{"id":24103,"date":"2024-10-15T11:54:19","date_gmt":"2024-10-15T06:24:19","guid":{"rendered":"https:\/\/samacheerkalvi.guide\/?p=24103"},"modified":"2024-10-16T09:54:35","modified_gmt":"2024-10-16T04:24:35","slug":"samacheer-kalvi-11th-maths-guide-chapter-6-ex-6-3","status":"publish","type":"post","link":"https:\/\/samacheerkalvi.guide\/samacheer-kalvi-11th-maths-guide-chapter-6-ex-6-3\/","title":{"rendered":"Samacheer Kalvi 11th Maths Guide Chapter 6 Two Dimensional Analytical Geometry Ex 6.3"},"content":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 11th Maths Guide<\/a> Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.3 Text Book Back Questions and Answers, Notes.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.3<\/h2>\n

Question 1.
\nShow that the lines 3x + 2y + 9 = 0 and 12x + 8y – 15 = 0 are parallel lines.
\nAnswer:
\nThe equations of the given lines are
\n3x + 2y + 9 = 0 ——- (1)
\n12x + 8y – 15 = 0 ——- (2)
\n\"Samacheer
\nm1<\/sub> = m2<\/sub>
\n\u2234 The given lines are parallel.<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nFind the equation of the straight line parallel to 5x – 4y + 3 = 0 and having x – intercept 3.
\nAnswer:
\nThe equation of any line parallel to 5x – 4y + 3 = 0 is
\n5x – 4y + k = 0 ……….. (1)
\nThe x – intercept of line (I) is obtained by putting
\ny = 0 in the equation.
\n(1) \u21d2 5x – 4(0) + k = 0
\n5x = – k \u21d2 x = \\(-\\frac{k}{5}\\)
\nGiven that the x – intercept in 3
\n\u2234 \\(-\\frac{k}{5}\\) = 3 \u21d2 k = – 15
\n\u2234 The equation of the required line is
\n5x – 4y – 15 = 0<\/p>\n

Question 3.
\nFind the distance between the line 4x + 3y + 4 = 0 and a point
\n(i) (- 2, 4)
\n(ii) (7, – 3)
\nAnswer:
\n(i) The distance between the line ax + by + c = 0 and the point (x1<\/sub>, y1<\/sub>) is d = \\(\\frac{a x_{1}+b y_{1}+c}{\\sqrt{a^{2}+b^{2}}}\\)
\nHere (x1<\/sub>, y1<\/sub>) = (- 2, 4) and the equation of the line is 4x + 3y + 4 = 0
\n\"Samacheer<\/p>\n

(ii) Given point (x1<\/sub>, y1<\/sub>) = (7, – 3)
\nGiven line 4x + 3y + 4 = 0
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nWrite the equation of the lines through the point (1, – 1)
\n(i) Parallel to x + 3y – 4 = 0
\n(ii) Perpendicular to 3x + 4y = 6
\nAnswer:
\n(i) Any line parallel to x + 3y – 4 = 0 will be of the form x + 3y + k = 0.
\nIt passes through (1,-1) \u21d2 1 – 3 + k = 0 \u21d2 k = 2
\nSo the required line is x + 3y + 2 = 0<\/p>\n

(ii) Any line perpendicular to 3x + 4y – 6 = 0 will be of the form 4x – 3y + k = 0.
\nIt passes through (1, -1) \u21d2 4 + 3 + k = 0 \u21d2 k = -7.
\nSo the required line is 4x – 3y – 7 = 0<\/p>\n

Question 5.
\nIf (- 4, 7 ) is one vertex of a rhombus and if the equation of one diagonal is 5x – y + 7 = 0 then find the equation of another diagonal.
\nAnswer:
\n\"Samacheer
\nIn a rhombus, the diagonal cut at right angles.
\nThe given diagonal is 5x – y + 7 = 0 and (- 4, 7) is not a point on the diagonal.
\nSo it will be a point on the other diagonal which is perpendicular to 5x – y + 7 = 0.
\nThe equation of a line perpendicular to 5x – y + 7 = 0 will be of the form x + 5y + k = 0.
\nIt passes through (-4, 7) \u21d2 -4 + 5(7) + k = 0 \u21d2 k = -31
\nSo the equation of the other diagonal is x + 5y – 31 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 6.
\nFind the equation of the lines passing through the point of intersection of the lines 4x – y + 3 = 0 and 5x + 2y + 7 = 0 and
\n(i) through the point (-1,2)
\n(ii) parallel to x – y + 5 =0
\n(iii) perpendicular to x – 2y + 1 =0
\nAnswer:
\nThe equation of the straight line passing through the point of intersection of the lines.
\n4x – y + 3 = 0 and 5x + 2y + 7 = 0 is
\n( 4x – y + 3) + \u03bb (5x + 2y + 7) = 0 ……… (1)<\/p>\n

