence locus of C is = a B( a,0) O D A(a,0) Ans. (A) Do ourself - 5 : (i) Find the locus of a variable point which is at a distance of units from the -ais. (ii) Find the locus of a point which is equidistant from both the aes. (iii) Find the locus of a point whose co-ordinates are given b = at, = at, where 't' is a parameter. 0. STRAIGHT LIN : Introduction : A relation between and which is satisfied b co-ordinates of ever point ling on a line is called equation of the straight line. Here, remember that ever one degree equation in variable and alwas represents a straight line i.e. a + b + c = 0 ; a & b 0 simultaneousl. ( a ) quation of a line parallel to -ais at a distance 'a' is = a or = a ( b ) quation of -ais is = 0 ( c ) quation of a line parallel to -ais at a distance 'b' is = b or = b ( d ) quation of -ais is = 0 NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Illustration : Prove that ever first degree equation in, represents a straight line. Solution :. SLOP OF LIN : Let a + b + c = 0 be a first degree equation in, where a, b, c are constants. Let P(, ) & Q(, ) be an two points on the curve represented b a + b + c = 0. Then a + b + c = 0 and a + b + c = 0 Let R be an point on the line segment joining P & Q Suppose R divides PQ in the ratio :. Then, the coordinates of R are, We have a b c = = 0 R, lies on the curve represented b a + b + c = 0. Thus ever point on the line segment joining P & Q lies on a + b + c = 0. Hence a + b + c = 0 represents a straight line. If a given line makes an angle (0 < 80, 90 ) with the positive direction of -ais, then slope of this line will be tan and is usuall denoted b the letter m i.e. m = tan. If A(, ) and B(, ) & then slope of line AB = (, ) (, ) 9

10 J-Mathematics Remark : (i) (ii) (iii) If = 90, m does not eist and line is parallel to -ais. If = 0, m = 0 and the line is parallel to -ais. Let m and m be slopes of two given lines (none of them is parallel to -ais) (a) (b) If lines are parallel, m = m and vice-versa. If lines are perpendicular, m m = and vice-versa. STANDARD FORMS OF QUATIONS OF A STRAIGHT LIN : ( a ) Slope Intercept form : Let m be the slope of a line and c its intercept on -ais. Then the equation of this straight line is written as : = m + c If the line passes through origin, its equation is written as = m ( b ) Point Slope form : If m be the slope of a line and it passes through a point (, ), then its equation is written as : = m( ) ( c ) Two point form : quation of a line passing through two points (, ) and (, ) is written as : = ( ) 0 or ( d ) Intercept form : If a and b are the intercepts made b a line on the aes of and, its equation is written as : + = a b B(0,b) (i) (ii) Length of intercept of line between the coordinate aes Area of triangle AOB OA.OB = ab a b ' b O a ' A(a,0) Illustration : The equation of the lines which passes through the point (, 4) and the sum of its intercepts on the aes is 4 is - (A) 4 = 4, = 7 (B) 4 + = 4, + = 7 (C) = 0, + + 7=0 (D) = 0, + 7 = 0 Solution : Let the equation of the line be...(i) a b This passes through (, 4), therefore 4...(ii) a b It is given that a + b = 4 b = 4 a. Putting b = 4 a in (ii), we get 4 a 4 a a a + 4 = 0 ( a 7) (a 6) = 0 a = 7, 6 For a = 7, b = 4 7 = 7 and for a = 6, b = 4 6 = 8 Putting the values of a and b in (i), we get the equations of the lines and 7 7 or + = 7 and 4 + = 4 Ans. (B) 6 8 Illustration : Two points A and B move on the positive direction of -ais and -ais respectivel, such that OA + OB = K. Show that the locus of the foot of the perpendicular from the origin O on the line AB is ( + )( + ) = K. 0 NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

