{"id":10641,"date":"2024-02-29T09:32:03","date_gmt":"2024-02-29T04:02:03","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=10641"},"modified":"2024-03-05T12:53:07","modified_gmt":"2024-03-05T07:23:07","slug":"ts-inter-first-year-maths-1b-locus-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-first-year-maths-1b-locus-important-questions\/","title":{"rendered":"TS Inter First Year Maths 1B Locus Important Questions"},"content":{"rendered":"

Students must practice these Maths 1B Important Questions<\/a> TS Inter First Year Maths 1B Locus Important Questions to help strengthen their preparations for exams.<\/p>\n

TS Inter First Year Maths 1B Locus Important Questions<\/h2>\n

Question 1.
\nFind the equation of the locus of a point that is at a distance 5 from (-2, 3), in the XOY plane. [Mar. ’02; May ’95]
\nSolution:
\nLet A (-2, 3) be the given point.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA = 5
\n\"TS
\nSquaring on both sides
\n\u21d2 13 + x2<\/sup> + y2<\/sup> + 4x – 6y = 25
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 6y = 25 – 13
\n\u21d2 x2<\/sup> + y2<\/sup> + 4x – 6y = 12
\n\u2234 The equation of locus of P(x, y) is x2<\/sup> + y2<\/sup> + 4x – 6y = 12.<\/p>\n

Question 2.
\nFind the equation of the locus of a point P such that the distance of P from the origin is twice the distance of P from A(1, 2). [Mar. ’12; May ’05, ’97]
\nSolution:
\nO(0, 0), A(1, 2) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is OP = 2PA
\n\"TS
\n\u21d2 x2<\/sup> + y2<\/sup> = 4[(x – 1)2<\/sup> + (y – 2)2<\/sup>]
\n\u21d2 x2<\/sup> + y2<\/sup> = 4(x2<\/sup> – 2x + 1 + y2<\/sup> – 4y + 4)
\n\u21d2 x2<\/sup> + y2<\/sup> = 4x2<\/sup> – 8x + 4 + 4y2<\/sup> – 16y + 16
\n\u21d2 3x2<\/sup> + 3y2<\/sup> – 8x – 16y + 20 = 0
\n\u2234 The equation of the locus of P(x, y) is 3x2<\/sup> + 3y2<\/sup> – 8x – 16y + 20 = 0.<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the equation of the locus of a point equidistant from A(2, 0) and the Y-axis. (May ’03)
\nSolution:
\nA(2, 0) is the given point.
\nLet P(x, y) be any point on the locus.
\n\"TS
\nThe distance of P from Y-axis = PN = |x|
\nThe given geometric condition is PA = |PN|
\n\\(\\sqrt{(x-2)^2+(y-0)^2}=|x|\\)
\nSquaring on both sides
\n\u21d2 \\(\\left(\\sqrt{(x-2)^2+(y-0)^2}\\right)^2=(|x|)^2\\)
\n\u21d2 (x – 2)2<\/sup> + (y)2<\/sup> = x2<\/sup>
\n\u21d2 x2<\/sup> – 4x + 4 + y2<\/sup> = x2<\/sup>
\n\u21d2 y2<\/sup> – 4x + 4 = 0
\n\u2234 The equation of the locus of P(x, y) is y2<\/sup> – 4x + 4 = 0.<\/p>\n

Question 4.
\nFind the equation of the locus of a point P, the square whose distance from the origin is 4 times its y-coordinate. [Mar. ’00]
\nSolution:
\nO(0, 0) is the origin.
\nLet P(x, y) be any point on the locus.
\nThe distance of P from the X-axis = y.
\n\"TS
\nThe given geometric condition is OP2<\/sup> = 4 PM
\n\u21d2 \\(\\left(\\sqrt{(x-0)^2+(y-0)^2}\\right)^2=4 y\\)
\n\u21d2 x2<\/sup> + y2<\/sup> = 4y
\n\u21d2 x2<\/sup> + y2<\/sup> – 4y = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> – 4y = 0.<\/p>\n

