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# Straight lines

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### Straight lines

1. 1. Straight lines . Line through two points The line through two distinct points (x1, y1) and (x2, y2) is given by (1) y = y1 + [ ]·(x - x1), where x1 and x2 are assumed to be different. In case they are equal, the equation is simplified to x = x1 and does not require a second point. Equation (1) can also be written as y - y1 = [ ]·(x - x1), or even as (x2 - x1)·(y - y1) = (y2 - y1)·(x - x1),where one does not have to worry whether x1 = x2 ornot. However, the simplest for me to remember is this (y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)
2. 2. which is not as universal is the one before.Intercept-interceptAssume a straight line intersects x-axis at (a, 0) and y-axis at (0, b). Then it is defined by the equation x/a + y/b = 1,which also can be written as xb + ya = ab.The latter form is somewhat more general as it allowseither a or b to be 0. a and b are defined as x-intercept and y-intercept of the linear function. Theseare signed distances from the points of intersection ofthe line with the axes.Point-slopeThe equation of a straight line through point (a, b) witha given slope of m is y = m(x - a) + b, or y - b = m(x - a).As a particular case, we haveSlope-intercept equation
3. 3. The equation of a line with a given slope m and the y-intercept b is y = mx + b.This is obtained from the point-slope equation bysetting a = 0. It must be understood that the point-slope equation can be written for any pointon the line, meaning that the equation in this form is not unique. The slope-intercept equation isunique because if the uniqueness for the line ofthe two parameters: slope and y-intercept..POINT-SLOPE FORMSuppose that we want to find the equation of a straight linethat passes through a known pointand has a known slope. Let (x,y) represent the coordinatesof any point on the line, and let (x,,y,)represent the coordinates of the known point. The slope isrepresented by m.Recalling the formula defining slope in terms of thecoordinates of two points, we have
4. 4. EXAMPLE: Write the equation of a line parallel to 3x - y - 2 = 0 and passing through the point (5,2). SOLUTION: The coefficients of x and y in the desired equation are the same as those in the given equation. Therefore, the equation is3x-y+D=0.EXAMPLE: Find the equation of a line passing throughthe point (2,3) and having a slope of 3.SOLUTION:The point-slope form may be used to find the equation ofa line through two known points.The values of x 1 , x 2 , y 1 , and Y2 are first used to find theslope of the line; then either knownpoint is used with the slope in the point-slope form.
5. 5. EXAMPLE: Find the equation of the line through thepoints ( - 3,4) and (4, - 2).SOLUTION:Letting (x,y) represent any point on the line and using ( -3,4) asUsing (4, - 2) as the known point will also give 7y + 6x =10 as the linear equation.SLOPE-INTERCEPT FORMAny line that is not parallel to the Y axis intersects the Yaxis at some point.The x coordinate of the point of intersection is 0, becausethe Y axis is vertical
6. 6. and passes through the origin. Let the y coordinate of thepoint of intersection berepresented by b. Then the point of intersection is (0,b), asshown in figure. The y coordinate, b, is called they intercept.The slope of the line in figure isThe value of y in this expression is y - b, where yrepresents the y coordinate of any point on the line. Thevalue of x is x - 0 = x, so
7. 7. Slope-intercept form.This is the standard slope-intercept form of a straight line.EXAMPLE: Find the equation of a line that intersects theY- axis at the point (0,3) and has a slope of 5/3.SOLUTION:PRACTICE PROBLEMS:Write equations for lines having points and slopes as follows:
8. 8. General equationA straight line is defined by a linear equation whosegeneral form is Ax + By + C = 0,where A, B are not both 0.Different forms of General form: (i) If B ≠0 , then y = it is slopeintercept form where m = -A/B and y-intercept = -C/B If B=0 , which is a vertical line whose slope isundefined and x-intercept is -C/A.(ii) If C ≠0, then x/a+y/b = 1 which is intercept formwhere a = x-intercept = -C/A , b = y-intercept= -C/B If C =0 , then it becomes a line passing throughorigin, has zero intercepts on the axes.