11 J-Mathematics Solution : Let the equation of AB be a b... (i) given, a + b = K... (ii) now, m AB m OM = ah = bk... (iii) (e) from (ii) and (iii), a = kk h k and b = hk h k from (i) (h k) (h k) k.k h.k as it passes through (h, k) h(h k) k(h k) (h + k)(h + k ) = Khk k.k h.k locus of (h, k) is ( + )( + ) = K. (0,b)B Normal form : If p is the length of perpendicular on a line from the origin, O M(h, k) A(a, 0) and the angle which this perpendicular makes with positive -ais, then the equation of this line is written as : cos + sin = p (p is alwas positive) where 0 < O p NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Illustration 4 : Find the equation of the straight line on which the perpendicular from origin makes an angle 0 with positive -ais and which forms a triangle of area Solution : NOA = 0 Let ON = p > 0, OA = a, OB = b In ONA, cos0 = ON p OA a or a = p p a and in ONB, cos60 = ON p p OB b b or b = p Area of OAB = ab = p (p) = p 50 ' sq. units with the co-ordinates aes. B b 60 p N 0 O a A ' p = 50 p = 5 or p = 5 Using cos + sin = p, the equation of the line AB is cos0 + sin0 = 5 or + = 0 ( f ) Parametric form : To find the equation of a straight line which passes through P L a given point A(h, k) and makes a given angle with the positive direction of the A r (, ) -ais. P(, ) is an point on the line LAL'. (h, k) L' Let AP = r, then h = r cos, k = r sin & h k = r is the cos sin O equation of the straight line LAL'. An point P on the line will be of the form (h + r cos, k + r sin), where r gives the distance of the point P from the fied point (h, k).

12 J-Mathematics Illustration 5 : quation of a line which passes through point A(, ) and makes an angle of 45 with ais. If this line meet the line + + = 0 at point P then distance AP is - (A) (B) (C) 5 (D) 5 Solution : Here =, = and = 45 hence from first two parts = + = 0 Co-ordinate of point P on this line is If this point is on line + + = 0 then r r cos 45 = sin 45 = r r r,. + = 0 r = ; r = Ans. (B) Illustration 6 : A variable line is drawn through O, to cut two fied straight lines L and L in A and A, respectivel. A point A is taken on the variable line such that m n m n. OA OA OA Show that the locus of A is a straight line passing through the point of intersection of L and L where O is being the origin. Solution : Let the variable line passing through the origin is ri cos sin Let the equation of the line L is p + q = quation of the line L is p + q = the variable line intersects the line (ii) at A and (iii) at A. Let OA = r. Then A = (r cos, r sin) A lies on L r = OA = p cos q sin Similarl, r = OA p cos q sin Let OA = r Let co-ordinate of A are (h, k) Now m n m n r OA OA (h, k) (rcos, rsin) m n m n r r r m + n = m(p rcos + q rsin) + n(p rcos + q rsin) n p h q k 0 m (p h + q k ) + n p q 0 m Therefore, locus of A is (p +q ) + L + L = 0 where = n m. This is the equation of the line passing through the intersection of L and L.... (i)... (ii)... (iii) Illustration 7 : A straight line through P(, ) cuts the pair of straight lines = 0 in Q and R. Find the equation of the line if PQ. PR = 0. Solution : Let line be r cos sin = rcos, = rsin... (i) NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

13 J-Mathematics Now, = 0... (ii) Taking intersection of (i) with (ii) and considering terms of r and constant (as we need PQ. PR = r. r = product of the roots) r (cos + sin + 4sin cos) + (some terms)r + 80 = 0 80 r.r = PQ. PR = cos 4 sin cos sin cos + 4sin cos + sin = 4 ( PQ. PR = 0) sin 4sincos + cos = 0 (sin cos)(sin cos) = 0 tan =, tan = hence equation of the line is + = ( + ) = and + = ( + ) + = 0. Illustration 8 : If the line 0 cuts the parabola = + at A and B, then find the value of PA.PB {where P (, 0)} Solution : Slope of line 0 is If line makes an angle with -ais, then tan = 60 0 r cos 60 sin 60 r r, be a point on the parabola = + O B A 60 P(, 0) then r r r r 4( + ) = 0 4 PA.PB = r r = 4 4 NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Do ourself - 6 : (i) Reduce the line + 5 = 0, (a) In slope- intercept form and hence find slope & Y-intercept (b) In intercept form and hence find intercepts on the aes. (c) In normal form and hence find perpendicular distance from the origin and angle made b the perpendicular with the positive -ais. (ii) Find distance of point A (, ) measured parallel to the line = 5 from the line = 0. ( g ) General form : We know that a first degree equation in and, a +b + c = 0 alwas represents a straight line. This form is known as general form of straight line. (i) a coeff. of Slope of this line b coeff. of (ii) Intercept b this line on -ais = c a and intercept b this line on -ais = c b (iii) To change the general form of a line to normal form, first take c to right hand side and make it positive, then divide the whole equation b a b.

