{"id":35086,"date":"2022-08-25T10:19:08","date_gmt":"2022-08-25T04:49:08","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35086"},"modified":"2022-08-25T10:19:08","modified_gmt":"2022-08-25T04:49:08","slug":"inter-1st-year-maths-1b-pair-of-straight-lines-solutions-ex-4c","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-1st-year-maths-1b-pair-of-straight-lines-solutions-ex-4c\/","title":{"rendered":"Inter 1st Year Maths 1B Pair of Straight Lines Solutions Ex 4(c)"},"content":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions<\/a> Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(c) will help students to clear their doubts quickly.<\/p>\n

Intermediate 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(c)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFind the equation of the lines joining the origin to the points of intersection of x\u00b2 + y\u00b2 = 1 and x + y = 1.
\nSolution:
\nThe given curves are
\nx\u00b2 + y\u00b2 = 1 …….. (1)
\nx + y = 1 ……….. (2)
\nHomogenising (1) with the help of (2) Combined equation of OA and OB is
\n\"Inter
\nx\u00b2 + y\u00b2 = (x + y)\u00b2
\n= x\u00b2 + y\u00b2 + 2xy
\ni.e., 2xy = 0 \u21d2 xy – 0<\/p>\n

Question 2.
\nFind the angle between the lines joining the origin to the points of intersection of y2 = x and x + y = 1.
\nSolution:
\nEquation of the centre is y\u00b2 = x …………. (1)
\nEquation of AB is x + y = 1 ………. (2)
\nHamogonsing (1) with the help of O
\nCombined equation of A & OB is
\ny\u00b2 = x(x + y) = x\u00b2 + xy
\nx\u00b2 + xy – y2 = 0
\na + b = 1 – 1 = 0
\nOA, OB an perpendicular
\n\u2234 \u2220AOB =90\u00b0<\/p>\n

\"Inter<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nShow that the lines joining the origin to the points of intersection of the curve x\u00b2 – xy + y\u00b2 + 3x + 3y – 2 = 0 and the straight line x – y – \u221a2 = 0 are mutually perpendicular.
\nSolution:
\n\"Inter
\nEquation of the curve is
\nx\u00b2 – xy + y\u00b2 + 3x + 3y – 2 = 0 ……….. (1)
\nEquation of AB is x – y – \u221a2 = 0
\nx – y = \u221a2
\n\\(\\frac{x-y}{\\sqrt{2}}\\) = 1 ………. (2)
\nHomogenising, (1) with the help of (2) combined equation of OA, OB is
\nx\u00b2 – xy + y\u00b2 + 3x.1 + 3y.1 – 2.1\u00b2 = 0
\nx\u00b2 – xy + y\u00b2 + 3(x + y)\\(\\frac{x-y}{\\sqrt{2}}\\) – 2\\(\\frac{(x-y)^{2}}{2}\\) = 0
\nx\u00b2 – xy + y\u00b2 + \\(\\frac{3}{\\sqrt{2}}\\)(x\u00b2 – y\u00b2) – (x\u00b2 – 2xy + y\u00b2) = 0
\nx\u00b2 – xy + y\u00b2 + \\(\\frac{3}{\\sqrt{2}}\\)x\u00b2 – \\(\\frac{3}{\\sqrt{2}}\\)y\u00b2 – x\u00b2 + 2xy – y\u00b2 = 0
\n\\(\\frac{3}{\\sqrt{2}}\\)x\u00b2 + xy – \\(\\frac{3}{\\sqrt{2}}\\)y\u00b2 = 0
\na + b = \\(\\frac{3}{\\sqrt{2}}\\) – \\(\\frac{3}{\\sqrt{2}}\\) = 0
\n\u2234 OA, OB are perpendicular.<\/p>\n