Question 5.
\nFind the equation of locus of a point P such that PA2<\/sup> + PB2<\/sup> = 2c2<\/sup>, where A = (a, 0), B = (-a, 0), and 0 < |a| < |c|. [Mar. ’00]
\nSolution:
\nA = (a, 0), B = (-a, 0) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA2<\/sup> + PB2<\/sup> = 2c2<\/sup>
\n\u21d2 \\(\\left[\\sqrt{(x-a)^2+(y-0)^2}\\right]^2\\) + \\(\\left[\\sqrt{(x+a)^2+(y-0)^2}\\right]^2\\) = 2c2<\/sup>
\n\u21d2 (x – a)2<\/sup> + y2<\/sup> + (x + a)2<\/sup> + y2<\/sup> = 2c2<\/sup>
\n\u21d2 x2<\/sup> – 2ax + a2<\/sup> + y2<\/sup> + x2<\/sup> + 2ax + a2<\/sup> + y2<\/sup> = 2c2<\/sup>
\n\u21d2 2x2<\/sup> + 2y2<\/sup> + 2a2<\/sup> = 2c2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> + a2<\/sup> = c2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> = c2<\/sup> – a2<\/sup>
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> = c2<\/sup> – a2<\/sup><\/p>\n

Question 6.
\nFind the equation of locus of P, if the line segment joining (2, 3) and (-1, 5) subtends a right angle at P. [Mar. ’13, ’05, ’03; May ’12, ’04, ’03, ’02]
\nSolution:
\nLet A(2, 3), B(-1, 5) are the given points.
\nLet P(x, y) be any point on the locus
\nThe given geometric condition is \u2220APB = 90\u00b0
\n\"TS
\nThen, PA2<\/sup> + PB2<\/sup> = AB2<\/sup>
\n\u21d2 \\(\\left[\\sqrt{(x-2)^2+(y-3)^2}\\right]^2+\\left[\\sqrt{(x+1)^2+(y-5)^2}\\right]^2\\) = \\(\\left[\\sqrt{(2+1)^2+(3-5)^2}\\right]^2\\)
\n\u21d2 (x – 2)2<\/sup> + (y – 3)2<\/sup> + (x + 1)2<\/sup> + (y – 5)2<\/sup> = (2 + 1)2<\/sup> + (3 – 5)2<\/sup>
\n\u21d2 x2<\/sup> + 4 – 4x + y2<\/sup> + 9 – 6y + x2<\/sup> + 1 + 2x + y2<\/sup> + 25 – 10y = 9 + 4
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 2x – 16y + 26 = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – x – 8y + 13 = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> – x – 8y + 13 = 0.<\/p>\n

\"TS<\/p>\n

Question 7.
\nFind the equation of the locus of P, if A = (4, 0), B = (-4, 0) and |PA – PB| = 4. [May ’13]
\nSolution:
\nA = (4, 0), B = (-4, 0) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is |PA – PB| = 4
\n\u21d2 PA – PB = \u00b14
\n\u21d2 PA = \u00b14 + PB
\nSquaring on both sides
\n\u21d2 (PA)2<\/sup> = (\u00b14 + PB)2<\/sup>
\n\u21d2 PA2<\/sup> = 16 + PB2<\/sup> \u00b1 8PB
\n\u21d2 (x – 4)2<\/sup> + (y – 0)2<\/sup> = 16 + (x + 4)2<\/sup> + (y – 0)2<\/sup> \u00b1 8PB
\n\u21d2 x2<\/sup> – 8x + 16 + y2<\/sup> = 16 + x2<\/sup> + 8x + 16y2<\/sup> \u00b1 8PB
\n\u21d2 -8x = 8x + 16 \u00b1 8PB
\n\u21d2 -8x – 8x – 16 = \u00b18PB
\n\u21d2 -16 – 16x = \u00b18PB
\n\u21d2 -2 – 2x = \u00b1PB
\nSquaring on both sides
\n\u21d2 (-2 – 2x)2<\/sup> = (\u00b1PB)2<\/sup>
\n\u21d2 4 + 8x + 4x2<\/sup> = PB2<\/sup>
\n\u21d2 4 + 4x2<\/sup> + 8x = (x + 4)2<\/sup> + (y – 0)2<\/sup>
\n\u21d2 4 + 4x2<\/sup> + 8x = x2<\/sup> + 8x + 16 + y2<\/sup>
\n\u21d2 3x2<\/sup> – y2<\/sup> – 12 = 0
\n\u2234 The equation of the locus of P(x, y) is 3x2<\/sup> – y2<\/sup> – 12 = 0.<\/p>\n