14 J-Mathematics. ANGL BTWN TWO LINS : ( a ) If be the angle between two lines : =m + c and = m + c, then Note : (i) 4 m m tan m m There are two angles formed between two lines but usuall the acute angle is taken as the angle between the lines. So we shall find from the above formula onl b taking positive value of tan. (ii) Let m, m, m are the slopes of three lines L = 0 ; L = 0 ; L = 0 where m > m > m then the interior angles of the ABC found b these formulas are given b, m m tana = m m m m ; tanb = m m m m & tanc = m m ( b ) If equation of lines are a +b +c =0 and a +b +c =0, then these line are - (i) Parallel a b c a b c (ii) Perpendicular a a +b b =0 (iii) Coincident a b c a b c Illustration 9 : If = 0 and 4 + k + 7 = 0 are two perpendicular lines then k is - (A) (B) 4 (C) (D) 4 4 Solution : m = m 4 = k Two lines are perpendicular if m m = 4 4 k k = Ans. (C) Illustration 0 : A line L passes through the points (, ) and (0, ) and another line M which is perpendicular to L passes through the point (0, /).The area of the triangle formed b these lines with -ais is - (A) 5/8 (B) 5/6 (C) 5/4 (D) 5/ Solution : quation of the line L is = quation of the line M is = /. ( ) = + If these lines meet -ais at P (0, /) and Q (0, ) then PQ = 5/. Also -coordinate of their point of intersection R = 5/4 P (0, /) (0, ) Q R (5/4, /4) area of the PQR = /6. 4 Ans. (B) Illustration : If the straight line k ( + + ) = 0 is parallel to -ais, then the value of k is - (A) (B) (C) (D) 4 Solution : A straight line is parallel to -ais, if its - coefficient is zero, i.e. 4 k = 0 i.e. k = 4 Ans. (D) 4. QUATION OF LINS PAR ALLL AND PRPNDICULAR TO A GIVN LIN : ( a ) quation of line parallel to line a + b + c = 0 a + b + = 0 ( b ) quation of line perpendicular to line a + b + c = 0 b a + k = 0 Here, k, are parameters and their values are obtained with the help of additional information given in the problem. NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

15 5. STRAIGHT LIN MAKING A GIVN ANGL WITH A LIN : J-Mathematics quations of lines passing through a point (, ) and making an angle, with the line =m+c is written as : m tan ( ) m tan Illustration : Find the equation to the sides of an isosceles right-angled triangle, the equation of whose hpotenuse is + 4 = 4 and the opposite verte is the point (, ). Solution : The problem can be restated as : Find the equations of the straight lines passing through the given point (, ) and making equal angles of 45 with the given straight line = 0. Slope of the line = 0 is m = /4. Do ourself - 7 : m m tan45 = ± m m, i.e., = ± m / 4 m 4 (, ) 45 m A = 7, and m B = = 4 Hence the required equations of the two lines are = m A ( ) and = m B ( ) 7 = 0 and 7 + = 6 A ns. (i) Find the angle between the lines + 7 = 0 and + 9 = 0. (ii) Find the line passing through the point (, ) and perpendicular to the straight line 4 = 0. ( i i i) Find the equation of the line which has positive -intercept 4 units and is parallel to the line 7 = 0. Also find the point where it cuts the -ais. ( i v) Classif the following pairs of lines as coincident, parallel or intersecting : (a) + = 0 & = 0 (b) + + = 0 & = 0 (c) + 5 = 0 & + 5 = 0 6. LNGTH OF PRPNDICULAR FROM A POINT ON A LIN : NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Length of perpendicular from a point (, ) on the line a + b + c = 0 is a b c a b In particular, the length of the perpendicular from the origin on the line a + b + c = 0 is Illustration : If the algebraic sum of perpendiculars from n given points on a variable straight line is zero then prove that the variable straight line passes through a fied point. Solution : Let n given points be ( i, i ) where i =,... n and the variable straight line is a + b + c = 0. Given that a b c a b n i i 0 a + b + cn = 0 i i i n P a i i a b c 0 Hence the variable straight line alwas passes through the fied point, n c. n n b i i. A ns. Illustration 4 : Prove that no line can be drawn through the point (4, 5) so that its distance from (, ) will be equal to. 5

