{"id":2763,"date":"2023-11-30T12:31:42","date_gmt":"2023-11-30T07:01:42","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=2763"},"modified":"2023-12-01T12:31:51","modified_gmt":"2023-12-01T07:01:51","slug":"samacheer-kalvi-11th-maths-solutions-chapter-6-ex-6-4","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-11th-maths-solutions-chapter-6-ex-6-4\/","title":{"rendered":"Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4"},"content":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4<\/h2>\n

Question 1.
\nFind the combined equation of the straight lines whose separate equations are x – 2y – 3 = 0 and x + y + 5 = 0.
\nSolution:
\nSeparate equations are x – 2y – 3 = 0; x + y + 5 = 0
\nSo the combined equation is (x – 2y – 3) (x + y + 5) = 0
\nx2<\/sup> + xy + 5x – 2y2<\/sup> – 2xy – 10y – 3x – 3y – 15 = 0
\n(i.e) x2<\/sup> – 2y2<\/sup> – xy + 2x – 13y – 15 = 0<\/p>\n

Question 2.
\nShow that 4x2<\/sup> + 4xy + y2<\/sup> – 6x – 3y – 4 = 0 represents a pair of parallel lines.
\nSolution:
\nComparing this equation with ax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0
\nwe get a = 4, h = \\(\\frac{4}{2}\\) = 2 , b = 1, g = – 3, f = – 3\/2, c = – 4
\nThe condition for the lines to be parallel is h2<\/sup> – ab = 0
\nNow h2<\/sup> – ab = 22<\/sup> – (4) (1) = 4 – 4 = 0
\nh2<\/sup> – ab = 0 \u21d2 The given equation represents a pair of parallel lines.<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nShow that 2x2<\/sup> + 3xy – 2y2<\/sup> + 3x + y + 1 = 0 represents a pair of perpendicular lines.
\nSolution:
\nComparing the given equation with the general form a = 2,h = 3\/2, b = -2,g= 3\/2, f = 1\/2 and c = 1
\nCondition for two lines to be perpendicular is a + b = 0. Here a + b = 2 – 2 = 0
\n\u21d2 The given equation represents a pair of perpendicular lines.<\/p>\n

Question 4.
\nShow that the equation 2x2<\/sup> – xy – 3y2<\/sup> – 6x + 19y – 20 = 0 represents a pair of intersecting lines. Show further that the angle between them is tan-1<\/sup>(5).
\nSolution:
\n\"Samacheer
\nThe given equation represents a pair of straight lines.
\n\"Samacheer<\/p>\n

Question 5.
\nProve that the equation to the straight lines through the origin, each of which makes an angle \u03b1 with the straight line y = x is x2<\/sup> – 2xy sec 2\u03b1 + y2<\/sup> = 0
\nSolution:
\nSlope of y = x is m = tan \u03b8 = 1
\n\u21d2 \u03b8 = 45\u00b0
\nThe new lines slopes will be
\nm = tan(45 + \u03b1) and m = tan (45 – \u03b1)
\n\u2234 The equations of the lines passing through the origin is given by
\ny = tan(45 + \u03b1)x and y = tan(45 – \u03b1)x
\n(i.e) y = tan(45 + \u03b1)x = 0 and y = tan(45 – \u03b1)x = 0
\nThe combined equation is [y – tan (45 + \u03b1)x] [y – tan (45 – \u03b1)x] = 0
\ny2<\/sup> + tan(45 + \u03b1)tan(45 – \u03b1)x2<\/sup> – xy[tan(45 – \u03b1) + tan(45 + \u03b1)] = 0
\n\"Samacheer
\n\"Samacheer
\nLet the equation of lines passes through the origin
\nSo the equations are y = m1<\/sub>x = 0 and y = m2<\/sub>x = 0
\nSo the combined equations is (y – m1<\/sub>x) (y – m2<\/sub>x) = 0
\n(i.e)y2 <\/sup>– xy(m1<\/sub> + m2<\/sub>) + m1<\/sub>m2<\/sub>x = 0
\n(i.e) y2 <\/sup>– xy(2sec \u03b1) + x2<\/sup>(1) = 0
\n(i.e) y2<\/sup> – 2xy sec 2\u03b1 + x2<\/sup> = 0<\/p>\n