Question 8.
\nFind the equation of the locus of a point, the sum of whose distances from (0, 2) and (0, -2) is 6. [Mar. ’16 (TS)]
\nSolution:
\nLet A = (0, 2), B = (0, -2) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA + PB = 6
\n\u21d2 PA = 6 – PB
\nSquaring on both sides
\n\u21d2 PA2<\/sup> = (6 – PB)2<\/sup>
\n\u21d2 PA2<\/sup> = 36 – 12PB + PB2<\/sup>
\n\u21d2 (x – 0)2<\/sup> + (y – 2)2<\/sup> = 36 + (x – 0)2<\/sup> + (y + 2)2<\/sup> – 12PB
\n\u21d2 x2<\/sup> + y2<\/sup> + 4 – 4y = 36 + x2<\/sup> + y2<\/sup> + 4 + 4y – 12PB
\n\u21d2 -4y = 36 + 4y – 12PB
\n\u21d2 12PB = 36 + 4y + 4y
\n\u21d2 12PB = 36 + 8y
\n\u21d2 3PB = 9 + 2y
\nSquaring on both sides
\n\u21d2 (3PB)2<\/sup> = (9 + 2y)2<\/sup>
\n\u21d2 9PB2<\/sup> = (9 + 2y)2<\/sup>
\n\u21d2 9[(x – 0)2<\/sup> + (y + 2)2<\/sup>] = 81 + 36y + 4y2<\/sup>
\n\u21d2 9[x2<\/sup> + y2<\/sup> + 4y + 4] = 81 + 36y + 4y2<\/sup>
\n\u21d2 9x2<\/sup> + 9y2<\/sup> + 36y + 36 = 81 + 36y + 4y2<\/sup>
\n\u21d2 9x2<\/sup> + 9y2<\/sup> + 36 – 81 – 4y2<\/sup> = 0
\n\u21d2 9x2<\/sup> + 5y2<\/sup> – 45 = 0
\n\u2234 The equation of the locus of P(x, y) is 9x2<\/sup> + 5y2<\/sup> – 45 = 0.<\/p>\n

Question 9.
\nA(2, 3) and B(-3, 4) are two given points. Find the equation of the locus of P, so that the area of the triangle PAB is 8.5. [Mar. ’11]
\nSolution:
\nA(2, 3), B(-3, 4) are the points given.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is an area of the triangle, PAB = 8.5
\n\"TS
\n\u2234 The equation of the locus of P(x, y) is (x + 5y) (x + 5y – 34) = 0
\n\u21d2 x2<\/sup> + 5xy – 34x + 5xy + 25y2<\/sup> – 170y = 0
\n\u21d2 x2<\/sup> + 10xy + 25y2<\/sup> – 34x – 170y = 0<\/p>\n

\"TS<\/p>\n

Question 10.
\nFind the equation of the locus of p, if the ratio of the distances from p to A(5, -4) and B(7, 6) is 2 : 3. [Mar. ’14, ’98; May ’08, ’01]
\nSolution:
\nLet A(5, -4), B(7, 6) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA : PB = 2 : 3
\n\u21d2 \\(\\frac{\\mathrm{PA}}{\\mathrm{PB}}=\\frac{2}{3}\\)
\n\u21d2 3PA = 2PB
\n\"TS
\n\u21d2 9[x2<\/sup> – 10x + 25 + y2<\/sup> + 16 + 8y] = 4[x2<\/sup> – 14x + 49 + y2<\/sup> – 12y + 36]
\n\u21d2 9x2<\/sup> – 90x + 225 + 9y2<\/sup> + 144 + 72y = 4x2<\/sup> – 56x + 196 – 4y2<\/sup> – 48y + 144
\n\u21d2 9x2<\/sup> – 90x + 225 + 9y2<\/sup> + 144 + 72y – 4x2<\/sup> + 56x – 196 – 4y2<\/sup> + 48y – 144 = 0
\n\u21d2 5x2<\/sup> + 5y2<\/sup> – 34x + 120y + 29 = 0
\n\u2234 The equation of the locus of P(x, y) is 5x2<\/sup> + 5y2<\/sup> – 34x + 120y + 29 = 0<\/p>\n