(i) Through the point (-1,2)
\nGiven that line(1) passes through the point (-1, 2)
\n(1) \u21d2 (4(-1) – 2 + 3) + \u03bb (5(- 1) + 2(2) + 7) = 0
\n(-4 – 2 + 3) + \u03bb (-5 + 4 + 7) = 0
\n– 3 + 6 \u03bb = 0 \u21d2 \u03bb = \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\)
\n\u2234 The equation of the required line is
\n(4x – y + 3) + \\(\\frac{1}{2}\\) (5x + 2y + 7) = 0
\n2(4x – y + 3) + (5x + 2y + 7) = 0
\n8x – 2y + 6 + 5x + 2y + 7 = 0
\n13x +13 = 0 \u21d2 x + 1 = 0<\/p>\n

(ii) Equation of a line parallel to x – y + 5 = 0 will be of the form x – y + k = 0.
\nIt passes through (-1, -1) \u21d2 -1 + 1 + k = 0 \u21d2 k = 0.
\nSo the required line is x – y = 0 \u21d2 x = y.<\/p>\n

(iii) Perpendicular to x – 2y + 1 = 0
\nGiven that the line (1) perpendicular to the line
\nx – 2y + 1 = 0 ………….. (3)
\n(1) \u21d2 (4x – y + 3) + \u03bb (5x + 2y + 7) = 0
\n4x – y + 3 + 5\u03bbx + 2\u03bby + 7\u03bb = 0
\n(4 + 5\u03bb)x + (2\u03bb – 1 )y + (3 + 7\u03bb) = 0 ……….. (4)
\nSlope of this line (3) = \\(-\\frac{4+5 \\lambda}{2 \\lambda-1}\\)
\nSlope of line (2) = \\(-\\frac{1}{-2}=\\frac{1}{2}\\)
\nGiven that line (3) and line (4) are perpendicular
\n\"Samacheer
\n4 + 5\u03bb = 2(2\u03bb – 1 )
\n4 + 5\u03bb = 4\u03bb – 2
\n\u03bb = – 6
\nSubstituting the value of \u03bb in equation (1) we have
\n(4x – y + 3) – 6 (5x + 2y + 7) = 0
\n4x – y + 3 – 30x – 12y – 42 = 0
\n-26x – 13y – 39 =0
\n2x + y + 3 = 0
\nwhich is the required equation.<\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nFind the equations of two straight line which are parallel to the line 12x + 5y + 2 = 0 and at a unit distance from the point (1, – 1).
\nAnswer:
\nThe equation of the given line is
\n12x + 5y + 2 = 0 ……… (1)
\nEquation of any line parallel to the line (1) is
\n12x + 5y + k = 0 ………. (2)
\nGiven that line (2) is at a unit distance from the point (1, – 1)
\n\"Samacheer
\nk = – 7 \u00b1 13
\nk = -7 + 13 or k = – 7 – 13
\nk = 6 or k = – 20
\n\u2234 The equation of the required lines are
\n12x + 5y + 6 = 0 and 12x + 5y – 20 = 0<\/p>\n

Question 8.
\nFind the equations of straight lines which are perpendicular to the line 3x + 4y – 6 = 0 and are at a distance of 4 units from (2, 1).
\nAnswer:
\nThe equation of the given line is
\n3x + 4y – 6 = 0 …………. (1)
\nThe equation of any line perpendicular to line (1) is
\n4x – 3y + k = 0 …………. (2)
\nGiven that this line is 4 units from the point (2, 1)
\n\"Samacheer
\n\u00b1 4 = \\(\\frac{5+k}{5}\\)
\n5 + k = \u00b1 20
\nk = \u00b1 20 – 5
\nk = 20 – 5 or k = -20
\nk = 15 or k = -25
\n\u2234 The equation of the required lines are
\n4x – 3y + 15 = 0 and 4x – 3y – 25 = 0<\/p>\n