Question 6.
\nFind the equation of the pair of straight lines passing through the point (1, 3) and perpendicular to the lines 2x – 3y + 1 = 0 and 5x + y – 3 = 0
\nSolution:
\nEquation of a line perpendicular to 2x – 3y + 1 = 0 is of the form 3x + 2y + k = 0.
\nIt passes through (1, 3) \u21d2 3 + 6 + k = 0 \u21d2 k = – 9
\nSo the line is 3x + 2y – 9 = 0
\nThe equation of a line perpendicular to 5x + y – 3 = 0 will be of the form x – 5y + k = 0.
\nIt passes through (1, 3) \u21d2 1 – 15 + k = 0 \u21d2 k = 14
\nSo the line is x – 5y + 14 = 0.
\nThe equation of the lines is 3x + 2y – 9 = 0 and x – 5y + 14 = 0
\nTheir combined equation is (3x + 2y – 9)(x – 5y + 14) = 0
\n(i.e) 3x2<\/sup> – 15xy + 42x + 2xy – 10y2<\/sup> + 28y – 9x + 45y – 126 = 0
\n(i.e) 3x2<\/sup> – 13xy – 10y2<\/sup> + 33x + 73y – 126 = 0<\/p>\n

Question 7.
\nFind the separate equation of the following pair of straight lines
\n(i) 3x2<\/sup> + 2xy – y2<\/sup> = 0
\n(ii) 6 (x – 1)2<\/sup> + 5(x – 1)(y – 2) – 4(y – 2)2<\/sup> = 0
\n(iii) 2x2<\/sup> – xy – 3y2<\/sup> – 6x + 19y – 20 = 0
\nSolution:
\n(i) Factorising 3x2<\/sup> + 2xy – y2<\/sup> we get
\n3x2<\/sup> + 3xy – xy – y2<\/sup> = 3x (x + y) – y (x + y)
\n= (3 x – y)(x + y)
\nSo 3x2<\/sup> + 2xy – y2<\/sup> = 0 \u21d2 (3x – y) (x + y) = 0
\n\u21d2 3x – y = 0 and x + y = 0<\/p>\n

\"Samacheer<\/p>\n

(ii) 6 (x – 1)2<\/sup> + 5 (x – 1)(y – 2) – 4(y – 2)2<\/sup> = 0
\n\u21d2 6(x2<\/sup> – 2x +1) + 5(xy – 2x – y + 2) – 4( y2<\/sup> – 4y + 4) = 0
\n(i.e) 6x2<\/sup> – 12x + 6 + 5xy – 10x – 5y + 10 – 4y2<\/sup> + 16y – 16 = 0
\n(i.e) 6x2<\/sup> + 5xy – 4y2<\/sup> – 22x + 11y = 0
\nFactorising 6x2<\/sup> + 5xy – 4y2<\/sup> we get
\n6x2<\/sup> – 3xy + 8xy – 4y2<\/sup> = 3x (2x – y) + 4y (2x – y)
\n= (3x + 4y)(2x – y)
\nSo, 6x2<\/sup> + 5xy – 4y2<\/sup> – 22x + 11y = (3x + 4y + l )(2x – y + m)
\nEquating coefficient of x \u21d2 3m + 21 = -22 …….. (1)
\nEquating coefficient of y \u21d2 4m – l = 11 ……. (2)
\nSolving (1) and (2) we get l = -11, m = 0
\nSo the separate equations are 3x + 4y – 11 = 0 and 2x – y = 0<\/p>\n

(iii) 2x2<\/sup> – xy – 3y2<\/sup> – 6x + 19y – 20 = 0
\nFactorising 2x2<\/sup> – xy – 3y2<\/sup> we get
\n2x2<\/sup> – xy – 3y2<\/sup> = 2x2<\/sup> + 2xy – 3xy – 3y2<\/sup>
\n= 2x(x + y) – 3y(x + y) = (2x – 3y) (x + y)
\n\u2234 2x2<\/sup> – xy – 3y2<\/sup> – 6x + 19y – 20 = (2x – 3y + l)(x + y + m)
\nEquating coefficient of x 2m + l = -6 ……. (1)
\nEquating coefficient of y -3m + l = 19 …….. (2)
\nConstant term -20 = lm
\nSolving (1) and (2) we get l = 4 and m = – 5 where lm = – 20.
\nSo the separate equations are 2x – 3y + 4 = 0 and x + y – 5 = 0<\/p>\n

Question 8.
\nThe slope of one of the straight lines ax2<\/sup> + 2hxy + by2<\/sup> = 0 is twice that of the other, show that 8h2<\/sup> = 9ab.
\nSolution:
\nax2<\/sup> + 2 hxy + by2<\/sup> = 0
\nWe are given that one slope is twice that of the other.
\nSo let the slopes be m and 2m.
\nNow sum of the slopes = m + 2m
\n\"Samacheer<\/p>\n

