{"id":22573,"date":"2020-02-06T11:24:48","date_gmt":"2020-02-06T05:54:48","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=22573"},"modified":"2021-07-02T15:25:46","modified_gmt":"2021-07-02T09:55:46","slug":"1st-puc-basic-maths-question-bank-chapter-17","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-question-bank-chapter-17\/","title":{"rendered":"1st PUC Basic Maths Question Bank Chapter 17 Straight Lines"},"content":{"rendered":"

Students can Download Basic Maths Chapter 17 Straight Lines Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Basic Maths Question Bank Chapter 17 Straight Lines<\/h2>\n

Question 1.
\nFind the slope of the line joining the points (3, -4) and (-7,3).
\nAnswer:
\n\"1st<\/p>\n

Question 2.
\nShow that the line joining points (2, -3) and (-5,1) is parallel to the line joining the points (7, -1) and (0,3) and perpendicular to the line joining the points (4,5) and (0, -2).
\nAnswer:
\nLet m1<\/sub> be the slope of the lien l1<\/sub> joining points (2, -3) and (0, 3)
\n\"1st
\nLet m2<\/sub> be the slope of the line l2<\/sub> going (7, -1) and (0,3)
\n\"1st
\n\u21d2 m1<\/sub> = m2<\/sub>
\n\u2234 L1<\/sub> and L2<\/sub> are parallel to each other.
\n\"1st<\/p>\n

\"1st<\/p>\n

Question 3.
\nL1<\/sub> and L3<\/sub> or l2<\/sub> l3<\/sub> are perpendicular to each other. The slope of a line is double the
\nslope of another line. If the tangent of the angle between the is \\(\\frac { 1 }{ 3 }\\). Find the slope of the lines.
\nAnswer:
\nIf slope of one line is m, then the slope of the other is 2m.
\nLet the angle between them \u03b8 then tan \u03b8 = \\(\\frac { 1 }{ 3 }\\)
\n\"1st
\n\u21d2 \\(\\frac { 1 }{ 3 }\\) = \\(\\frac{m}{1+2 m^{2}}\\)
\n\u21d2 1 + 2m2<\/sup> = 3m
\n\u21d2 2m2<\/sup> – 3m + x = 0
\n\u21d2 (2m – 1)(m – 1 ) = 0
\n\u2234 2m – 1 = 0 and m – 1 = 0
\nm = \\(\\frac { 1 }{ 2 }\\) m = 1
\n\u2234 Slope are \\(\\frac { 1 }{ 2 }\\), and 1.<\/p>\n

On standard form of straight lines:<\/span><\/p>\n

Question 1.
\nFind the equation of line passing through (-3, 5) with slope \\(\\frac { -1 }{ 5 }\\)
\nAnswer:
\nLet (x1<\/sub> y1<\/sub>) = (-3, 5). slope m = \\(\\frac { -1 }{ 5 }\\)
\nThe equation of line is y – y1<\/sub> = m(x – x1<\/sub>)
\n= y – 5 = \\(\\frac { -1 }{ 5 }\\) (x + 3)
\n\u21d2 5y – 25 = -x – 3
\n\u21d2 x + 5y – 22 = 0<\/p>\n

\"1st<\/p>\n

Question 2.
\nFind the equation line passing through (-3, 2) and (11, -1).
\nAnswer:
\nLet (x1<\/sub>, y1<\/sub>) = (-3,2), (x2<\/sub>, y2<\/sub>) = (11,-1)
\n\"1st
\n\u21d2 \\(\\frac{y-2}{x+3}=\\frac{-3}{14}\\)
\n\u21d2 14y – 28 = -3x – 9
\n\u21d2 3x + 14y – 19 = 0 is the equation<\/p>\n

