{"id":17475,"date":"2022-05-24T22:30:09","date_gmt":"2022-05-24T17:00:09","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=17475"},"modified":"2022-05-16T15:43:53","modified_gmt":"2022-05-16T10:13:53","slug":"mcq-questions-for-class-11-maths-chapter-10","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-11-maths-chapter-10\/","title":{"rendered":"MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers"},"content":{"rendered":"

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers<\/a> during preparation and score maximum marks in the exam. Students can download the Straight Lines Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 10 Straight Lines Objective Questions.<\/p>\n

Straight Lines Class 11 MCQs Questions with Answers<\/h2>\n

Students are advised to solve the Straight Lines Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Straight Lines Class 11 with answers will boost your confidence thereby helping you score well in the exam.<\/p>\n

Explore numerous MCQ Questions of Straight Lines Class 11 with answers provided with detailed solutions by looking below.<\/p>\n

Question 1.
\nIn a \u0394ABC, if A is the point ( 1, 2) and equations of the median through B and C are respectively x + y = 5 and x = 4, then B is
\n(a) (1, 4)
\n(b) (7, -2)
\n(c) none of these
\n(d) (4, 1)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) (7, -2)
\nThe equation of median through B is x + y = 5
\nThe point B lies on it.
\nLet the coordinates of B are (x1<\/sub>, 5 – x1<\/sub>)
\nNow CF is a median through C,
\nSo coordiantes of F i.e. mid-point of AB are
\n((x1<\/sub> + 1)\/2, (5 – x1<\/sub> + 2)\/2)
\nNow since this lies on x = 4
\n\u21d2 (x1<\/sub> + 1)\/2 = 4
\n\u21d2 x1<\/sub> + 1 = 8
\n\u21d2 x1<\/sub> = 7
\nHence, the cooridnates of B are (7, -2)<\/p>\n<\/details>\n

\n

Question 2.
\nThe equation of straight line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0
\n(a) y – x + 1 = 0
\n(b) y – x – 1 = 0
\n(c) y – x + 2 = 0
\n(d) y – x – 2 = 0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) y – x – 1 = 0
\nGiven straight line is: x + y + 1 = 0
\n\u21d2 y = -x – 1
\nSlope = -1
\nNow, required line is perpendicular to this line.
\nSo, slope = -1\/-1 = 1
\nHence, the line is
\ny – 2 = 1 \u00d7 (x – 1)
\n\u21d2 y – 2 = x – 1
\n\u21d2 y – 2 – x + 1 = 0
\n\u21d2 y – x – 1 = 0<\/p>\n<\/details>\n

\n

Question 3.
\nThe points (-a, -b), (0, 0), (a, b) and (a\u00b2, ab) are
\n(a) vertices of a square
\n(b) vertices of a parallelogram
\n(c) collinear
\n(d) vertices of a rectangle<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) collinear
\nLet the four points are P(-a, -b), O(0, 0), Q(a, b) and R(a\u00b2, ab)
\nNow,
\nm1<\/sub> = slope of OP = b\/a
\nm2<\/sub> = slope of OQ = b\/a
\nm3<\/sub> = slope of OR = b\/a
\nSince m1<\/sub> = m2<\/sub> = m3<\/sub>
\nSo, the points O, P, Q, R are collinear.<\/p>\n<\/details>\n

\n

Question 4.
\nThe equation of the line through the points (1, 5) and (2, 3) is
\n(a) 2x – y – 7 = 0
\n(b) 2x + y + 7 = 0
\n(c) 2x + y – 7 = 0
\n(d) x + 2y – 7 = 0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 2x + y – 7 = 0
\nGiven, points are: (1, 5) and (2, 3)
\nNow, equation of line is
\ny – y1<\/sub> = {(y2<\/sub> – y1<\/sub>)\/(x2<\/sub> – x1<\/sub>)} \u00d7 (x – x1<\/sub>)
\n\u21d2 y – 5 = {(3 – 5)\/(2 – 1)} \u00d7 (x – 1)
\n\u21d2 y – 5 = (-2) \u00d7 (x – 1)
\n\u21d2 y – 5 = -2x + 2
\n\u21d2 2x + y – 5 – 2 = 0
\n\u21d2 2x + y – 7 = 0<\/p>\n<\/details>\n

\n

Question 5.
\nThe slope of a line which passes through points (3, 2) and (-1, 5) is
\n(a) 3\/4
\n(b) -3\/4
\n(c) 4\/3
\n(d) -4\/3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) -3\/4
\nGiven, points are (3, 2) and (-1, 5)
\nNow, slope m = (5 – 2)\/(-1 – 3)
\n\u21d2 m = -3\/4
\nSo, the slope of the line is -3\/4<\/p>\n<\/details>\n