9. 9. NORMAL FORMMethods for determining the equation of a line usually dependupon some knowledgeof a point or points on the line. Lets now consider a methodthat does not require advance knowledge concerning any of the lines points. All that isknown about the line is its perpendicular distance from the origin and the angle between the perpendicular andthe X axis, where the angle ismeasured counterclockwise from the positive side of the Xaxis. line AB is a distance p away from the origin, and line OMforms an angle Ѳ(the Greek letter theta) with the X- axis. We select any pointP(x,y) on line AB and develop the
10. 10. Normal form.Equation of line AB in terms of the x and y of P. Since Prepresents any point on the line,the x and y of the equation will represent every point on theline and therefore will represent the line itself.PR is constructed perpendicular to OB at point R. NR isdrawn parallel to AB and PN is parallel to OB.PS is perpendicular to NR and to AB. A right angle is formedby angles NRO and PRN.Triangles ONR and OMB are similar right triangles.Therefore, angles NRO and MBO are equaland are designated as Ѳ’. Since Ѳ + Ѳ’ = 90 intriangle OMB and angle NRO is equal to Ѳ’,then angle PRN equals Ѳ. Finally, the x distance of point P isequal to OR, and the y distance of P is equal to PR.
11. 11. To relate the distance p to x and y, we reason as follows:This final equation is the normal form. The word "normal" inthis usage refers to the perpendicular relationshipbetween OM and AB. "Normal" frequently means"perpendicular" in mathematical and scientific usage. The distance p is always considered to be positive, and Ѳ isany angle between 0 and 360.EXAMPLE: Find the equation of the line that is 5 units awayfrom the origin, if the perpendicularfrom the line to the origin forms an angle of 30 from thepositive side of the X- axis.SOLUTION.
12. 12. DISTANCE FROM A POINT TO A LINEWe must often express the distance from a point to a line interms of the coefficients inthe equation of the line. To do this, we compare the two formsof the equation of a straight line, as follows:General equation: Ax + By + C = 0Normal form:The general equation and the normal form represent the samestraight line.Therefore, A (the coefficient of x in the general form) isproportional to cos Ѳ (the coefficient of x in the normal form).By similar reasoning, B is proportional to sin 0, and C isproportional to -p. Recalling
13. 13. that quantities proportional to each other form ratiosinvolving a constant of proportionality,let k be this constant. Thus, we haveSquaring both sides of these two expressions and thenadding, we have YORWE can prove perp. distance from appoint p(x1,y1) onA line Ax+By+C=0 is
14. 14. Y R (0, -C/B) Ax+By+C=0 P( x1,y1) ( -C/A ,0) Q XThe coefficients in the normal form, expressed in terms of A,B, and C, are as follows:The sign of is chosen so as to make p (adistance) always positive.The conversion formulas developed in the foregoingdiscussion are used in finding
15. 15. the distance from a point to a line. Let p represent thedistance of line LK from the origin.To find d, the distance from point P, to line LK, weconstruct a lineDistance from a point to a line.through P, parallel to LK. The distance of this line from theorigin is OS, and the difference between OS and p is d.We obtain an expression for d, based on the coordinates of P1,as follows:and
16. 16. Returning to the expressions for sin Ѳ , cos Ѳ, and p in termsof A, B, and C (the coefficients in the general equation), wehaveIn the formula for d, the denominator in each of theexpressions is the same. Therefore, we may combine terms asfollows:We use the absolute value, since d is a distance, and thus avoidany confusion arising from the ± radical.Note that the absolute value,| | of a number is defined asfollows:andThat is, for the positive number 2,|2|=2For the negative number -2,
17. 17. The absolute value of isEXAMPLE: Find the distance from the point (2,1) to theline 4x+2y+7=0.SOLUTION:PRACTICE PROBLEMS:In each of the following problems point to the line:find the distance from the1. (5,2), 3x - y + 6 = 02. (-2,5), 3x + 4y - 9 = 0
