{"id":7388,"date":"2023-11-02T04:21:30","date_gmt":"2023-11-01T22:51:30","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=7388"},"modified":"2023-11-03T17:01:23","modified_gmt":"2023-11-03T11:31:23","slug":"ts-inter-2nd-year-maths-2a-solutions-chapter-1-ex-1d","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2a-solutions-chapter-1-ex-1d\/","title":{"rendered":"TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Ex 1(d)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions<\/a> Chapter 1 Functions Ex 1(d) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(d)<\/h2>\n

I.
\nQuestion 1.
\ni) Find the equation of the perpendicular bisector of the line segment joining the points 7 + 7i, 7 – 7i.
\nii) Find the equation of the straight line joining the points – 9 + 6i, 11 – 4i in the Argand plane.
\nSolution:
\ni) z1<\/sub> = 7 + 7i
\nz2<\/sub> = 7 – 71
\nA (7, 7) B (7, – 7)<\/p>\n

\"TS<\/p>\n

M(7, 0)<\/p>\n

Slope of AB = \\(\\frac{7+7}{7-7}\\) \u2192 \u221e
\nLine \u22a5 to AB slope is zero,
\ny = 0 is line.<\/p>\n

ii) A (- 9, 6) B (11, – 4)
\nSlope of AB = \\(\\frac{6+4}{-9-11}\\)
\n= \\(\\frac{10}{-20}=\\frac{-1}{2}\\)
\nEquation of line AB,
\ny – 6 = \\(\\frac{- 1}{2}\\) (x + 9)
\n2y – 12 = – x – 9
\nx + 2y = 3.<\/p>\n

\"TS<\/p>\n

Question 2.
\nIf z = x + ily and if the point Pin the Argand plane represents z, then describe geometrically the locus of z satisfying the equations
\ni) |z – 2 – 3i| = 5
\nii) 2|z – 2| = |z – 1|
\niii) im z2<\/sup> = 4
\niv) Arg \\(\\left(\\frac{z-1}{z+1}\\right)=\\frac{\\pi}{4}\\)
\nSolution:
\ni) |z – 2 – 3i| = 5
\n|(x – 2) + (y – 3)i| = 5
\n(x – 2)2<\/sup> + (y – 3)2<\/sup> = 25
\nx2<\/sup> + y2<\/sup> – 4x – 6y + 4 + 9 – 25 = 0
\nx2<\/sup> + y2<\/sup> – 4x – 6y – 12 = 0<\/p>\n

ii) 2|z – 2| = |z – 1|
\n4(z – 2) (- 2) = (z – 1) (\\(\\overline{\\mathbf{z}}\\) – 1)
\n4z\\(\\overline{\\mathbf{z}}\\) – 8z – 8\\(\\overline{\\mathbf{z}}\\) + 16 = z\\(\\overline{\\mathbf{z}}\\) – z – \\(\\overline{\\mathbf{z}}\\) + 1
\n3z\\(\\overline{\\mathbf{z}}\\) – 7z – 7\\(\\overline{\\mathbf{z}}\\) + 15 = 0
\n3(x2<\/sup> + y2<\/sup>) – 7(2x) + 15 = 0.<\/p>\n

iii) Im (z2<\/sup>) = 4
\nIm (z2<\/sup>) = 4
\nz = x + iy
\nz2<\/sup> = (x + iy)2<\/sup>
\nz2<\/sup> = x2<\/sup> – y2<\/sup> + 2xyi
\nIm(z2<\/sup>) = 2xy
\n2xy = 4
\nxy = 2 rectangualr hyperbola<\/p>\n

iv) Arg \\(\\left(\\frac{z-1}{z+1}\\right)=\\frac{\\pi}{4}\\)<\/p>\n

\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nShow that the points in the Argand diagram represented by the complex numbers 2 + 2i, – 2 – 2i, – 2\u221a3 + 2\u221a3i are the vertices of an equilateral triangle..
\nSolution:
\nA(2, 2), B(- 2, – 2), C (- 2\u221a3 + 2\u221a3)
\nAB = \\(\\sqrt{(2+2)^2+(2+2)^2}\\) = 4\u221a2
\nBC = \\(\\sqrt{(-2+2 \\sqrt{3})^2+(-2-2 \\sqrt{3})^2}\\)
\nBC = \\(\\sqrt{4+12-8 \\sqrt{3}+4+12+8 \\sqrt{3}}\\)= 4\u221a2
\nAC = \\(\\sqrt{\\left(2+2 \\sqrt{3}^2\\right)+(2-2 \\sqrt{3})^2}\\) = 4\u221a2
\nAB = AC = BC
\n\u2206 ABC is equilateral.<\/p>\n

Question 4.
\nFind the eccentricity of the ellipse whose equtaion is | z – 4 | + |z – \\(\\frac{12}{5}\\)| = 10
\nSolution:
\nSP + S’P = 2a
\nS (4, 0) S\u2019(\\(\\frac{12}{5}\\), 0)
\n2a = 10
\na = 5
\nSS’ = 2ae
\n4 – \\(\\frac{12}{5}\\) = 2 \u00d7 5e
\n\\(\\frac{8}{5}\\) = 10e
\ne = \\(\\frac{4}{25}\\).<\/p>\n

\"TS<\/p>\n

II.
\nQuestion 1.
\nIf \\(\\frac{z_3-z_1}{z_2-z_1}\\) is a real number, show that the points represented by the complex numbers z1<\/sub>, z2<\/sub>, z3<\/sub> are collinear.
\nSolution:<\/p>\n