Question 11.
\nA(1, 2), B(2, -3) and C(-2, 3) are three points. A point p moves such that PA2<\/sup> + PB2<\/sup> = 2PC2<\/sup>. Show that the equation to the locus of P is 7x – 7y + 4 = 0. [Mar. ’19 (T.S); Mar. ’17 (AP); May ’15 (A.P); ’07]
\nSolution:
\nA(1, 2), B(2, -3), C(-2, 3) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA2<\/sup> + PB2<\/sup> = 2PC2<\/sup>
\n\u21d2 [(x – 1)2<\/sup> + (y – 2)2<\/sup>] + [(x – 2)2<\/sup> + (y + 3)2<\/sup>] = 2[(x + 2)2<\/sup> + (y – 3)2<\/sup>]
\n\u21d2 x2<\/sup> – 2x + 1 + y2<\/sup> + 4 – 4y + x2<\/sup> + 4 – 4x + y2<\/sup> + 9 + 6y = 2[x2<\/sup> + 4 + 4x + y2<\/sup> + 9 – 6y]
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 6x + 2y + 18 = 2x2<\/sup> + 8 + 8x + 2y2<\/sup> + 18 – 12y
\n\u21d2 -6x + 2y = 8 + 8x – 12y
\n\u21d2 8 + 8x – 12y + 6x – 2y = 0
\n\u21d2 14x – 14y + 8 = 0
\n\u21d2 7x – 7y + 4 = 0
\n\u2234 The equation of the locus of P(x, y) is 7x – 7y + 4 = 0.<\/p>\n

Some More Maths 1B Locus Important Questions<\/h3>\n

Question 12.
\nFind the equation of the locus of a point that is at a distance 5 from A(4, -3).
\nSolution:
\nA(4, -3) is the given point.
\nLet P(x, y) be any point on the locus.
\n\"TS
\nThe given geometric condition is PA = 5
\n\u21d2 \\(\\sqrt{(x-4)^2+(y+3)^2}\\) = 5
\nSquaring on both sides
\n\u21d2 \\(\\left(\\sqrt{(x-4)^2+(y+3)^2}\\right)^2\\) = (5)2<\/sup>
\n\u21d2 (x – 4)2<\/sup> + (y + 3)2<\/sup> = 25
\n\u21d2 x2<\/sup> – 8x + 16 + y2 + 9 + 6y = 25
\n\u21d2 x2<\/sup> + y2 – 8x + 6y = 0
\n\u2234 The equation of locus of P(x, y) is x2<\/sup> + y2<\/sup> – 8x + 6y = 0<\/p>\n

Question 13.
\nFind the equation of the locus of a point P, if the distance of P from A(3, 0) is twice the distance of P from B(-3, 0).
\nSolution:
\nA(3, 0), B(-3, 0) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA = 2PB
\n\"TS
\n\u21d2 (x – 3)2<\/sup> + (y – 0)2<\/sup> = 4[(x + 3)2<\/sup> + (y – 0)2<\/sup>]
\n\u21d2 x2<\/sup> – 6x + 9 + y2<\/sup>\u00a0<\/sup>= 4[x2<\/sup> + 6x + 9 + y2<\/sup>]
\n\u21d2 x2<\/sup> – 6x + 9 + y2<\/sup> = 4×2 + 24x + 36 + 4y2<\/sup>
\n\u21d2 4x2<\/sup> + 24x + 36 + 4y2 – x2<\/sup> + 6x – 9 – y2<\/sup> = 0
\n\u21d2 3x2<\/sup> + 3y2<\/sup> + 30x + 27 = 0
\n\u21d2 x2<\/sup> + y2<\/sup>\u00a0+ 10x + 9 = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> + 10x + 9 = 0.<\/p>\n

\"TS<\/p>\n

Question 14.
\nThe ends of the hypotenuse of a right-angled triangle are (0, 6) and (6, 0). Find the equation of the locus of its third vertex. [June ’10; Mar. ’08; Sept. ’00]
\nSolution:
\nLet A(0, 6), B(6, 0) are the given points.
\nLet P(x, y) be the third vertex
\n\"TS
\nThe given geometric condition is \u2220APB = 90\u00b0
\n\u21d2 PA2<\/sup> + PB2<\/sup> = AB2<\/sup>
\n\u21d2 \\(\\left[\\sqrt{(x-0)^2+(y-6)^2}\\right]^2+\\left[\\sqrt{(x-6)^2+(y-0)^2}\\right]^2\\) = \\(\\left[\\sqrt{(0-6)^2+(6-0)^2}\\right]^2\\)
\n\u21d2 x2<\/sup> + (y – 6)2<\/sup> + (x – 6)2<\/sup> + y2<\/sup> = (-6)2<\/sup> + (6)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> – 12y + 36 + x2<\/sup> – 12x + 36 + y2<\/sup> = 36 + 36
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 12x – 12y + 72 = 72
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 12x – 12y = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 6x – 6y = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> – 6x – 6y = 0.<\/p>\n