18. 18. ANSWERS: PARALLEL AND PERPENDICULARLINES ( GENERAL FORM) The general equation of a straight line is often written withcapital letters for coefficients, as follows: Ax+By+C=0 These literal coefficients, as they are called, represent the numerical coefficients encountered in a typical linear equation. Suppose we are given two equations that are duplicates except for the constant term, as follows: Ax +By+C=0 Ax +By+D=0 By placing these two equations in slope-intercept form, we can show that their slopes are equal, as follows:
19. 19. y=Thus, the slope of each line is -A/B.Since lines having equal slopes are parallel, we reach thefollowing conclusion:In any two linear equations, if the coefficients of thex and y terms are identical in value and sign, then the linesrepresented by these equations are parallel. Since the line passes through (5,2), the values x = 5 and y = 2 must satisfy the equation. Substituting these, we have 3(5) - (2) + D = 0 D= -13 Thus, the required equation is 3x – y - 13 =0 A situation similar to that prevailing with parallel lines involves perpendicular lines. For example, consider the equations
20. 20. Ax +By+C=0Bx - Ay+D=0Transposing these equations into the slope-interceptform, we havey=Since the slopes of these two lines are negativereciprocals, the lines are perpendicular.The conclusion derived from the foregoing discussionis as follows:If a line is to be perpendicular to a given line, thecoefficientsof x and y in the required equation are found byinterchangingthe coefficients of x and y in the given equationand changing the sign of one of them.EXAMPLE: Write the equation of a lineperpendicular to the line x + 3y + 3 = 0 and having a yintercept of 5.SOLUTION: The required equation is3x-y+D=0
21. 21. Notice the interchange of coefficients and the changeof sign. At the point where the line crosses the Y axis,the value of x is 0 and the value of y is 5. Therefore,the equation is3(0) - (5) + D = 0 D=5The required equation is 3x - y+5=0PRACTICE PROBLEMS:1. Find the equation of the line whose perpendicularforms an angle of 135 from the positive side of the Xaxis and whose perpendicular distance is V-2-unitsfrom the origin.Find the equations of the following lines:2. Through (1,1) and parallel to 5x - 3y = 9.3. Through (- 3,2) and perpendicular to x + y = 5.ANSWERS:
22. 22. SUMMARY (STRAIGHT LINES)The following are the major topics covered in this chapter:1. Distance between two points:where (x,,y,) and (x2,y2) are given points on a line.2. Division of a line segment:where k is the desired proportion of the distance between points(x1,y1) and(x 2 ,y 2 ) and (x,y) is the desired point.3. Midpoint of a line segment:4. Inclination: The angle of inclination is the angle the line crossing
23. 23. the X- axis makes with the positively directed portion of the X axis, such that 0 ° < α <180.5. Slope:The slope of a horizontal line is zero.The slope of a vertical line is meaningless.6. Slopes of parallel lines: Slopes are equal orwhere m1 and m2 are the slopes of the lines L, and L 2, respectively.7. Slopes of perpendicular lines: Slopes are negative reciprocals or8. Acute angle between two lines: ( arctan means tan-1 )
24. 24. However, if one line, L 2 , is parallel to the Y axis and the other, L1, has apositive slope, then
25. 25. If L 2 is parallel to the Y axis and L1, has a negative slope, then9. Obtuse angle between two lines:where + is the acute angle between the two lines.10. Point-slope form of a straight line:11. Slope-intercept form of a straight line:y = mx + bwhere b is the y- intercept.12. Normal form of a straight line:where p is the lines perpendicular distance from the origin and Ѳ is the
26. 26. anglebetween the perpendicular and the X- axis.13. Parallel lines:In any two linear equations, if the coefficients of the x and y terms areidentical in value and sign,then the lines represented by these equations are parallel; that is,Ax+By+C=0 and Ax+By+D=0are parallel lines.Perp. distant b/w two ii lines.
27. 27. 14. Perpendicular lines:If a line is to be perpendicular to a given line, the coefficients of x and yin the required equations are found by interchanging the coefficients of x and yin the givenequation and changing the sign of one of them; that is,Ax+By+C=0 and Bx-Ay+D=0are perpendicular lines.