Question 9.
\nThe slope of one of the straight lines ax2<\/sup> + 2hxy + by2<\/sup> = 0 is three times the other, show that 3h2<\/sup> = 4ab.
\nSolution:
\nLet the slopes be m and 3m.
\n\"Samacheer<\/p>\n

Question 10.
\nA \u2206OPQ is formed by the pair of straight lines x2<\/sup> – 4xy + y2<\/sup> = 0 and the line PQ. The equation of PQ is x + y – 2 = 0. Find the equation of the median of the triangle \u2206OPQ drawn from the origin O.
\nSolution:
\nEquation of pair of straight lines is x2<\/sup> – 4xy + y2<\/sup> = 0 ….. (1)
\nEquation of the given line is x + y – 2 = 0 \u21d2 y = 2 – x ……… (2)
\nOn solving (1) and (2) we get x2<\/sup> – 4x (2 – x) + (2 – x)2<\/sup> = 0
\n(i.e) x2<\/sup> – 8x + 4x2<\/sup> + 4 + x2<\/sup> – 4x = 0
\n(i.e) 6x2<\/sup> – 12x + 4 = 0
\n(\u00f7 by 2) 3x2<\/sup> – 6x + 2 = 0
\n\"Samacheer
\nMid point of PQ is
\n\"Samacheer<\/p>\n

Question 11.
\nFind p and q, \u00a1f the following equation represents a pair of perpendicular lines 6x2<\/sup> + 5xy – py2<\/sup> + 7x + qy – 50
\nSolution:
\n6x2<\/sup> + 5xy – py2<\/sup> + 7x + qy – 50
\nThe given equation represents a pair of perpendicular lines
\n\u21d2 coefficient of x2<\/sup> + coefficient of y2<\/sup> = 0
\n(i.e) 6 – p = 0 \u21d2 p = 6
\nNow comparing the given equation with the general form
\nax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0
\nwe get a = 6, b = -6 and c = -5, f = q\/2, g = 7\/2 and h = 5\/2
\nThe condition for the general form to represent a pair of straight lines is abc + 2fgh – af2<\/sup> – bg2<\/sup> – ch2<\/sup> = 0
\n\"Samacheer<\/p>\n

Question 12.
\nFind the value of k, if the following equation represents a pair of straight lines. Further, find whether these lines are parallel or intersecting, 12x2<\/sup> + 7xy – 12y2<\/sup> – x + 7y + k = 0.
\nSolution:
\nComparing the given equation with the general form ax2<\/sup> + 2hxy + by2<\/sup> + 2gx + 2fy + c = 0
\nwe get a = 12, b = -12, c = k, f = 7\/2, g = – 1\/2, h = 7\/2
\nHere a + b = 0 \u21d2 the given equation represents a pair of perpendicular lines
\nTo find k: The condition for the given equation to represent a pair of straight lines is abc + 2fgh – af2<\/sup> – bg2<\/sup> – ch2<\/sup> = 0
\n\"Samacheer
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 13.
\nFor what value of k does the equation 12x2<\/sup> + 2kxy + 2y2<\/sup>, + 11x – 5y + 2 = 0 represent two straight lines.
\nSolution:
\n12x2<\/sup> + 2 kxy + 2y2<\/sup> + 11x – 5y + 2 = 0
\nComparing this equation with the general form we get
\n\"Samacheer
\n4k2<\/sup> + 55k + 175 = 0
\n4k2<\/sup> + 20k + 35k + 175 = 0
\n4k(k + 5) + 35(k + 5) = 0
\n(4k + 35) (k + 5) = 0
\nk = -5 or -35\/4<\/p>\n

Question 14.
\nShow that the equation 9x2<\/sup> – 24xy + 16y2<\/sup> – 12x + 16y – 12 = 0 represents a pair of parallel lines. Find the distance between them.
\nSolution:
\nComparing the given equation with ax2<\/sup> + 2kxy + by2<\/sup> = 0 we get a = 9, h = -12, b = 16.
\nNow h2<\/sup> = (-12)2<\/sup> = 144, ab = (9) (16) = 144
\nh2<\/sup> = ab \u21d2 The given equation represents a pair of parallel lines.
\nTo find their separate equations:
\n9x2<\/sup> – 24xy + 16y2<\/sup> = (3x – 4y)2<\/sup>
\nSo, 9x2<\/sup> – 24xy +16y2<\/sup> – 12x + 16y – 12 = (3x – 4y + l )(3x – 4y + m)
\nHere coefficient of x \u21d2 3m + 3l = -12 \u21d2 m + l = -4
\ncoefficient of y \u21d2 -4m – 4l = 16 \u21d2 m + l = -4
\nConstant term l m = -12
\nNow l + m = -4 and lm = -12 \u21d2 l = -6 and m = 2
\nSo the separate equations are 3x – 4y – 6 = 0 and 3x – 4y + 2 = 0
\n\"Samacheer<\/p>\n