16 J-Mathematics Solution : Suppose, if possible. quation of line through (4, 5) with slope of m is + 5 = m( 4) m 4m 5 = 0 m( ) 4m 5 Then m 6m 8 = m On squaring, (6m + 8) = 44(m + ) 4(m + 4) = 44(m + ) (m + 4) = 6(m + ) 7m 4m + 0 = 0... (i) Since the discriminant of (i) is ( 4) = 584 which is negative, there is no real value of m. Hence no such line is possible. 7. DISTANC BTWN TWO PAR ALLL LINS : ( a ) The distance between two parallel lines a + b + c =0 and a+b+c =0 is (Note : The coefficients of & in both equations should be same) c a c p p ( b ) The area of the parallelogram = s in, where p & p are distances between two pairs of opposite sides & is the angle between an two adjacent sides. Note that area of the parallelogram bounded b b the lines = m + c, = m + c and = m + d, = m + d is given b ( c c ) ( d d ) m m Illustration 5 : Three lines + + = 0, + 7 = 0 and 4 = 0 form sides of two squares. Find the equation of remaining sides of these squares. Solution : Distance between the two parallel lines is The equations of sides A and C are of the form + k = 0. Since distance between sides A and B = distance between sides Do ourself - 8 : 6 A + + = 0 B + 7 = 0 4 = 0 k 4 k 4 B and C 5 5 k = 6, Hence the fourth sides of the two squares are (i) + 6 = 0 (ii) 4 = 0. A ns. (i) Find the distances between the following pair of parallel lines : (a) + 4 =, + 4 = (b) = 0, = 0 (ii) Find the points on the -ais such that their perpendicular distance from the line = is 'a', a, b > 0. a b (iii) Show that the area of the parallelogram formed b the lines + a = 0, a = 0, + a = 0 and a = 0 is a 5 square units. C NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

17 J-Mathematics 8. POSITION OF TWO POINTS WITH RSPCT TO A GIVN LIN : Let the given line be a + b + c = 0 and P(, ), Q(, ) be two points. If the epressions a + b + c and a + b + c have the same signs, then both the points P and Q lie on the same side of the line a + b + c = 0. If the quantities a + b + c and a + b + c have opposite signs, then the lie on the opposite sides of the line. Illustration 6 : Let P (sin, cos) (0 ) be a point and let OAB be a triangle with vertices (0, 0),, 0 and 0,. Find if P lies inside the OAB. Solution : quations of lines along OA, OB and AB are = 0, = 0 and + = respectivel. Now P and B will lie on the same side of = 0 if cos > 0. Similarl P and A will lie on the same side of = 0 if sin > 0 and P and O will lie on the same side of + = if sin + cos <. Hence P will lie inside the ABC, if sin > 0, cos > 0 and sin + cos <. Now sin + cos < sin( + 4 ) < i.e. 0 < + 4 < / or 4 9. CONCURRNCY OF LINS : Since sin > 0 and cos > 0, so 0 < < B(0, (/)) P O A( (/), 0) or 5 < <. 4 NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 a b c ( a ) Three lines a + b + c = 0; a + b + c = 0 and a + b + c = 0 are concurrent, if a b c 0 a b c ( b ) To test the concurrenc of three lines, first find out the point of intersection of an two of the three lines. If this point lies on the remaining line (i.e. coordinates of the point satisf the equation of the line) then the three lines are concurrent otherwise not concurrent. Illustration 7 : If the lines a + b + p = 0, cos + sin p = 0 (p 0) and sin cos = 0 are concurrent Solution : and the first two lines include an angle 4, then a + b is equal to - (A) (B) (C) 4 (D) p Since the given lines are concurrent, a b p cos sin p 0 sin cos 0 a cos b sin 0... (i) 7

18 J-Mathematics As a + b + p = 0 and cos sin p 0 include an angle. 4 a cos tan b sin 4 a cos b sin a sin + bcos = ± (bsin + acos) a sin + bcos = ± [from (i)]... (ii) Squaring and adding (i) & (ii), we get a + b =. Ans. (B) Do ourself - 9 : (i) amine the positions of the points (, 4) and (, 6) w.r.t. 4 = 8 (ii) If (, 9), (, ) and (, ) are the vertices of a triangle, then prove that the origin lies inside the triangle. (iii) Find the equation of the line joining the point (, 9) and the point of intersection of lines = 0 (iv) and 4 5 = 0. Find the value of, if the lines 4 = 0, 8 = 0 and + = 0 are concurrent. 0. RFLCTION OF A POINT : Let P(, ) be an point, then its image with respect to ( a ) -ais is Q(, ) ( b ) -ais is R(, ) ( c ) origin is S(, ) ( d ) line = is T(, ) (e) Reflection of a point about an arbitrar line : The image (h,k) of a point P(, ) about the line a + b + c = 0 is given b following formula. h k ( a b c ) a b a b and the foot of perpendicular () from a point (, ) on the line a + b + c = 0 is given b following formula. a b c a b a b. TR ANSFORMATION OF A XS ( a ) Shifting of origin without rotation of aes : Let P (, ) with respect to aes OX and OY. Let O' (, ) is new origin with respect to aes OX and OY and let P (', ') with respect to aes O'X' and O'Y', where OX and O'X' are parallel and OY and O'Y' are parallel. Then = ' +, = ' + or ' =, ' = 8 R(, ) S(, ) Y O O' P Y' Y (, l ) ' ' = T(,) P(,) O Q(, ) a+b+c=0 P(,) (',') X' X Q (h,k) Thus if origin is shifted to point (, ) without rotation of aes, then new equation of curve can be obtained b putting + in place of and + in place of. X NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