Question 9.
\nFind the equation of a straight line parallel to 2x + 3y = 10 and which is such that the sum of its intercepts on the axes is 15.
\nAnswer:
\nThe equation of the given line is
\n2x + 3y = 10 ………….. (1)
\nThe equation of any line parallel to (1) is
\n2x + 3y = k …………. (2)
\n\"Samacheer
\nGiven that the sum of the intercepts of the line (2) on the axes is 15
\n\"Samacheer
\n\u2234 The equation of the required line is 2x + 3y = 18<\/p>\n

\"Samacheer<\/p>\n

Question 10.
\nFind the length of the perpendicular and the coordinates of the foot of the perpendicular from (-10, -2) to the line x + y – 2 = 0.
\nAnswer:
\nThe coordinate of the foot of the perpendicular from the point (x1<\/sub>, y1<\/sub>) on the line ax + by + c = 0 is
\n\"Samacheer
\n\u2234 The coordinate of the foot of the perpendicular from the point (- 10, – 2) on the line x + y – 2 = 0 is
\n\"Samacheer
\nx + 10 = y + 2 = \\(\\frac{14}{2}\\)
\nx + 10 = y + 2 = 7
\nx + 10 = 7, y + 2 = 7
\nx = – 3, y = 5
\n\u2234 The required foot of the perpendicular is (- 3, 5).
\nLength of the perpendicular
\n\"Samacheer<\/p>\n

Question 11.
\nIf p1<\/sub> and p2<\/sub> are the lengths of the perpendiculars from the origin to the straight lines x sec \u03b8 + y cosec \u03b8 = 2a and x cos \u03b8 – y sin \u03b8 = a cos 2\u03b8, then prove that p1<\/sub>2<\/sup> + p2<\/sub>2<\/sup> = a2<\/sup>.
\nAnswer:
\nGiven P1<\/sub> is the length of the perpendicular from the origin to the straight line
\nx sec \u03b8 + y cosec \u03b8 – 2a = 0
\n\"Samacheer
\nAlso given P2<\/sub> is the length of the perpendicular from the origin to the straight line
\nx cos \u03b8 – y sin \u03b8 – a cos 2\u03b8 = 0
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 12.
\nFind the distance between the parallel lines
\n(i) 12x + 5y = 7 and 12x + 5y + 7 = 0
\n(ii) 3x – 4y + 5 = 0 and 6x – 8y – 15 = 0
\nAnswer:
\n(i) 12x + 5y = 7 and 12x + 5y + 7 = 0
\nThe equation of the given lines are
\n12x + 5y – 7 = 0 ……….. (1)
\n12x + 5y + 7 = 0 ……….. (2)
\nThe distance between the parallel lines
\nax + by + c1<\/sub> = 0 and ax + by + c2<\/sub> = 0 is
\nThe equation of any line parallel to (1) is
\n\"Samacheer
\nThe distance cannot be negative
\n\u2234 Required distance = \\(\\frac{14}{13}\\)<\/p>\n

(ii) 3x – 4y + 5 = 0 and 6x – 8y -15 = 0
\nThe equation of the given lines are
\n3x – 4y + 5 = 0 ……….. (1)
\n6x – 8y – 15 = 0
\n3x – 4y – 1 = 0 ………… (2)
\nThe distance between the parallel lines (1) and (2) is
\n\"Samacheer<\/p>\n

Question 13.
\nFind the family of straight lines
\n(i) Perpendicular
\n(ii) Parallel to 3x + 4y – 12
\nAnswer:
\n(i) Equation of lines perpendicular to 3x + 4y – 12 = 0 will be of the form 4x – 3y + k = 0, k \u2208 R
\n(ii) Equation of lines parallel to 3x + 4y – 12 = 0 will be of the form 3x + 4y + k = 0, k \u2208 R<\/p>\n

\"Samacheer<\/p>\n

Question 14.
\nIf the line joining two points A (2, 0) and B (3, 1) is rotated about A in an anticlockwise direction through an angle of 15\u00b0 , then find the equation of the line in the new position.
\nAnswer:
\n\"Samacheer
\nSlope of the line AB
\nm = tan \u03b8 = \\(\\frac{1-0}{3-2}\\)
\ntan \u03b8 = 1
\n\u03b8 = 45\u00b0
\n\u2234 The line AB makes an angle 45\u00b0 with x-axis.
\nGiven that the line AB is rotated through an angle of 15\u00b0 about the point A in the anticlockwise direction.
\n\u2234 The angle made by the new line AB’ is 45\u00b0 + 15\u00b0 = 60\u00b0
\nSlope of the new line AB’ is m1<\/sub> = tan 60\u00b0 = \u221a3
\n\u2234 The equation of the new line AB’ is the equation of the straight line passing through the point A (2, 0) and having slope m1<\/sub> = \u221a3
\ny – 0 = \u221a3 (x – 2)
\ny = \u221a3x – 2\u221a3
\n\u221a3x – y – 2\u221a3 = 0<\/p>\n