Question 3.
\nFind the ratio in which the line join (1, 2) and (4, 3) is divided by the line joining the points (2,3) and (4, 1).
\nAnswer:
\nLet A = (2,3) B = (4, 1)
\n\u2234 the equation of line is a given by \\(\\frac{y-3}{x-2}=\\frac{1-3}{4-2}\\)
\n\u21d2 \\(\\frac{y-3}{x-2}=\\frac{-2}{2}\\) = -1
\n\u21d2 y – 3 = -x + 2
\n\u2234 x + y – 5 = 0 is the equation of line
\nLet c = (1,2) and D(4,3) cut the line AB at P(x ,y) is the ratio 2 : 1 then
\n\"1st
\n\"1st
\n\u21d2 4x + 1 + 3r + 2 – 5(r + 1) = 0
\n\u21d2 2r – 2 = 0
\n\u21d2 r = 1
\n\u2234 the required ratio is 1 : 1<\/p>\n

Question 4.
\nExpress 9x – 4y – 13 = 0 in the intercept from
\nAnswer:
\nConsider 9x – 4y = 13 = \\(\\frac{9}{13} x-\\frac{4}{13} y\\) = 1
\n\"1st
\n\u2234 x – intercept = \\(\\frac{13}{9}\\)
\ny – intercept = \\(\\frac{-13}{9}\\)<\/p>\n

\"1st<\/p>\n

Question 5.
\nIf P \u00a1s the length of the perpendicular from the origin on a line, whose x and y intercepts are respectively a and b then show that = \\(\\frac{1}{p^{2}}=\\frac{1}{a^{2}}+\\frac{1}{b^{2}}\\)
\nAnswer:
\nLet the line any the axis at A and y – axis at B
\nThen by data OA = a and OB = b from the figure
\n\"1st
\nArea of OAB = \\(\\frac { 1 }{ 2 }\\) (O.A) (O.B) = \\(\\frac { 1 }{ 2 }\\) ab
\nor = \\(\\frac { 1 }{ 2 }\\) (AB).P
\n\u21d2 \\(\\frac { 1 }{ 2 }\\) ab = \\(\\frac { 1 }{ 2 }\\) (AB).P \u21d2 a2<\/sup>b2<\/sup> = (AB)2<\/sup>. p2<\/sup>
\nAlso AB2<\/sup> = OA2<\/sup> + OB2<\/sup> \u21d2 AB2<\/sup> = a2<\/sup> + b2<\/sup> (\u2234 because a2<\/sup>b2<\/sup> = (a2<\/sup> + b2<\/sup>) . p2<\/sup>
\n\"1st<\/p>\n

Intersection of two lines and Concurrency of Lines<\/span><\/p>\n

Question 1.
\nFind the equation of line parallel to the y-axis and drawn through the point of intersection of x – 7y + 5 = 0 and 3x + y – 7 = 0.
\nAnswer:
\nThe equation of line through the point of intersection of the given lines is of the form x – 7y + 5 + k(3x + y – 7) = 0
\n\u2234 (1 + 3k)x + (k -7)y + 5 – 7k = 0 (1)
\nSince the lines are parallel to y-axis, coefficient of y = 0
\ni.e., k – 7 =0
\n\u2234 k = 7 substitute in (1) we get
\n22x – 44 = 0
\n\u2234 x – 2 = 0 is required equation<\/p>\n

Question 2.
\nFind the equation of the line through the intersection of 5x – 3y = 1 and 2x + 3y – 23 = 0 and perpendicular to the line whose equation is 5x – 3y -1 = 0 and perpendicular to the line whose equation is 5x – 3y – 1 = 0.
\nAnswer:
\nWe have 5x – 3y – 1 = 0 ….. (1)
\nand 2x + 3y – 23 = 0 ….. (2) .
\n\u2234 the equation will be in the form
\n(5x – 3y – 1) + k(2x + 3y – 23) = 0
\n\u21d2 (5 + 2k)x + 3(k – 1)y – (23k + 1) = 0 …… (3)
\n\"1st
\n\u2234 Slope of line = \\(\\frac{5+2 k}{3(k-1)}\\)
\nAlso 5x – 3y – 1 = 0
\n\u21d2 y = \\(\\frac{5}{3} x \\frac{1}{3}\\) …….. (1)
\n\u2234slope = \\(\\frac{5}{3}\\)
\nGiven (4) and (5) are perpendicular
\n\"1st
\n\u21d2 k = -34. Substituting the value of k in (3), we get
\n(5x – 3y – 1) -34 (2x + 3y – 25) = 0
\n\u21d2 5x – 68x – 3y – 102y – 1 + 782 = 0
\n\u21d2 63x + 105y – 781 = 0<\/p>\n