28. 28. 15. Distance from a point to a line:where A, B, and C are the coefficients of the general equation of a line Ax+ By + C = 0 and(x,,y,) are the coordinates of the point.SYMMETRIC FORM AND PARAMETRIC EQUATIONS OF ALINE The equation of the straight line passing through (x1,y1) and makingan angle Ѳ wih the positive direction of x-axis is = = r , where r is the distant of the point (x,y) on the linefrom the point (x1,y1).REMARK: If P(x,y) be a point at a distant of r units from a givenpoint Q( x1,y1), then x = x1 + r cosѲ and y = y1 + rsinѲ EXAMPLE: If the straight line drawn through the point P( ,2) andmaking an angle п/6 with the x-axis meets the line x – 4y + 8=0 at Q, find the length of PQ.SOLUTION: Let the line through P making an angle п/6 with the x-axismeets the line x – 4y + 8=0 at Q. By above formula Q = ( + r cosп/6 , 2 + r sinп/6) lies onequation ⇨ r =6.
29. 29. ADDITIONAL PRACTICE PROBLEMS1. Find the distance between P1 (- 3, - 2) and P2(-7,I).2. Find the distance between P1 (- 3/4, - 2) and P2(1, - 1/2). 3.Find the coordinates of a point 115 of the way from P1(- 2,0) to P2(3, -5).4. Find the midpoint of the line between P1(-8/3,4/5) and P2(- 4/3,6/5).5. Find the slope of the line joining P1(4,6) and P2(-4,6).6. Find the slope of the line parallel to the line joining P1(7,4) andP2(4,7).7. Find the slope of the line perpendicular to the line joining P1(8,1) andP2(2,4).8. Find the obtuse angle between the two lines which have m, = 7 andm2 = - 3 for slopes.9. Find the obtuse angle between the Y axis and a line with a slope of m= -1/4.10. Find the equation of the line through the points ( - 6,5) and (6,5).11. Find the equation of the line whose y intercept is (0,0) and whoseslope is 4.12. Find the slope and y intercept of the line whose equation is4y+8x=7.13. Find the equation of the line that is 3/2 units away from the origin,
30. 30. if the perpendicular from the line to the origin forms an angle of 210 0 from the positive side of the X axis. 14. Find the equation of the line through (2,3) and perpendicular to 3x- 2y=7. 15. Find the equation of the line through (2,3) and parallel to 3x-2y=7. 16. Find the distance from the point (3, - 5) to the line 2x+y+4=0. 17. Find the distance from the point (3, - 4) to the line 4x +3y= 10 ANSWERS.
31. 31. ASSIGNMENT Question 1 In what ratio is the line joining the points A(4,4)and B(7,7) divide by P(-1,-1)? [Hint: use section formula after assuming ratio k:1 , k= -5/8] Question 2 Determine the ratio in which the line 3x+y – 9 =0 divides the segment joining the points (1,3) and (2,7). [ Hint: use section formula , k=3/4]**Question 3 The area of a triangle is 5. Two of its verticesare (2,1) and (3,-2). Third vertex is (x,y) where y = x+3. Find The co-ordinates of the third vertex. [Let the vertices are A(x,y), B(2,1), C(3,-2) , Area of triangleABC = ½ |3x+y – 7|=5 , in case (i) x=7/2,y=13/2 In case (ii) x=-3/2 , y=3/2] Question 4 Find the equation of the straight lines which passthrough the origin and trisectthe intercept of line 3x+4y=12 b/w the axes. [ Hint: Let the line AB be trisected at P and Q, then AP : PB =1:2 , A(4,0) , B(0,3) BY using section formula we get P(8/3 , 1) AQ : QB = 2 : 1 ⇨ Q=(4/3 ,2) then equation of line OP andOQ passing through (0,0) is 3x – 8y =0 and 3x – 2y =0 resp.]
32. 32. Question 5 If the straight line drawn through the point P(2,3) and making an angleп/4 with the x-axis meets the line x + y + 1=0 at Q, find the length of PQ. [ answer is r =3 ]** Question 6 Find the distant of the point (2,5) from the line3x+4y+4=0 measured parallel to a line having slope ¾. [Hint: tanѲ= ¾ ⇨ sinѲ =3/5 , cosѲ=4/5 , equation passingthrough A(2,5) by symmetric form is X= 2+(4/5)r , y= 5+(3/5)r they lie on a given line ⇨ r = -5 Question 7 The line segment joining A(2,3), B(-3,5) isextended through each end by a length equal to its originallength. Find the co-ordinates of the new ends. [ Hint: answer is x= 7, y= 1 and α = -8 , =7 ( ) (-3,5) (2,3) (x,y) C B A D ] Question 8 The line segment joining A (6,3) to B (-1,-4) isdoubled in length by having added to each end. Find the co-ordinates Of the new ends.