Question 15.
\nFind the locus of the third vertex of a right-angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4). [Mar. ’13(old); Mar. ’18 (TS)]
\nSolution:
\nLet A(4, 0), B(0, 4) are the given points.
\nLet P(x, y) be the third vertex.
\nThe given geometric condition is \u2220APB = 90\u00b0
\n\u21d2 PA2<\/sup> + PB2<\/sup> = AB2<\/sup>
\n\"TS
\n\u21d2 (x – 4)2<\/sup> + y2<\/sup> + x2<\/sup> + (y – 4)2<\/sup> = (4)2<\/sup> + (-4)2<\/sup>
\n\u21d2 x2<\/sup> – 8x + 16 + y2<\/sup> + x2<\/sup> + y2<\/sup> – 8y + 16 = 16 + 16
\n\u21d2 2x2<\/sup> + 2y2<\/sup> – 8x – 8y = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 4y – 4x = 0
\n\u21d2 x2<\/sup> + y2<\/sup> – 4x – 4y = 0
\n\u2234 The equation of the locus of P(x, y) is x2<\/sup> + y2<\/sup> – 4x – 4y = 0.<\/p>\n

Question 16.
\nFind the equation of the locus of a point, the difference of whose distances from (-5, 0) and (5, 0) is 8.
\nSolution:
\nLet A(-5, 0), B(5, 0) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is |PA – PB| = 8
\n\u21d2 PA – PB = \u00b18
\n\u21d2 PA = \u00b18 + PB
\nSquaring on both sides
\n\u21d2 PA2<\/sup> = (\u00b18 + PB)2<\/sup>
\n\u21d2 PA2<\/sup> = (\u00b18)2<\/sup> + PB2<\/sup> + 2(\u00b18) PB = 64 + PB2<\/sup> \u00b1 16PB
\n\u21d2 (x + 5)2<\/sup> + (y – 0)2<\/sup> = 64 + (x – 5)2<\/sup> + (y – 0)2<\/sup> \u00b1 16PB
\n\u21d2 x2<\/sup> + 25 + 10x + y2<\/sup> = 64 + x2<\/sup> – 10x + 25 + y2<\/sup> \u00b1 16PB
\n\u21d2 10x = 64 – 10x \u00b1 16PB
\n\u21d2 10x – 64 + 10x = \u00b116PB
\n\u21d2 20x – 64 = \u00b116PB
\n\u21d2 5x – 16 = \u00b14PB
\nSquaring on both sides
\n\u21d2 (5x – 16)2<\/sup> = (\u00b14PB)2<\/sup>
\n\u21d2 25x2<\/sup> + 256 – 160x = 16PB2<\/sup>
\n\u21d2 25x2<\/sup> + 256 – 160x = 16[(x – 5)2<\/sup> + (y – 0)2<\/sup>]
\n\u21d2 25x2<\/sup> + 256 – 160x = 16[x2<\/sup> + 25 – 10x + y2<\/sup>]
\n\u21d2 25x2<\/sup> + 256 – 160x = 16x2<\/sup> + 400 – 160x + 16y2<\/sup>
\n\u21d2 25x2<\/sup> + 256 – 16x2<\/sup> – 400 – 16y2<\/sup> = 0
\n\u21d2 9x2<\/sup> – 16y2<\/sup> – 144 = 0
\n\u21d2 9x2<\/sup> – 16y2<\/sup> = 144
\n\u2234 The equation of the locus of P(x, y) is 9×2 – 16y2 = 114.<\/p>\n