19 J-Mathematics ( b ) Rotation of aes without shifting the origin : Let O be the origin. Let P (, ) with respect to aes OX and OY and let P (', ') with respect to aes OX' and OY' where X'OX = YOY' = where is measured in anticlockwise direction. then and = ' cos ' sin = ' sin + ' cos ' = cos + sin ' = sin + cos Y' O Y ' P(,) (',') ' The above relation between (, ) and (', ') can be easil obtained with the help of following table X' X Old New ' cos sin ' sin cos Illustration 8 : Through what angle should the aes be rotated so that the equation 9 be changed to + 5 = 5? Solution : Let angle be then replacing (, ) b ( cos sin, sin + cos) then becomes 9( cos sin) cos sin sin cos 7( sin cos ) = 0 ma (9cos sin cos + 7sin ) + ( 9sin cos cos + 7 sin cos) + (9cos + sin cos + 7cos ) = 0 On comparing with + 5 = 5 (coefficient of = 0) We get 9sin cos cos + 7 sin cos = 0 or sin = cos or tan = = tan(80 60 ) or = 0 = 60 NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Do ourself - 0 : ( i ) The point (4, ) undergoes the following transformations, then the match the correct alternatives : Column-I Column-II (A) Reflection about -ais is (p) (4, ) (B) Reflection about -ais is (q) ( 4, ) (C) Reflection about origin is (r) (D) Reflection about the line = is (s) ( 4, ) () Reflection about the line = 0 is (t) (, 4) 59, 5 5 (ii) On what point must the origin be shifted, if the coordinates of a point (4, 5) become (, 9). (iii) If the aes be turned through an angle tan (in anticlockwise direction), what does the equation 4 = a become?. QUATION OF BISCTORS OF ANGLS BTWN TWO LINS : If equation of two intersecting lines are a +b +c =0 and a + b +c =0, then equation of bisectors of the angles between these lines are written as : a b c a b c a b a b...(i) 9

20 J-Mathematics ( a ) quation of bisector of angle containing origin : If the equation of the lines are written with constant terms c and c positive, then the equation of the bisectors of the angle containing the origin is obtained b taking positive sign in (i) ( b ) quation of bisector of acute/obtuse angles : To find the equation of the bisector of the acute or obtuse angle : (i) let be the angle between one of the two bisectors and one of two given lines. Then if tan < i.e. < 45 i.e. < 90, the angle bisector will be bisector of acute angle. (ii) See whether the constant terms c and c in the two equation are +ve or not. If not then multipl both sides of given equation b to make the constant terms positive. Determine the sign of a a + b b If sign of aa + b b For obtuse angle bisector For acute angle bisector + use+sign in eq. () use sign in eq. () use sign in eq. () use+sign in eq. () i.e. if a a + b b > 0, then the bisector corresponding to + sign gives obtuse angle bisector (iii) a b c a b c a b a b Another wa of identifing an acute and obtuse angle bisector is as follows : Let L = 0 & L = 0 are the given lines & u = 0 and u = 0 are the bisectors between L = 0 & L = 0. Take a point P on an one of the lines L = 0 or L = 0 and drop perpendicular on u = 0 & u = 0 as shown. If, p < q u is the acute angle bisector. p > q u is the obtuse angle bisector. p = q the lines L & L are perpendicular. Note : quation of straight lines passing through P(, ) & equall inclined with the lines a + b + c = 0 & a + b + c = 0 are those which are parallel to the bisectors between these two lines & passing through the point P. Illustration 9 : For the straight lines = 0 and = 0, find the equation of the Solution : (i) (ii) (iii) bisector of the obtuse angle between them. bisector of the acute angle between them. bisector of the angle which contains origin. (iv) bisector of the angle which contains (, ). quations of bisectors of the angles between the given lines are = 0 and = 0 If is the acute angle between the line = 0 and the bisector = 0, then tan = NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