Question 15.
\nShow that the equation 4x2<\/sup> + 4xy + y2<\/sup> – 6x – 3y – 4 = 0 represents a pair of parallel lines. Find the distance between them.
\nSolution:
\n4x2<\/sup> + 4xy + y2<\/sup> – 6x – 3y – 4 = 0
\na = 4,
\nb = 1,
\nh = 4\/2 = 2
\nh2<\/sup> – ab = 22<\/sup> – (4) (1) = 4 – 4 = 0
\n\u21d2 The given equation represents a pair of parallel lines.
\nTo find the separate equations 4x2<\/sup> + 4xy + y2<\/sup> = (2x + y)2<\/sup>
\nSo, 4x2<\/sup> + 4xy + y2<\/sup> – 6x – 3y – 4 = (2x + y + l )(2x + y + m)
\nCoefficient of x \u21d2 2m + 2l = -6 \u21d2 l + m = – 3 ……. (1)
\nCoefficient of y \u21d2 l + m = – 3 ……… (2)
\nConstant term \u21d2 l m = – 4 ……… (3)
\nNow l + m = -3 and lm = -4 \u21d2 l = -4, m = 1
\nSo the separate equations are 2x + y + 1 = 0 and 2x + y – 4 = 0
\n\"Samacheer<\/p>\n

Question 16.
\nProve that one of the straight lines given by ax2<\/sup> + 2hxy + by2<\/sup> = 0 will bisect the angle between the co-ordinate axes if (a + b)2<\/sup> = 4h2<\/sup>.
\nSolution:
\nLet the slopes be l and m
\n\u2235 One line bisects the angle between the coordinate axes \u21d2 \u03b8 = 45\u00b0
\nSo tan \u03b8 = 1
\nThe slopes are l and m
\n\"Samacheer<\/p>\n

Question 17.
\nIf the pair of straight lines x2<\/sup> – 2kxy – y2<\/sup> = 0 bisect the angle between the pair of straight lines x2<\/sup> – 2lxy – y2<\/sup> = 0, show that the later pair also bisects the angle between the former.
\nSolution:
\nGiven that x2<\/sup> – 2kxy – y2<\/sup> = 0 …….. (1)
\nBisect the angle between the lines x2<\/sup> – 11xy – y2<\/sup> = 0 …… (2)
\n\"Samacheer
\nx2<\/sup> – 2kxy – y2<\/sup> = 0<\/p>\n

Question 18.
\nProve that the straight lines joining the origin to the points of intersection of 3x2<\/sup> + 5xy – 3y2<\/sup> + 2x + 3y = 0 and 3x – 2y – 1 = 0 are at right angles.
\nSolution:
\nHomogenizing the given equations 3x2<\/sup> + 5xy – 3y2<\/sup> + 2x + 3y = 0 and 3x – 2y – 1 = 0
\n(i.e) 3x – 2y = 1.
\nWe get (3x2<\/sup> + 5xy – 3y2<\/sup>) + (2x + 3y)( 1) = 0
\n(i.e) (3x2<\/sup> + 5xy – 3y2<\/sup>) + (2x + 3y)(3x – 2y) = 0
\n3x2<\/sup> + 5xy – 3y2<\/sup> + bx2<\/sup> – 4xy + 9xy – 6y2<\/sup> = 0
\n9x2<\/sup> + 10xy – 9y2<\/sup> = 0
\nCoefficient of x2<\/sup> + coefficient of y2<\/sup> = 9 – 9 = 0
\n\u21d2 The pair of straight lines are at right angles.<\/p>\n

Samacheer Kalvi 11th Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Ex 6.4 Additional Questions Solved<\/h3>\n