\"TS<\/p>\n

Question 17.
\nFind the equation of the locus of P, if A = (2, 3), B = (2, -3), and PA + PB = 8. [Mar. ’08, ’03]
\nSolution:
\nLet A = (2, 3) and B = (2, -3) are the given points..
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA + PB = 8
\n\u21d2 PA = 8 – PB
\nSquaring on both sides
\n\u21d2 PA2<\/sup> = (8 – PB)2<\/sup>
\n\u21d2 PA2<\/sup> = 64 – 16PB + PB2<\/sup>
\n\u21d2 (x – 2)2<\/sup> + (y – 3)2<\/sup> = (x – 2)2<\/sup> + (y + 3)2<\/sup> + 64 – 16PB
\n\u21d2 x2<\/sup> – 4x + 4 + y2<\/sup> – 6y + 9 = x2<\/sup> – 4x + 4 + y2<\/sup> + 6y + 9 + 64 – 16PB
\n\u21d2 -6y = 6y + 64 – 16PB
\n\u21d2 16PB = 6y + 6y + 64
\n\u21d2 16PB = 12y + 64
\n\u21d2 4PB = 3y + 16
\nSquaring on both sides
\n\u21d2 16PB2<\/sup> = (3y + 16)2<\/sup>
\n\u21d2 16PB2<\/sup> = 9y2<\/sup> + 256 + 96y
\n\u21d2 16[(x – 2)2<\/sup> + (y + 3)2<\/sup>] = 9y2<\/sup> + 256 + 96y
\n\u21d2 16[x2<\/sup> – 4x + 4 + y2<\/sup> + 9 + 6y] = 9y2<\/sup> + 256 + 96y
\n\u21d2 16x2<\/sup> – 64x + 64 + 16y2<\/sup> + 144 + 96y – 9y2<\/sup> – 256 – 96y = 0
\n\u21d2 16x2<\/sup> + 7y2<\/sup> – 64x – 48 = 0
\n\u2234 The equation of the locus of point P(x, y) is 16x2<\/sup> + 7y2<\/sup> – 64x – 48 = 0.<\/p>\n

Question 18
\nA(5, 3) and B(3, -2) are two fixed points. Find the equation of the locus of P, so that the area of triangle PAB is 9. [Mar. ’09, ’06]
\nSolution:
\nLet A(5, 3), B(3, -2) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is Area of the triangle PAB = 9
\n\u21d2 \\(\\frac{1}{2}\\left|\\begin{array}{ll}
\nx_1-x_2 & y_1-y_2 \\\\
\nx_1-x_3 & y_1-y_3
\n\\end{array}\\right|\\) = 9
\n\u21d2 \\(\\frac{1}{2}\\left|\\begin{array}{ll}
\nx-5 & y-3 \\\\
\nx-3 & y+2
\n\\end{array}\\right|\\) = 9
\n\u21d2 \\(\\left|\\begin{array}{ll}
\nx-5 & y-3 \\\\
\nx-3 & y+2
\n\\end{array}\\right|\\) = 18
\n\u21d2 |(x – 5)(y + 2) – (x – 3) (y – 3)| = 18
\n\u21d2 |xy – 5y + 2x – 10 – xy + 3x + 3y – 9| = 18
\n\u21d2 |5x – 2y – 19| = 18
\n\u21d2 (5x – 2y – 19) = \u00b118
\n(-) 5x – 2y – 19 = -18
\n5x – 2y – 19 + 18 = 0
\n5x – 2y – 1 = 0
\n(+) 5x – 2y – 19 = 18
\n5x – 2y – 19 – 18 = 0
\n5x – 2y – 37 = 0
\n\u2234 The equation of the locus of point P(x, y) is (5x – 2y – 1) (5x – 2y – 37) = 0
\n\u21d2 25x2<\/sup> – 10xy – 185x – 10xy + 4y2<\/sup> + 74y – 5x + 2y + 37 = 0
\n\u21d2 25x2<\/sup> – 20xy + 4y2<\/sup> – 190x + 76y + 37 = 0<\/p>\n