21 Hence (i) bisector of the obtuse angle is = 0 (ii) bisector of the acute angle is = 0 (iii) bisector of the angle which contains origin (iv) L (, ) = = 4 > 0 L (, ) = = 8 > 0 +ve sign will give the required bisector, = 0. Alternative : J-Mathematics Making c and c positive in the given equation, we get = 0 and = 0 Since a a + b b = 0 6 = 56 < 0, so the origin will lie in the acute angle. Hence bisector of the acute angle is given b Similarl bisector of obtuse angle is = 0. Illustration 0 : A ra of light is sent along the line = 0. Upon reaching the line mirror 5 = 0, the ra is reflected from it. Find the equation of the line containing the reflected ra. Solution : Let Q be the point of intersection of the incident ra and the line mirror, then = 0 & 5 = 0 on solving these equations, we get = & = Since P(, ) be a point lies on the incident ra, so we can find the image of the point P on the reflected ra about the line mirror (b propert of reflection). Let P'(h, k) be the image of point P about line mirror, then h k ( 4 5) h = 4 and k =. NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 4 So P', Then equation of reflected ra will be 4 ( ) ( + ) = 9 + = 0 is the required equation of reflected ra.. FA MILY OF LINS : If equation of two lines be P a + b + c = 0 and Q a + b + c =0, then the equation of the lines passing through the point of intersection of these lines is : P + Q = 0 or a + b + c + (a + b + c ) = 0. The value of is obtained with the help of the additional informations given in the problem.

22 J-Mathematics Illustration : Prove that each member of the famil of straight lines (sin + 4cos) + (sin 7cos) + (sin + cos) = 0 ( is a parameter) passes through a fied point. Solution : The given famil of straight lines can be rewritten as ( + + )sin + (4 7 + )cos = 0 or, (4 7 + ) + tan( + + ) = 0 which is of the form L + L = 0 Hence each member of it will pass through a fied point which is the intersection of = 0 and + + = 0 i.e., 9 9. Do ourself - : (i) Find the equations of bisectors of the angle between the lines 4 + = 7 and = 0. Also find which of them is (a) the bisector of the angle containing origin (b) the bisector of the acute angle. (ii) Find the equations of the line which pass through the point of intersection of the lines 4 = and 5 + = 0 and is equall inclined to the ais. (iii) Find the equation of the line through the point of intersection of the lines 4 + = 0 & 5 + = 0 and perpendicular to the line = PAIR OF STRAIGHT LINS : ( a ) Homogeneous equation of second degree : Let us consider the homogeneous equation of nd degree as a + h + b = 0...(i) which represents pair of straight lines passing through the origin. Now, we divide b, we get a + h b 0 m (sa) then a + hm + bm = 0...(ii) h if m & m are the roots of equation (ii), then m m, m b m = a b and also, 4h 4a b b a b These line will be : m (m m ) 4m m m tan m m m m h a b = ± a b (i) Real and different, if h ab > 0 (ii) Real and coincident, if h ab = 0 (iii) Imaginar, if h ab < 0 (iv) The condition that these lines are : () At right angles to each other is a + b = 0. i.e. coefficient of ² + coefficient of ² = 0. () Coincident is h² = ab. () quall inclined to the aes of is h = 0. i.e. coefficient of = 0. NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

23 J-Mathematics Homogeneous equation of nd degree a + h + b whose equations are = 0 alwas represent a pair of straight lines NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 b h h ab = m & = m These straight lines passes through the origin. and m + m = h b ; m m = a b Note : A homogeneous equation of degree n represents n straight lines passing through origin. ( b ) The combined equation of angle bisectors : The combined equation of angle bisectors between the lines represented b homogeneous equation of nd degree is given b Note :, a b, h 0. a b h (i) If a = b, the bisectors are = 0 i.e. = 0, + = 0 (ii) If h = 0, the bisectors are = 0 i.e. = 0, = 0. (iii) The two bisectors are alwas at right angles, since we have coefficient of + coefficient of = 0 ( c ) General quation and Homogeneous quation of Second Degree : (i) (ii) (iii) The general equation of second degree a + h + b + g + f + c = 0 represents a pair of straight lines, if =abc+fgh af bg ch = 0 i.e. If be the angle between the lines, then Obviousl these lines are tan h ab a b () Parallel, if = 0, h = ab or if h = ab and bg = af a h g h b f 0 g f c () Perpendicular, if a + b =0 i.e. coeff. of + coeff. of = 0. Pair of straight lines perpendicular to the lines a + h + b = 0 and through origin are given b b h + a = 0. (iv) The product of the perpendiculars drawn from the point (, ) on the lines a + h + b = 0 is a h b ( a b ) 4 h ( v ) The product of the perpendiculars drawn from the origin to the lines a + h + b + g + ƒ + c = 0 is ( a b ) 4 h Illustration : If = 0 represents a pair of straight lines, then is equal to - (A) 4 (B) (C) (D) Solution : Here a =, b =, c =, f = 8, g = 5/, h = 5 Using condition abc + fgh af bg ch = 0, we have () ( ) + ( 8) (5/) ( 5) (64) (5/4) + (5) = = 0 00 = 00 = Ans. (C) c