Question 1.
\nFind the angle between the pair of straight lines given by
\n(a2<\/sup> – 3b2<\/sup> )x2<\/sup> + 8ab xy + (b2<\/sup> – 3a2<\/sup>)y2<\/sup> =0 .
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nShow that 9x2<\/sup> + 24xy + 16y2<\/sup> + 21x + 28y + 6 = 0 represents a pair of parallel straight lines and find the distance between them.
\nSolution:
\n9x2<\/sup> + 24xy + 16y2<\/sup> + 21x + 28y + 6 = 0
\nHere a = 9.6,
\nb = 16,
\ng = \\(\\frac{21}{2}\\),
\nf = 14,
\nc = 6,
\nh = 12
\nh2<\/sup> – ab = (12)2<\/sup> – 9(16) = 144 – 144 = 0
\n\u2234 The lines are parallel.
\n9x2<\/sup> + 24xy + 16y2<\/sup> = (3x + 4y)(3x + 4y)
\nLet 9x2<\/sup> + 24xy + 16y2<\/sup> + 21x + 28y + 6 = (3x + 4y + l)(3x + 4y + m)
\nEquating the coefficients of x and constant term
\n3l + 3m = 21
\nlm = 6
\nSolving we get, l = 1 or 6
\nm = 6 or 1
\n\u2234 The separate equations are 3x + 4y + 1 = 0 and 3x + 4y + 6 = 0
\n\"Samacheer<\/p>\n

Question 3.
\nIf the equation 12x2<\/sup> – 10xy + 2y2<\/sup> + 14x – 5y + c = 0 represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle
\nbetween them.
\nSolution:
\n12x2<\/sup> – 10xy – 2y2<\/sup> + 14x – 5y + c = 0
\nax2<\/sup> + 2hxy + by2<\/sup> +2gx + 2fy – c = 0
\nHere a = 12,
\nb = 2,
\ng = 7,
\nf = 5\/2,
\nc = c,
\nh = -5
\naf2<\/sup> + bg2<\/sup> + ch2<\/sup> – 2fgh – abc = 0 is the condition
\n\"Samacheer
\nThe equation is 12x2<\/sup> – 10y + 2y2<\/sup> + 14x – 5y + 2 = 0
\n12x2<\/sup> – 10xy + 2y = (3x – y)(4x – 2y)
\nLet 12x2<\/sup> – 10y + 2y2<\/sup> + 14x – 5y + 2(3x – y + l)(4x – 2y + m)
\nSo that 4l + 3m = 14 , -2l – m = -5
\n\"Samacheer<\/p>\n

Question 4.
\nFor what value of k does 12x2<\/sup> + 7xy + ky2<\/sup> + 13x – y + 3 = 0 represents a pair of straight lines? Also write the separate equations.
\nSolution:
\n12x2<\/sup> + 7xy + ky2<\/sup> + 13x – y + 3 = 0
\na = 12,
\nh = \\(\\frac{7}{2}\\),
\nf = \\(-\\frac{1}{2}\\) ,
\nc = 3
\naf2<\/sup> + bg2<\/sup> + ch2<\/sup> – abc – 2fgh = 0
\n\"Samacheer
\n\u21d2 12 + 169k + 147 – 144k + 91 = 0
\n25k = – 250 \u21d2 k = -10
\nThe equation is 12x2<\/sup> + 7xy – 10y2<\/sup> + 13x – y + 3 = 0
\nTo find separate equations: 12x2<\/sup> + 7xy – 10y2<\/sup> = (3x – 2y)(4x + 5y)
\nLet 12x2<\/sup> + 7xy – 10y2<\/sup> + 13x – y + 3 = 0(3x – 2y + l)(4x + 5y + m)
\nEquating the coefficient of x \u21d2 4l + 3m = 13 …… (1)
\nEquating the coefficient of y \u21d2 5l – 2m = -1 …… (2)
\n(1) \u00d7 2 \u21d2 8l + 6m = 26
\n(2) \u00d7 3 \u21d2 15l – 6m = -3
\n23l = 23 \u21d2 l = 1
\n4 + 3 m = 13
\n3 m = 9 \u21d2 m = 3
\nThe separate equations are 3x – 2y + 1 = 0 and 4x + 5y + 3 = 0<\/p>\n

\"Samacheer<\/p>\n

Question 5.
\nShow that 3x2<\/sup> + 10xy + 8y2<\/sup> + 14x + 22y + 15 = 0 represents a pair of straight lines and the angle between them is tan-1<\/sup>\\(\\left(\\frac{2}{11}\\right)\\)
\nSolution:
\n3x2<\/sup> + 10xy + 8y2<\/sup> + 14x + 22y + 15 = 0
\na = 3,
\nA = 5,
\nb = 8,
\ng = 7,
\nf = 11,
\nc = 15
\nThe condition is af2<\/sup> + bg2<\/sup> + ch2<\/sup> – abc – 2fgh = 0
\n3(11)2<\/sup> + 8(7)2<\/sup> + 15 (5)2<\/sup> – (3)(8)(15) – 2(11)(7)(5) = 363 + 392 + 375 – 360 – 770 = 0
\nThe angle between the pair of straight line is given by
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

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