33. 33. [ Hint : given CB = ½ BA , AD = ½ BA ∴ B divides CAinternally in 1:2 and A divides BD internally in 2:1 C( -9/2 ,-15/2) , D( 19/2 , 13/2)] Question 9 Two opposite vertices of a square are (3,4) and(1,-1). Find the co-ordinates of other vertices. [ Let A(3,4) , C(1,-1) then slope of AC = 5/2 , M midpoint of AC & BD =(2, 3/2) Let m is the slope of a line making an angle of 450 withAC i.e., m is slope of lines AD or CD. USE formula of angle b/w two lines ⇨ m= -7/3, 3/7then equations of AD , CD are 7x+3y – 33=0 , 3x – 7y – 10 =0 and D(9/2 ,1/2) ,M ismid point of BD ⇨ B(-1/2 , 5/2)] Question 10 (i) Find the co-ordinates of the orthocenter ofthe whose angular points are (1,2), (2,3), (4,3). (ii) Find the co-ordinates of the circumcenterof the whose angular points are (1,2), (3,-4), (5,-6). [ answer (i) (1,6) (ii) (11,2)] Question 11 Find the equation of the line through theintersection of the lines x -3y+1=0 and 2x+5y -9=0 and whose distant from the origin is .
34. 34. [ Hint: (x -3y+1) +k(2x+5y -9)=0 -------(1) , then finddistant from (0,0) on the line (1) is ⇨ k=7/8 ,put in (1) Answer is 2x+y – 5=0] Question 12 The points (1,3) and (5,1) are the oppositevertices of a rectangle.The other two vertices lie on the line y= 2x+c. Find c and the remaining vertices. [Hint: Use mid point formula , it lies on BD ∴ c = -4 , useM(3,2) (by mid point D( C(5,1) y=2x+c M (3,2) A(1,3) B(X,2X-4)(AB)2 + (BC)2 = (AC)2 ⇨ x=4 or 2 ∴ B(4,4) then D (2,0), ifB(2,0) then D(4,4)]Question 13 The consecutive sides of a parallelogram are4x+5y=0 and 7x+2y=0. If the equation of one diagonal be11x+7y=9, find the equation of other diagonal .
35. 35. [ Hint: D C 11x+7y=9 P 7x+2y=0 O 4x+5y=0 BB(5/3,-4/3) , D(-2/3,7/3) by solving equations of OB & BD andOD & BD resp. Then find point P (1/2,1/2) & equation of OC i.e, OP isy=x.] Question 14 One side of a rectangle lies along the line4x+7y+5=0. Two of vertices are (-3,1) & (1,1). Find the equation of other three sides. [ Hint: D slope=-4/7 C(1,1) Slope=7/4 slope=7/4 A(-3,1) 4x+7y+5=0 B
36. 36. Equation of BC is 7x – 4y -3=0 , equation of AD & CD are7x – 4y +25=0 & 4x+7y=11=0 resp.]Question 15 The extremities of the base of an isosceles ∆are the points (2a,0) & (0,a). The equation of the one of the sides is x=2a. Find the equation of the other two sides and the area of the ∆.[ Hint: y C B(0,a) x+2y-2a=0 o A(2a,0) Xby solving CA2 = CB2 ⇨ Y=(5a)/2 i.e, C is (2a,5a/2),equation of BC is 3x – 4y+4a=0 & area of ∆ACB is 5a2/2sq.units.] Question 16 One side of a square is inclined to x-axis at anangle α and one of its extremities is at origin.