24 J-Mathematics Illustration : Show that the two straight lines equation are such that the difference of their slopes is. (tan cos ) tan sin 0 represented b the Solution : The given equation is (tan cos ) tan sin 0... (i) and general equation of second degree a + h + b = 0 Comparing (i) and (ii), we get a = tan + cos h = tan and b = sin Let separate lines of (ii) are = m and = m... (ii) where m = tan and m = tan h tan therefore, m m b sin and m.m tan cos sin m m m m 4m m 4 tan 4(tan cos ) tan 4 tan sin sin = tan sin tan cos sin sin = sin sec tan cos sin cos sin sin sin = A ns. Illustration 4 : If pairs of straight lines p = 0 and q = 0 be such that each pair bisects the angle between the other pair, prove that pq =. Solution : According to the question, the equation of the bisectors of the angle between the lines p = 0 is q = 0... (i)... (ii) 4 The equation of bisectors of the angle between the lines (i) is ( ) p p + p = 0 q Since (ii) and (iii) are identical, comparing (ii) and (iii), we get pq = p p Do ourself - : (i) Prove that the equation = 0 represents two lines passing through the origin. Also find their equations. (ii) If the equation + k = 6 represents a pair of lines, find the value of k. (iii) If the equation c = 0 represents a pair of lines, find their equations. Also find the angle between the two lines. (iv) Find the point of intersection and the angle between the lines given b the equation : = 0. NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

25 J-Mathematics 5. QUATIONS OF LINS JOINING TH POINTS OF INTRSCTION OF A LIN AND A CURV TO TH ORIGIN : ( a ) Let the equation of curve be : a + h + b + g + f + c = 0...(i) P and straight line be Q l + m + n = 0...(ii) Now joint equation of line OP and OQ joining the origin and points of intersection P and Q can be obtained b making the equation (i) homogenous with the help of equation of the line. Thus required equation is given b a +h+b +(g+f) m m n + c 0 n O (an + gln + cl ) + (hn + gmn + fln + clm) + (bn + fmn + cm ) = 0...(iii) All points which satisf (i) and (ii) simultaneousl, will satisf (iii) ( b ) An second degree curve through the four points of intersection of f(, ) = 0 & = 0 is given b f(, ) + = 0 where f(, ) = 0 is also a second degree curve. Illustration 5 : The chord 6 8p of the curve p + = 4 subtends a right angle at origin then find the value of p. Solution : p is the given chord. Homogenizing the equation of the curve, we get, p 4( p)+( p) = 0 (4p + 8p) + (p + ) 4 4 p = 0 NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 Now, angle at origin is 90 coefficient of + coefficient of = 0 4p + 8p + p + = 0 4p + 9p + = p =. 8 8 Do ourself - : (i) Find the angle subtended at the origin b the intercept made on the curve = 0 b the line =. (ii) Find the equation of the lines joining the origin to the points of intersection of the curve = 0 and the line + =. 5

26 J-Mathematics Miscellaneous Illustration : Illustration 6 : ABCD is a variable rectangle having its sides parallel to fied directions (sa m). The vertices B and D lie on = a and = a and A lies on the line = 0. Find the locus of C. Solution : Let A be (, 0), B(a, ) and D be ( a, 4 ). We are given AB and AD have fied directions and hence their slopes are constants. i.e. m & m (sa) C(h,k) a m and 4 a Further, mm =. Since ABCD is a rectangle. a m and 4 a m m. D( a, ) 4 A(,0) B(a, ) 4 The mid point of BD is 0, and mid point of AC is h k,, where C is taken to be (h, k). This gives h = and k = + 4. So C is (, + 4 ). Also, a m and 4 a m gives m(k ) = a + = m(k m(a )) = a + mk m (a ) = a + m (a + h) mk + a h = 0 (m )h mk = (m + )a ( m )h + mk = (m + )a ( m ) + m = (m + )a The locus of C is ( m ) + m = (m +)a. ANSWRS FOR DO YOURSLF : (i) PQ = 4 ; (ii) = 6 or = 0 (iii), 7 : (i) (a) (,) ; ( b ) (7,6) ; (ii) ( a ) : (internall) ; ( b ) 9 : 4 (eternall) ; (c) 8 : 7 (internall) 5 : (i) (a), ; ( b ) 9 5,, 5 0, (c) (4, 6) 4 : (i) 5 square units ; (ii) square units ; ( v ) 5 : (i) = ; (ii) = ± ; (iii) = 4a 6 : (i) ( a ) = 5,, 5 ; ( b ) ( 5 / ) (5 / ) 6, 5 5, ; 5 5 (c),, tan ; (ii) / units 7 : (i) = 5 or 45 ; (ii) + 4 = 8 ; (iii) + = 0, ( 6, 0) (iv) ( a ) Coincident, (b) Parallel, (c) Intersecting a b a b,0 b 8 : (i) ( a ) (b) /0 ; (ii) 9 : (i) opposite sides of the line; (iii) + = ; (iv) = 7 0 : (i) (A) (p), (B) (s), (C) (q), (D) (t), () (r), ; (ii) (7, 4) ; (iv) 4 = a : (i) + = 0 & + = 0; ( a ) + = 0 ; ( b ) + = 0 (ii) + =, = ; (iii) = 0 : (i) = 0 & 4 = 0; (ii) k =, or 7 6 ; (iii) 5 = 0 & + + = 0 ; : (i) 4 tan ; (ii) 5 = 0; 7 9 tan 4 ; (iv),, ; NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65