Question 15.
\nA ray of light coming from the point (1, 2) is reflected at a point A on the x – axis and it passes through the point (5, 3). Find the coordinates of point A.
\nAnswer:
\n\"Samacheer
\nLet P(1, 2) and (5, 3) are the given points.
\nBy the property of reflection,
\n\u2220XAB = \u2220OAP = \u03b8
\n(Angle of incidence = Angle of reflection)
\nSlope of the line OA (x – axis) m1<\/sub> = 0
\nSlope of the line joining the points P (1, 2) and A (x, 0)
\nSlope of AP, m2<\/sub> = \\(\\frac{2-0}{1-x}=\\frac{2}{1-x}\\)
\nSlope of the line joining the points B (5, 3 ) and A (x, 0)
\nSlope of AP, m3<\/sub> = \\(\\frac{3-0}{5-x}=\\frac{3}{5-x}\\)
\n\"Samacheer
\nFrom equations (1) and (2)
\n\\(\\frac{3}{5-x}\\) = –\\(\\frac{2}{1-x}\\)
\n3(1 – x) = – 2 (5 – x)
\n3 – 3x = – 10 + 2x
\n2x + 3x = 10 + 3
\n5x = 13 \u21d2 x = \\(\\frac{13}{5}\\)
\n\u2234 The required point A is \\(\\left(\\frac{13}{5}, 0\\right)\\)<\/p>\n

\"Samacheer<\/p>\n

Question 16.
\nA line is drawn perpendicular to 5x = y + 7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 10. sq. units.
\nAnswer:
\n\"Samacheer
\nLet the given line be PQ whose equation is
\n5x = y + 7
\n5x – y – 7 = 0 ——- (1)
\nLet AB be the line perpendicular to the line PQ such that the area of the triangle OAB is 10 units.
\nThe equation of the line AB is
\n– x – 5y + k = 0
\nx + 5y – k = 0
\nx + 5y = k ——- (2)
\n\"Samacheer
\n\u2234 A is (k, 0) and B is (0, \\(\\frac{\\mathrm{k}}{5}\\))
\nOA = k and OB = \\(\\frac{\\mathrm{k}}{5}\\)
\nArea of \u2206OAB = \\(\\frac{1}{2}\\) \u00d7 OA \u00d7 OB
\n= \\(\\frac{1}{2}\\) \u00d7 k \u00d7 \\(\\frac{\\mathrm{k}}{5}\\)
\nGiven area of \u2206 OAB = 10
\n\u2234 \\(\\frac{\\mathrm{k}^{2}}{10}\\) = 10
\nk2<\/sup> = 100 \u21d2 k = \u00b110
\n\u2234 The required equation of the straight line is
\nx + 5y = \u00b110<\/p>\n

Question 17.
\nFind the image of the point (- 2, 3) about the line x + 2y – 9 = 0.
\nAnswer:
\nThe image of the point (x1<\/sub>, y1<\/sub>) about the line ax + by + c = 0 is
\n\"Samacheer
\n\u2234 The image of the point (- 2, 3) about the line x + 2y – 9 = 0 is
\n\"Samacheer
\n\u2234 The required point is (0, 7)<\/p>\n

\"Samacheer<\/p>\n

Question 18.
\nA photocopy store charges \u20b9 1.50 per copy for the first 10 copies and \u20b9 1.00 per copy after the 10th<\/sup> copy. Let x be the number of copies, and y be the total cost of photo coping.
\n(i) Draw graph the cost as x goes from 0 to 50 copies
\n(ii) Find the cost of making 40 copies.
\nAnswer:
\n(i) Draw graph of the cost as x goes from 0 to 50 copies:
\nLet x represent the number of copies and y represent the cost of photocopying.
\nGiven photo copying charge for 1 copy is Rs. 1.50 for the first 10 copies and Rs. 1.00 per copy after the 10th<\/sup> copy.
\n\u2234 The relation connecting the number of copies and cost of photocopying charge is given by
\ny = 1.50x, 0 \u2264 x \u2264 10
\ny = 10(1.50) + (x – 10) (1)
\ny = 15 + x – 10
\ny = x + 5 ………… (1)
\n\"Samacheer
\nThe graph for 0 to 50 copies:
\n\"Samacheer
\nWhen x = 10, y = 1.50 \u00d7 x
\n\u21d2 y = 1.50 \u00d7 10 = 15
\nThe corresponding point is (10 , 15)
\nWhen x = 20, y = x + 5
\n\u21d2 y = 20 + 5 = 25
\nThe corresponding point is (20 , 25)
\nWhen x = 30, y = x + 5
\n\u21d2 y = 30 + 5 = 35
\nThe corresponding point is (30, 35)
\nWhen x = 40, y = 40 + 5 = 45
\nThe corresponding point is (40, 45 )
\nWhen x = 50, y = 50 + 5 = 55
\nThe corresponding point is (50, 55 )
\nThe cost of 40 copies is the value of y
\nWhen x = 40 , y = 40 + 5 = 45 rupees
\nCost of 40 copies = 45 rupees<\/p>\n