37. 37. If the sides of the square is 4, find the equations ofthe diagonals of the square. [ Hint: Take ∆OLA , find A as OL/4=cosα & AL/4=sinαand in ∆OMC ,find point C, then find equation of OB & AC( by using mid pointof OB & AC) equations of OB & AC arex(cosα+sinα) – y(cosα – sinα)=0 , x(cosα - sinα) + y(cosα + sinα)=4 resp Y B(h,k)(-4sinα, 4cosα) C 4 4 A(4cosα, 4sinα) M O LAngle COM=900-α , angle AOL=α ]Question 17 Prove that the diagonals of the //gm. Formedby the four lines. x/a + y/b = 1 ……(i) , x/b + y/a = 1 …….(ii) , x/a +y/b = -1 ……(iii) , x/b + y/a = -1 ……..(iv) are perp. to eachother.[HINT:
38. 38. ( , )D line (iii) C ( , ) Slope=1 Line(iv) Slope=-1 line(ii) ( , )A line(i) B( , )Find all co-ordinates & for perpendicularity of diagonals showproduct of slopes of AC & BD = -1 ] Question 18 On the portion of the line x+3y – 3 =0 which is intercepted b/w the co-ordinates axes, a square is constructed on the side of the line away from the origin. Find the co-ordinates of the intersection of its diagonals. Also find the equations of its sides. [HINT: P is the mid point of BD Y C D(4,3) (0,1)B 45 P(2,2) X+3y-3=0 A(3,0)
39. 39. angle ABD=450 , use formula tan 450 =| | ⇨ m =1/2 or -2 , equations of BD ,AC, CD AD & BC are x-2y+2=0, 2x+y-6=0, x+3y-13=0, 3x-y-9=0 & 3x-y+1=0 resp.]Question 19 The hypotenuse of a right angled ∆ has its endsat the points (1,3) and (-4,1). Find the equation of the legs (perpendicular sides) of thetriangle. [ Hint: The legs (perpendicular sides) of the triangle , wecan consider as to x-axis & y-axis resp. line to x-axis isy=k It passes through (-4,1) ⇨ k=1 ∴ y=1 similarily x=1 as x=kpasses through (1,3)]Question 20 If one diagonal of a square is along the line 8x-15y=0 and one of its vertex is at (1,2), then find the equationof sides of the square passing through this vertex.
40. 40. [ Hint: use formula tan 450 = ⇨ m1 = 23/7 , (1,2) A B m1 8x-15y=0 450 m2 =8/15 D C Equations of AD & AB ( Perp. to each other) are 23x-7y-9=0 , 7x+23y-53=0]Question 21 Find the reflection of (4,-13) about the line5x+y+6=0. [Hint: use mid point formula & concept of product ofslopes of perp. lines , then answer is (-1,-14)]General equationA straight line is defined by a linear equation whosegeneral form is Ax + By + C = 0,
41. 41. where A, B are not both 0.The coefficients A and B in the general equation are thecomponents of vector n = (A, B) normal to the line. The pair r = (x, y) can belooked at in two ways: as a point or as aradius-vector joining the origin to that point. The latterinterpretation shows that a straight lineis the locus of points r with the property r·n = const.That is a straight line is a locus of points whose radius-vector has a fixed scalar productwith a given vector n, normal to the line. To see whythe line is normal to n, take two distinct but otherwise arbitrary points r1 and r2 on the line, sothat r1·n = r2·n.But then we conclude that (r1 - r2)·n = 0.In other words the vector r1 - r2 that joins the twopoints and thus lies on the line is perpendicular to n.Normalized equation
42. 42. The norm ||n|| of a vector n = (A, B) is defined via ||n||2 = A2 + B2 and has the property that, for any non-trivial vector n, n/||n|| is a unit vector, i.e., || n/||n|| || = 1. Note that the line defined by a general equation would not change if the equation were to be multiplied by a non-zero coefficient. This property can be used to keep the coefficient A non-negative. It can also be used to normalize the equation by dividing it by ||n||. As a result, in a normalized equation Ax + By + C = 0, A2 + B2 = 1. (In the applet, the coefficients of the normalized equation are rounded to up to 6 digits, for which reason the above identity may only hold approximately.)The normalized equation is conveniently used indetermining the distance from a point to a line Parametric equation A line through point r0 = (a, b) parallel to vector u = (u, v) is given by