\"1st<\/p>\n

Question 3.
\nIf lines whose equation are y = m1<\/sub> + c1<\/sub>y = m2<\/sub>x + c2<\/sub> and y = m5<\/sub>x + c3<\/sub> meet is a point then prove that m1<\/sub>( (c2<\/sub> – c3<\/sub>) + m2<\/sub> (c3<\/sub> – c1<\/sub>) + m3<\/sub>(c1<\/sub> – c2<\/sub>) = 0
\nAnswer:
\nThe equation of the given lines are
\nm1<\/sub>x – y + c1<\/sub> = 0 ……. (1)
\nm2<\/sub>x – y + c2<\/sub> = 0 …… (2)
\nm3<\/sub>x – y + c3<\/sub> = 0 .. (3)
\n\"1st
\nThe three lines will be concurrent if the point of intersection of (1) and (2) lies on (3)
\n\"1st
\n\u21d2 m3<\/sub>(c1<\/sub> – c2<\/sub>) – (m2<\/sub>c1<\/sub> – m1<\/sub>c2<\/sub>) + c3<\/sub> (m2<\/sub> – m1<\/sub>) =0, simplify,
\n\u21d2 m1<\/sub>(c2<\/sub> – c3<\/sub>) + m2<\/sub>(c3<\/sub> – c1<\/sub>) + m3<\/sub>(c1<\/sub> – c2<\/sub>) =0<\/p>\n

Question 4.
\nFor what values of k are the lines x – 2y + 1 = 0, 2x – 5y + 3 = 0 and 5x – 9y + k = 0 are concurrent.
\nAnswer:
\nconsider
\nx – 2y + 1 = 0 ……. (1)
\n2x – 5y + 3 = 0 ….. (2)
\n5x – 9y + k = 0 …….. (3)
\nSolve (1) and (2) \\(\\frac{x}{-6+5}=\\frac{-y}{3-2}=1\\)
\n\u21d2 \\(\\frac{x}{-1}=\\frac{-y}{-1}=\\frac{1}{-1}\\)
\n\u21d2 x = 1, y = 1
\npoint of intersection of (1) and (2) is (1,1),
\n\u21d2 (1, 1) lies on (3) = 5 – 9 + k = 0 = k = 4<\/p>\n

Angle between the lines, Length of perpendicular<\/span><\/p>\n

Question 1.
\nFind the angle between the lines 2x + 3y – 4 = 0 and 3x – 2y + 5 =0
\nAnswer:
\nBy data
\n2x + 3y – 4 = 0 … (1)
\n3x – 2y + 5 = 0 …… (2)
\n\u2234 Slope of (1) = \\(\\frac{-a}{b}=\\frac{-2}{3}\\) = m1<\/sub>
\n\"1st
\n\u2234 the angle between the lines is 90\u00b0<\/p>\n

\"1st<\/p>\n

Question 2.
\nDetermine the position of the points (2, 1) and (-1, 1) w.r.t the line 4x – ly + 1 = 0.
\nAnswer:
\n4(2) -7(1) = -7 + 1 = 2 > 0
\n4( -1) – 7(1) = – 4 – 7 + 2 = -10 < 0
\nSince the two points are opposite in sign the two points lie an either sides of the given line 4x – 7y + 1 = 0<\/p>\n

Question 3.
\nIf points (9,8) and (-3,3) are equidistant from the line 5x + 2y + 7 = 0 find ‘a’.
\nAnswer:
\nDistance from (a, 8) to the line 5x + 2y + 7 = 0
\n\"1st
\nGiven that the distance are equal
\n\"1st
\n\u21d2 5a + 23 = 2
\nor
\n5a + 23 = -2
\n\u21d2 a = \\(\\frac{-21}{5}\\) and a = -5<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Chapter 17 Straight Lines Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Basic Maths Question Bank Chapter 17 Straight Lines Question 1. Find the slope of …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22573"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=22573"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22573\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=22573"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=22573"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=22573"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}