\"Samacheer<\/p>\n

Question 19.
\nFind at least two equations of the straight lines in the family of file lines y = 5x + b for which b and the x – coordinate of the point of intersection of the lines with 3x – 4y = 6 are integers.
\nAnswer:
\nThe equations of the given straight lines are
\ny = 5x + b ……….. (1)
\n3x – 4y = 6 ……….. (2)
\nTo find atleast two equations from the family y = 5x + b for which b is an integer and x – coordinate of the point of intersection of (1) and (2) is an integer. Solving (1)and (2) using equation (1) inequation (2) (2) \u21d2 3x – 4 (5x + b) = 6
\n3x – 20x – 4b = 6
\n-17x = 6 + 4b
\n\"Samacheer
\nThe corresponding equation of the line is = 5x + 7
\nWhen b = – 10, we have x = \\(\\frac{6-40}{-17}\\)
\n= \\(\\frac{-34}{-17}\\) = 2
\nThe corresponding equation of the line is y = 5x – 10
\nThus y = 5x + 7 and y = 5x – 10 are the two straight lines belonging to the family such that b is an integer and the x – coordinate of the point of intersection with the line (2) is an integer.<\/p>\n

Question 20.
\nFind all the equations of the straight lines in the family of the lines y = mx – 3 for which m and the x – coordinate of the point of intersection of the lines with x – y = 6 are integers.
\nAnswer:
\nThe equations of the given lines are
\ny = mx – 3 ………. (1)
\nx – y = 6 ………. (2)
\nSolving equations (1) and (2)
\n(2) \u21d2 x – (mx – 3 ) = 6
\nx – mx + 3 = 6
\nx (1 – m) = 3
\nx = \\(\\frac{3}{1-\\mathrm{m}}\\) …….. (3)
\nFrom equation (3) let us find the values of x and m for which they are integers. The only values of m for which , x is an integer are m = 0, 2, -2
\nWhen m = 0, x = \\(\\frac{3}{1-0}\\) = 3
\nThe corresponding equation is
\ny = 0 . x – 3
\ny + 3 = 0<\/p>\n

\"Samacheer<\/p>\n

When m = 2, x = \\(\\frac{3}{1-2}\\) = \\(\\frac{3}{-1}\\) = – 3
\nThe corresponding equation is y = -2x + 3
\n2x + y – 3 = 0<\/p>\n

When m = – 2, x = \\(\\frac{3}{1+2}\\) = \\(\\frac{3}{3}\\) = 1
\nThe corresponding equation is
\ny = – 2 x + 3
\n2x + y – 3 = 0
\n\u2234 The required equations of the lines are
\ny + 3 = 0 , 2x – y – 3 = 0 and
\n2x + y – 3 = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

Tamilnadu State Board New Syllabus\u00a0Samacheer Kalvi 11th Maths Guide Pdf Chapter 6 Two Dimensional Analytical Geometry Ex 6.3 Text Book Back Questions and Answers, Notes. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.3 Question 1. Show that the lines 3x + 2y + 9 = 0 and 12x + …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":false,"jetpack_social_options":{"image_generator_settings":{"template":"highway","enabled":false},"version":2}},"categories":[6],"tags":[],"class_list":["post-24103","post","type-post","status-publish","format-standard","hentry","category-class-11"],"jetpack_publicize_connections":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/24103"}],"collection":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/comments?post=24103"}],"version-history":[{"count":1,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/24103\/revisions"}],"predecessor-version":[{"id":41265,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/posts\/24103\/revisions\/41265"}],"wp:attachment":[{"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/media?parent=24103"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/categories?post=24103"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalvi.guide\/wp-json\/wp\/v2\/tags?post=24103"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}