43. 43. (x, y) = (a, b) + t·(u, v),where t is any real number. In the vector form, we have r = r0 + t·u,where r = (x, y).Implicit equationA line through point r0 = (a, b) perpendicular tovector n = (m, n) is given by m(x - a) + n(y - b) = 0,or if we take r = (x, y), a generic point on the line, wesee that n·(r - r0) = 0,where dots indicates the scalar product of two vectors.A function whose graph is a straight line is linear andcontinuous.A continuous linear function must have theform f(x) = ax. Discontinuous linearfunctions look dreadful.To be more specific, I am going to discuss real valuedfunctions of one real variable,
44. 44. i.e. f: R R, where R is, as usual, the set of all realnumbers. Such a function is calledlinear provided the following condition holds:(*) For every two real x1 and x2, f(x1 + x2) = f(x1) + f(x2)Assuming that the function f is also continuous I plan toshow that f(x) = ax for some real a.Please note that if indeed f(x) = ax then a = f(1) whichprovides a starting point for the proof.But first let me note that (*) contains an unknownwhich, as we are going to establish,is equal to f(x) = ax. In other words, (*) serves as anexample of a functional equation –an equation whose unknown is a function.ProofThe proof proceeds in several steps. 1. x is 0. f(0) = f(0 + 0) = f(0) + f(0) = 2f(0). Therefore f(0) = 2f(0) and finally f(0) = 0. 2. x is negative.
45. 45. Let x be negative, e.g., let x + y = 0, where y is positive; so that -x = y. Then 0 = f(0) = f(x + y) = f(x) + f(y). Therefore f(-x) = f(y) = -f(x).3. x is an integer. We have f(2) = f(1 + 1) = f(1) + f(1) = 2f(1). By induction, assume f(k - 1) = (k - 1)f(1). Then f(k) = f(1 + (k-1)) = f(1) + (k-1)f(1) = kf(1). Lets denote a = f(1). We have shown that for all integers n, f(n) = an.4. x is rational First of all, for any integer n≠0, we have 1 = n/n. Then, as before, a = f(1) = f(n/n) = nf(1/n).Hence, f(1/n) = a/n = a(1/n). For p = m/n we similarly have f(p) = f(m/n) = mf(1/n) = m·a/n = a(m/n) = ap.5. x is irrational Any irrational number r can be approximated by a sequence of rational numbers pi. The closerpi is to r, the closer api is to ar. However, since api = f(pi) and
46. 46. assuming f continuous we must necessarily get f(r) = ar.Continuity of the function is quite essential as itspossible to show [Ref. 1, 2] that the graph of any discontinuous solution to (*) is dense in theplane R2. For the sake of reference, the graph of a function f: R R is defined as a set ofpairs (x, y), i.e. elements of R2 such that y = f(x).Formally,graph(f) = {(x, y)∈R2: y = f(x)}.RemarkGenerally speaking, a function that satisfies (*) iscalled additive. The function that satisfies f(x) = axfor some a is said to be homogeneous.A function is said to be linear if its both additive and homogeneous. We have justshown that a continuous additivefunction is necessarily linear.The graph of a linear function is a straight line whose(linear) equation may be obtained in different forms depending on the manner in whichthe line is defined.
47. 47. HOT SKILLS QUESTIONS ** Question 1 Show that the lines 4x+y-9=0 , x-2y+3=0, 5x-y- 6=0 make equal intercepts on any line of gradient 2. Question 2 (i) If the lines p1x+q1y=1, p2x+q2y=1 and p3x+q3y=1 be concurrent , show that the points (p1,q1), (p2,q2), (p3,q3) are collinear. (ii) Find the eqns. Of the lines passing through the point of intersection of the lines x+3y+4=0 and 3x+y+4=0 and equally inclined to the axes.Question 3 The line 2x-y=5 turns about the point on it, wherethe ordinate is equal to the abscissa through an angle of 450 inthe anticlockwise direction. Find the equation of the line inthe new position. Question 4 A line 4x+y=1 through the point A(2,-7) meets the line BC whose equation is 3x-4y+1=0 at the point B. Find the eqn. of the line AC so that AB=AC. Question 5 Straight lines 3x+4y=5 and 4x-3y=15 intersect at A. Points B & C are chosen on these lines such that AB=AC. Find the possible eqns. Of BC passing through the point (1,2).