Question 2.
\nFind the values of k, if the lines joining the origin to the points of intersection of the curve 2x\u00b2 – 2xy + 3y\u00b2 + 2x – y – 1 = 0 and the line x + 2y = k are muiually perpendicular.
\nSolution:
\nGiven equation of the curve is
\nS \u2261 2x\u00b2 – 2xy + 3y\u00b2 + 2x – y – 1 = 0 ……….. (1)
\nEquation of AB is x + 2y = k
\n\"Inter
\n\\(\\frac{x+2y}{k}\\) = 1
\nHomogenising, (1) with the help of (2), combined equation of OA, OB is
\n2x\u00b2 – 2xy + 3y\u00b2 + 2x.1 – y.1 – 1\u00b2 = 0
\n2x\u00b2 – 2xy + 3y\u00b2 + 2x\\(\\frac{x+2y}{k}\\) – y\\(\\frac{x+2y}{k}\\) – \\(\\frac{(x+2y)^{2}}{k^{2}}\\) = 0
\nMultiplying with k\u00b2
\n2k\u00b2x\u00b2 – 2k\u00b2xy + 3k\u00b2y\u00b2 + 2kx(x + 2y) – ky(x + 2y) – (x + 2y)\u00b2 = 0
\n2k\u00b2x\u00b2 – 2k\u00b2xy + 3k\u00b2y\u00b2 + 2kx\u00b2 + 4kxy – kxy – 2ky\u00b2 – x\u00b2 – 4xy – 4y\u00b2 = 0
\n(2k\u00b2 + 2k – 1)x\u00b2 + (-2k\u00b2 + 3k – 4) xy + (3k\u00b2 – 2k – 4) y\u00b2 = 0
\nSince OA, OB are perpendicular.
\nCo-efficient x\u00b2 + Co-efficient of y\u00b2 = 0
\n2k\u00b2 + 2k – 1 + 3k\u00b2 – 2k – 4 = 0
\n5k\u00b2 = 5 \u21d2 k\u00b2 = 1
\n\u2234 k = \u00b1 1<\/p>\n