Question 19.
\nFind the equation of the locus of a point, which forms a triangle of area 2 with the points A(1, 1) and B(-2, 3). [May ’13(old)]
\nSolution:
\nLet A(1, 1), B(-2, 3) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is Area of the triangle PAB = 2
\n\u21d2 \\(\\frac{1}{2}\\left|\\begin{array}{ll}
\nx_1-x_2 & y_1-y_2 \\\\
\nx_1-x_3 & y_1-y_3
\n\\end{array}\\right|\\) = 2
\n\u21d2 \\(\\frac{1}{2}\\left|\\begin{array}{ll}
\nx-1 & y-1 \\\\
\nx+2 & y-3
\n\\end{array}\\right|\\) = 2
\n\u21d2 |(x – 1)(y – 3) – (y – 1) (x + 2)| = 4
\n\u21d2 |xy – 3x – y + 3 – xy – 2y + x + 2| = 4
\n\u21d2 |-2x – 3y + 5 | = 4
\n\u21d2 -2x – 3y + 5 = \u00b14
\n(+) -2x – 3y + 5 = 4
\n2x + 3y – 5 + 4 = 0
\n2x + 3y – 1 = 0
\n(-) -2x – 3y + 5 = -4
\n2x + 3y – 5 – 4 = 0
\n2x + 3y – 9 = 0
\n\u2234 The equation of the locus of P(x, y) is (2x + 3y – 1) (2x + 3y – 9) = 0.
\n\u21d2 4x2<\/sup> + 6xy – 18x + 6xy + 9y2<\/sup> – 27y – 2x – 3y + 9 = 0
\n\u21d2 4x2<\/sup> + 12xy + 9y2<\/sup> – 20x – 30y + 9 = 0<\/p>\n

\"TS<\/p>\n

Question 20.
\nIf the distance from p to the points (2, 3) and (2, -3) are in the ratio 2 : 3, then find the equation of the locus of P. [May ’15 (TS)]
\nSolution:
\nLet A(2, 3), B(2, -3) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA : PB = 2 : 3
\n\"TS
\n\u21d2 9[x2<\/sup> + 4 – 4x + y2<\/sup> + 9 – 6y] = 4[x2<\/sup> + 4 – 4x + y2<\/sup> + 9 + 6y]
\n\u21d2 9x2<\/sup> + 36 – 36x + 9y2<\/sup> + 81 – 54y – 4x2<\/sup> + 16 – 16x + 4y2<\/sup> + 36 + 24y
\n\u21d2 9x2<\/sup> + 36 – 36x + 9y2<\/sup> + 81 – 54y – 4x2<\/sup> – 16 + 16x – 4y2<\/sup> – 36 – 24y = 0
\n\u21d2 5x2<\/sup> + 5y2<\/sup> – 20x – 78y + 65 = 0
\n\u2234 The equation of the locus of P(x, y) is 5x2<\/sup> + 5y2<\/sup> – 20x – 78y + 65 = 0.<\/p>\n

Question 21.
\nFind the equation of the locus of a point that is equidistant from the coordinate axes.
\nSolution:
\nP(x, y) is any point on the locus.
\nLet M, and N are the projections drawn from P to the X-axis and Y-axis respectively.
\nThe distance of P from the X-axis is |PM| = |y|
\nThe distance of P from the Y-axis is |PN| = |x|
\nThe given geometric condition is |PM| = |PN|
\n|y| = |x|
\n\"TS
\nSquaring on both sides |y| = |x|
\n\u21d2 y2<\/sup> = x2<\/sup>
\n\u21d2 x2<\/sup> – y2<\/sup> = 0
\n\u2234 The equation of the locus of P(x, y) is x2 – y2 = 0.<\/p>\n

\"TS<\/p>\n

Question 22.
\nFind the equation of the locus of a point that is equidistant from points A(-3, 2) and B(0, 4).
\nSolution:
\nA(-3, 2) and B(0, 4) are the given points.
\nLet P(x, y) be any point on the locus.
\nThe given geometric condition is PA = PB
\n\"TS
\n\u21d2 (x + 3)2<\/sup> + (y – 2)2<\/sup> = (x – 0)2<\/sup> + (y – 4)2<\/sup>
\n\u21d2 x2<\/sup> + 6x + 9 + y2<\/sup> – 4y + 4 = x2<\/sup> + y2<\/sup> + 16 – 8y
\n\u21d2 13 + 6x – 4y = 16 – 8y
\n\u21d2 13 + 6x – 4y – 16 + 8y = 0
\n\u21d2 6x + 4y – 3 = 0
\n\u2234 The equation of the locus of P(x, y) is 6x + 4y – 3 = 0.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Locus Important Questions to help strengthen their preparations for exams. TS Inter First Year Maths 1B Locus Important Questions Question 1. Find the equation of the locus of a point that is at a distance 5 from (-2, 3), in the … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10641"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=10641"}],"version-history":[{"count":1,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10641\/revisions"}],"predecessor-version":[{"id":10657,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10641\/revisions\/10657"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=10641"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=10641"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=10641"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}