27 J-Mathematics XRCIS-0 CHCK YOUR GRASP SLCT TH CORRCT ALTRNATIV (ONLY ON CORRCT ANSWR). If (, 4) and ( 6, 5) are the etremities of a diagonal of a parallelogram and (, ) is its third verte, then its fourth verte is - (A) (, 0) (B) (, ) (C) (0, ) (D) ( 5, 0). The ratio in which the line joining the points (, 4) and ( 5, 6) is divided b -ais - (A) : (B) 6 : 4 (C) : (D) none of these. The circumcentre of the triangle with vertices (0, 0), (, 0) and (0, 4) is - (A) (, ) (B) (, /) (C) (/, ) (D) none of these 4. The mid points of the sides of a triangle are (5, 0), (5, ) and (0, ), then orthocentre of this triangle is - (A) (0, 0) (B) (0, 4) (C) (0, 0) (D) 5. Area of a triangle whose vertices are (a cos, b sin), ( a sin, b cos ) and ( a cos, b sin ) is - (A) a b sin cos (B) a cos sin (C) ab (D) ab 6. The point A divides the join of the points ( 5,) and (,5) in the ratio k : and coordinates of points B and C are (,5) and (7, ) respectivel. If the area of ABC be units, then k equals - (A) 7,9 (B) 6,7 (C) 7,/9 (D) 9,/9 7. If A(cos, sin), B (sin, cos), C (,) are the vertices of a ABC, then as varies, the locus of its centroid is - (A) = 0 (B) = 0 (C) ( + ) 4 + = 0 (D) none of these 8. The points with the co-ordinates (a, a), (b, b) & (c, c) are collinear- (A) for no value of a, b, c (B) for all values of a, b, c (C) if a, c, b are in H.P. 5 (D) if a, c, b are in H.P A stick of length 0 units rests against the floor and a wall of a room. If the stick begins to slide on the floor then the locus of its middle point is - (A) + =.5 (B) + = 5 (C) + = 00 (D) none,8 NOD6 \ :\Data\04\Kota\J-Advanced\SMP\Maths\Unit#05\ng\.Straight line.p65 0. The equation of the line cutting an intercept of units on negative -ais and inclined at an angle tan 5 to the -ais is - (A) = 0 (B) 5 = 5 (C) = 0 (D) none of these. The equation of a straight line which passes through the point (, 5) such that the portion of it between the aes is divided b the point in the ratio 5 :, internall (reckoning from -ais) will be - (A) + = 0 (B) + + = 0 (C) + 7 = 0 (D) + 8 = 0 8. The points 0,, (, ) and (8, 0) are vertices of- [IIT-J 986] (A) an obtuse angled triangle (B) an acute angled triangle (C) a right angled triangle (D) an isosceles triangle. The straight lines + = 0, + 4 = 0, + 4 = 0 form a triangle which is- [IIT-J 98] (A) isosceles (B) equilateral (C) right angled (D) none of these 4. The co-ordinates of the vertices P, Q, R & S of square PQRS inscribed in the triangle ABC with vertices A (0, 0), B (, 0) & C (, ) given that two of its vertices P, Q are on the side AB are respectivel : (A), 0 4,, 0 8,, 8 8 &, 4 8 (C) (, 0),, 0,, &, (B) (D), 0,, 0 4,, 4 4 &, 4, 0, 9, 0 4, 9, 4 4 &, 4 7