Question 3.
\nFind the angle between the lines joining the origin to the points of intersection of the curve x\u00b2 + 2xy + y\u00b2 + 2x + 2y – 5 = 0 and the line 3x – y + 1 = 0
\nSolution:
\n\"Inter
\nEquation of the curve is
\nx\u00b2 + 2xy + y\u00b2 + 2x + 2y- 5 = 0 …….. (1)
\nEquation of AB is 3x – y + 1 = 0
\ny – 3x = 1…………. (2)
\nHomogenising (1) with the help of (2), combined equation of OA, OB is
\nx\u00b2 + 2xy + y\u00b2 + 2x.1 + 2y.1 – 5.1\u00b2 = 0
\nx\u00b2 + 2xy + y\u00b2 + 2x(y – 3x) + 2y(y – 3x) – 5 (y – 3x)\u00b2 = 0
\nx\u00b2 + 2xy + y\u00b2 + 2xy – 6x\u00b2 + 2y\u00b2 – 6xy – 5(y\u00b2 + 9x\u00b2 – 6xy) = 0
\n– 5x\u00b2 – 2xy + 3y\u00b2 – 5y\u00b2 – 45x\u00b2 + 30 xy = 0
\n-50x\u00b2 + 28xy – 2y\u00b2 = 0
\ni.e, 25x\u00b2 – 14xy + y\u00b2 = 0
\nSuppose \u03b8 is the angle between OA and OB
\n\"Inter<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nFind the condition for the chord lx + my = 1 of the circle x\u00b2 + y\u00b2 = a\u00b2 (whose centre is the origin) to subtend a right angle at the origin.
\nSolution:
\n\"Inter
\nEquation of the circle is x\u00b2 + y\u00b2 = a\u00b2 ………. (1)
\nEquation of AB is lx + my = 1 ……….. (2)
\nHomogenising (1) with the help of (2)
\nGombined equation of OA, OB is
\nx\u00b2 + y\u00b2 = a\u00b2.l\u00b2
\nx\u00b2 + y\u00b2 = a\u00b2 (lx + my)\u00b2
\n= a\u00b2(l\u00b2x\u00b2 + m\u00b2y\u00b2 + 2lmxy)
\n= a\u00b2l\u00b2x\u00b2 + a\u00b2m\u00b2y\u00b2 + 2a\u00b2lmxy
\ni.e., a\u00b2l\u00b2x\u00b2 + 2a\u00b2lmxy + a\u00b2m\u00b2y\u00b2 – x\u00b2 – y\u00b2 = 0
\n(a\u00b2l\u00b2 – 1) x\u00b2 + 2a\u00b2 lmxy + (a\u00b2m\u00b2 – 1)y\u00b2 = 0
\nSince OA, OB are perpendicular
\nCo-efficient of x\u00b2 + co-efficient of y\u00b2 = 0
\na\u00b2l\u00b2 – 1 + a\u00b2m\u00b2 – 1 = 0
\na\u00b2(l\u00b2 + m\u00b2) = 2
\nThis is the required condition.<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the condition for the lines joining the origin to the points of intersection of the circle x\u00b2 + y\u00b2 = a\u00b2 and the line lx + my = 1 to coincide.
\nSolution:
\n\"Inter
\nEquation of the circle is
\nx\u00b2 + y\u00b2 = a\u00b2 ………. (1)
\nEquation of AB is lx + my = 1 ……….. (2)
\nHomogenising (1) with the help of (2)
\nCombined equation of OA, OB is
\nx\u00b2 + y\u00b2 = a\u00b2.l\u00b2
\n= a\u00b2(lx + my)\u00b2
\n= a\u00b2(l\u00b2x\u00b2 + m\u00b2y\u00b2 + 2lmxy)
\ni.e., x\u00b2 + y\u00b2 = a\u00b2l\u00b2x\u00b2 + a\u00b2m\u00b2y\u00b2 + 2a\u00b2lmxy
\n(a\u00b2l\u00b2 – 1)x\u00b2 + 2a\u00b2lmxy + (a\u00b2m\u00b2 – 1)y\u00b2 = 0
\nSince OA, OB are coincide
\n\u21d2 h\u00b2 = ab
\na4<\/sup> l\u00b2m\u00b2 = (a\u00b2l\u00b2 – 1)(a\u00b2m\u00b2 – 1)
\na4<\/sup> l\u00b2m\u00b2 = a4<\/sup> l\u00b2m\u00b2 – a\u00b2l\u00b2 – a\u00b2 m\u00b2 + 1
\n\u2234 a\u00b2l\u00b2 – a\u00b2m\u00b2 +1=0
\na\u00b2 (l\u00b2 + m\u00b2) = 1
\nThis is the required condition.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nWrite down the equation of the pair of straight lines joining the origin to the points of intersection of the line 6x – y + 8 = 0 with the pair of straight lines 3x\u00b2 + 4xy – 4y\u00b2 – 11x + 2y + 6 = 0. Show that the lines so obtained make equal angles with the coordinate axes.
\nSolution:
\nGiven pair of lines is
\n3x\u00b2 + 4xy – 4y\u00b2 – 11x + 2y + 6 = 0 ……….. (1)
\nGiven line is
\n6x – y + 8 = 0 \u21d2 \\(\\frac{6x-y}{-8}\\) = 1
\n\u21d2 \\(\\frac{y-6x}{8}\\) = 1
\nHomogenising (1) w. r. t. (2)
\n3x\u00b2 + 4xy – 4y\u00b2 – (11x – 2y) (\\(\\frac{y-6x}{8}\\)) = 6(\\(\\frac{y-6x}{8}\\))\u00b2 = 0
\n64[3x\u00b2 + 4xy – 4y\u00b2] – 8[11xy – 66x\u00b2 – 2y\u00b2 + 12xy] + 6[y\u00b2 + 36x\u00b2 – 12xy] = 0
\n936x\u00b2 + 256 xy – 256 xy – 234y\u00b2 = 0
\n\u2234 468x\u00b2 – 117y\u00b2 = 0 …….. (3)
\n\u21d2 4x\u00b2 – y\u00b2 = 0
\nis eq. of pair of lines joining the origin to the point of intersection in.
\nThe eq. pair of angle bisectors to (3) is
\nh(x\u00b2 – y\u00b2) – (a – b)xy = 0
\n0(x\u00b2 – y\u00b2) – f4 – l)xy = 0
\n\u21d2 xy = 0
\nx = 0 or y = 0 [Eqs. is of coordinate axes]
\n\u2234 The pair of lines are equally inclined to the co-ordinate axes.<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(c) will help students to clear their doubts quickly. Intermediate 1st Year Maths 1B Pair of Straight Lines Solutions Exercise 4(c) I. Question 1. Find the equation of the lines joining the origin to the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35086"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35086"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35086\/revisions"}],"predecessor-version":[{"id":35087,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35086\/revisions\/35087"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35086"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35086"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35086"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}