48. 48. Question 6 A ray of light is sent along the line x-2y-3=0.Upon reaching the line 3x-2y-5=0, the ray is reflected from it. Find the eqn. of the line containing thereflected ray.Question 7 The eqns. Of two sides of a ∆ are 3x-2y+6=0 and4x+5y=20 and orthocenter is (1,1). Find eqn. of third side.Question 8 The eqns. Of the perp. bisectors of the sides AB& AC of ∆ ABC are x-y+5=0 and x+2y =0 resp. If the point A is (1, -2) , find the eqn. of the line BC.Question 9 Prove that the length of perps. From points (m2,2m), (mn, m+n) and (n2, 2n) to the line xcosѲ+ysinѲ + P=0 Where P = sin2Ѳ/cosѲ form G.P. Question 10 Find the distant of the point (1,2) from thestraight line with slope 5 and passing through the point Of intersection of x+2y=5 and x-3y = 7 ANSWERS WITH HINTSAnswer 1 B A C
49. 49. Eqn. of line with gradient 2 is y = 2x +c , solve it with giveneqns. We will have points B, A, C are (1 - , 2 - ) , ( - , 3+ ) , ( 2+ , 4+ ) resp. ⇨ AB = AC.Answer 2 (i) If lines are concurrent then= 0 , points are collinear. (ii) ( x+3y+4)+ k(3x+y+4) = 0 ……..(1) , takeslope of eqn.(1) = - = ± 1 ( tan450 or tan1350) then put the value of k = 1, -1 in eqn. (1) andeqns are x-y=0 and 4x+4y+8=0 .Answer 3 Let P (h,h) on the line AB is 2x-y =5 ∴ P(5,5) ,CD is another line passing through P and makingAn angle 450 with AB ∴ slope of CD = tan (Ѳ+450) = -3 ( slopeof AB is 2) by using angle b/w two lines So eqn. of CD will be 3x+y-20=0.
50. 50. Answer 4 A(2,-7) 4X+Y-1=0 α α B 3x-4y+1=0 C BY using angle b/w two lines formula ,weget=slope of AC= m=- 52/89 ( = ) Eqn. of AC is 52x+89y+519=0.Answer 5 same as Q. 4 Let m be the slope of BC , it will be1/7, -7 (angle b/w AB & BC =angle b/w BC & AC) Eqns of BC are 7x+y-9=0 , x-7y+13=0.Answer 6 L A 3x-2y-5=0 M α Ѳ Ѳ β x – 2y-3=0 P N QA(1,-1) by solving LM & PA and
51. 51. Tanα = tanβ ⇨ = ⇨ m= 29/2 ∴eqn. of AQ is 29x-2y-31=0.Answer 7 Let eqns. of AB & AC are given and H(1,1) beorthocenter A N M 3x-2y+6=0 H 4x+5y=20 B L CEqns. Of BM & CN are 5x-4y-1=0 , 2x+3y-5=0 by using pointslope form , and B(-13, -33/2) , C(35/2, -10) , Then eqn. of BC will be 26x-122y-1675=0.Answer 8 same as Q.7 by using point slope form & midpoint formula , we will get Eqns. Of AB & AC are x+y=-1 , 2x-y=4 and B(-7, 6)& C(11/5, 2/5) , then find eqn. of BC by two point form .Answer 9 USE perp. distant from a point A( m2, 2m) onthe line xcosѲ+ysinѲ + sin2Ѳ/cosѲ =0 is
52. 52. p1 = | (m2 cosѲ + 2msinѲ +sin2Ѳ/cosѲ)/ |= similarly find p2 , p3 , then show p1 . p3 = (p2 )2.Answer 10 same formula of perp. distant from a point tothe line i.e., P = length of perp. from (1,2) to line 25x-5y-147=0 ( bysolving given eqns.) is 132/ .