\"TS<\/p>\n

Arg \\(\\left(\\frac{z_1-z_3}{z_1-z_2}\\right)\\) = 0 then \\(\\frac{z_1-z_3}{z_1-z_2}\\) is real \u03b8 = 0.
\n\u2234 z1<\/sub>, z2<\/sub>, z3<\/sub> are collinear.<\/p>\n

Question 2.
\nShow that the four points in the Argand plane represented by the complex numbers 2 + i, 4 + 3i, 2 + 5i, 3i are vertices of a square.
\nSolution:
\nA (2, 1), B (4, 3), C (2, 5), D (0, 3)
\nAB = \\(\\sqrt{(4-2)^2+(3-1)^2}\\) = 2\u221a2
\nBC = \\(\\sqrt{(4-2)^2+(3-5)^2}\\) = 2\u221a2
\nCD = \\(\\sqrt{(2-0)^2+(5-3)^2}\\) = 2\u221a2
\nAD = \\(\\sqrt{(2-0)^2+(1-3)^2}\\) = 2\u221a2
\nSlope of AB = \\(\\frac{5-3}{2-4}\\)= 1
\nSlope of BC = \\(\\frac{3-1}{4-2}\\) = 1
\nAB \u22a5 BC
\nBC \u22a5 CD
\n\u21d2 ABCD is a square.<\/p>\n

Question 3.
\nShow that die points in the Argand plane represented by the complex numbers – 2 + 7i, – \\(\\frac{-3}{2}\\) + \\(\\frac{1}{2}\\)i, 4 – 3i, \\(\\frac{7}{2}\\) (1 + i) are the vertices of a rhombus.
\nSolution:<\/p>\n

\"TS<\/p>\n

AC \u22a5 BD
\n\u2234 ABCD is rhombus.<\/p>\n

\"TS<\/p>\n

Question 4.
\nShow that the points in the Argand diagram represented by the complex numbers z1<\/sub>, z2<\/sub>, z3<\/sub> are collhitear if and only if there exists three real numbers p, q, r not all zero satisfying p + qz2<\/sub> + rz3<\/sub> = 0 and p + q + r = 0.
\nSolution:
\npz1<\/sub> + qz2<\/sub> + rz3<\/sub> = 0
\npz1<\/sub> + qz2<\/sub> = – rz3<\/sub>
\n\\(\\left(\\frac{p z_1+q z_2}{p+q}\\right)\\) (p + q) = – rz3<\/sub>
\nNow p + q = – r
\n\\(\\left(\\frac{p z_1+q z_2}{p+q}\\right)\\) = z3<\/sub>
\n\u21d2 z3<\/sub> divides z1<\/sub> and z2<\/sub> is q : p ratio.
\n\u2234 z1<\/sub>, z2<\/sub>, z3<\/sub> are collinear.<\/p>\n

Question 5.
\nThe points P, Q denote the complex numbers z1<\/sub>, z2<\/sub> in the Argand diagram. O is
\norigin. If z1<\/sub>\\(\\overline{\\mathbf{z}}_2\\) + \\(\\overline{\\mathbf{z}}_1\\)z2<\/sub> = 0 tlien show that \u2220POQ = 90\u00b0.
\nSolution:
\nz1<\/sub>\\(\\overline{\\mathbf{z}}_2\\) + \\(\\overline{\\mathbf{z}}_1\\)z2<\/sub> = 0
\n\\(\\frac{\\mathbf{z}_1 \\overline{\\mathbf{z}}_2+\\overline{\\mathbf{z}}_1 \\mathbf{z}_2}{\\mathbf{z}_2 \\overline{\\mathrm{z}}_2}\\) = 0
\n\u21d2 Real of \\(\\frac{\\mathrm{z}_1}{\\mathrm{z}_2}\\) = 0
\n\\(\\left(\\frac{\\mathrm{z}_1}{\\mathrm{z}_2}+\\frac{\\overline{\\mathrm{z}}_1}{\\mathrm{z}_2}\\right)\\)
\nImaginary part of (\\(\\frac{\\mathrm{z}_1}{\\mathrm{z}_2}\\)) is k.
\n\\(\\left(\\frac{\\mathrm{z}_1}{\\mathrm{z}_2}\\right)+\\left(\\frac{\\overline{\\mathrm{z}}_1}{\\mathrm{z}_2}\\right)\\) = 0 or
\n\\(\\frac{\\mathrm{z}_1}{\\mathrm{z}_2}\\) is purely imaginary.
\n\\(\\frac{\\mathrm{z}_1}{\\mathrm{z}_2}\\) = ki
\n\u21d2 Arg\\(\\frac{\\mathrm{z}_1}{\\mathrm{z}_2}\\) = \\(\\frac{\\pi}{2}\\).<\/p>\n

\"TS<\/p>\n

Question 6.
\nThe complex number z has argument \u03b8 0 < \u03b8 < \\(\\frac{\\pi}{2}\\) and satisfy the equation |z – 3i| = 3. Then prove that (cot \u03b8 – \\(\\frac{6}{z}\\)) = 1.
\nSolution:
\n(x2<\/sup>) + (y – 3)2<\/sup> = 9
\nx + y = 0
\nx + y = 6y<\/p>\n

\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2A Solutions Chapter 1 Functions Ex 1(d) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2A Solutions Chapter 1 Complex Numbers Exercise 1(d) I. Question 1. i) Find the equation of the perpendicular bisector of the line segment joining the points 7 … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7388"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=7388"}],"version-history":[{"count":2,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7388\/revisions"}],"predecessor-version":[{"id":7443,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/7388\/revisions\/7443"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=7388"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=7388"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=7388"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}