\n

Question 6.
\nThe ratio of the 7th to the ( n – 1)th mean between 1 and 31, when n arithmetic means are inserted between them, is 5 : 9. The value of n is
\n(a) 15
\n(b) 12
\n(c) 13
\n(d) 14<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 14
\nLet the A.P. are 1, A1<\/sub>, A2<\/sub>, A3<\/sub> …… Am<\/sub>, 31
\na = 1, an<\/sub> = 31 and n = m + 2
\nNow, an<\/sub> = a + (n – 1)d
\n\u21d2 31 = 1 + (m + 2 – 1)d
\n\u21d2 30 = (m + 1)d
\n\u21d2 d = 30\/(m + 1)
\nAgain, A7<\/sub> = a + 7d = 1 + 7[30\/(m + 1)] …………….. 1
\nand Am-1<\/sub> = a + (m – 1)d = 1 + (m – 1)[30\/(m + 1)] ………. 2
\nFrom equation 1 and 2, we get
\nA7<\/sub>\/Am-1<\/sub> = 5\/9
\n\u21d2 1 + 7[30\/(m + 1) \/ 1 + (m – 1)[30\/(m + 1)] = 5\/9
\n\u21d2 [m + 1 + 7(30)] \/ [m + 1 + 30 m – 30] = 5\/9
\n\u21d2 [m + 211] \/ [31 m – 29] = 5\/9
\n\u21d2 9[m + 211] = 5[31 m – 29]
\n\u21d2 9 m + 1899 = 155 m – 145
\n\u21d2 146 m = 2044
\n\u21d2 m = 2044\/146
\n\u21d2 m = 14
\nSo, the value of m is 14<\/p>\n<\/details>\n

\n

Question 7.
\nThe ortho centre of the triangle formed by lines xy = 0 and x + y = 1 is :
\n(a) (0, 0)
\n(b) none of these
\n(c) ( 1\/2, 1\/2)
\n(d) ( 1\/3, 1\/3)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) (0, 0)
\nGiven lines are:
\nxy = 0 and x + y = 1
\n\u21d2 x = 0, y = 0 and x + y = 1
\n\"MCQ
\nIn a triangle OAB, OA and OB are the altitudes which intersect at O.
\nSo, the required orthocentre is (0, 0)<\/p>\n<\/details>\n

\n

Question 8.
\nTwo lines a1<\/sub> x + b1<\/sub> y + c1<\/sub> = 0 and a2<\/sub> x + b2<\/sub> y + c2<\/sub> = 0 are parallel if
\n(a) a1<\/sub> \/a2<\/sub> = b1<\/sub> \/b2<\/sub> \u2260 c1<\/sub> \/c2<\/sub>
\n(b) a1<\/sub> \/a2<\/sub> \u2260 b1<\/sub> \/b2<\/sub> = c1<\/sub> \/c2<\/sub>
\n(c) a1<\/sub> \/a2<\/sub> \u2260 b1<\/sub> \/b2<\/sub> \u2260 c1<\/sub> \/c2<\/sub>
\n(d) a1<\/sub> \/a2<\/sub> = b1<\/sub> \/b2<\/sub> = c1<\/sub> \/c2<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) a1<\/sub> \/a2<\/sub> = b1<\/sub> \/b2<\/sub> \u2260 c1<\/sub> \/c2<\/sub>
\nTwo lines a1<\/sub> x + b1<\/sub> y + c1<\/sub> = 0 and a2<\/sub> x + b2<\/sub> y + c2<\/sub> = 0 are parallel if
\na1<\/sub> \/a2<\/sub> = b1<\/sub> \/b2<\/sub> \u2260 c1<\/sub> \/c2<\/sub><\/p>\n<\/details>\n

\n

Question 9.
\nIf the line x\/a + y\/b = 1 passes through the points (2, -3) and (4, -5), then (a, b) is
\n(a) a = 1 and b = 1
\n(b) a = 1 and b = \u22121
\n(c) a = \u22121 and b = 1
\n(d) a = \u22121 and b = -1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) a = \u22121 and b = -1
\nGiven equation of the line is x\/a + y\/b = 1
\n\u21d2 bx + ay = ab
\nIt is given that this line passes through (2, -3)
\n\u21d2 b(2) + a(-3) = ab
\n\u21d2 2b – 3a = ab ——– (1)
\nIt also passes through (4, -5)
\n\u21d2 4b – 5a = ab ——– (2)
\nOn solving equation (1) and (2), we get
\na = -1 and b = -1<\/p>\n<\/details>\n

\n

Question 10.
\nThe angle between the lines x – 2y = y and y – 2x = 5 is
\n(a) tan-1<\/sup> (1\/4)
\n(b) tan-1<\/sup> (3\/5)
\n(c) tan-1<\/sup> (5\/4)
\n(d) tan-1<\/sup> (2\/3)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) tan-1<\/sup> (5\/4)
\nGiven, lines are:
\nx – 2y = 5 ………. 1
\nand y – 2x = 5 ………. 2
\nFrom equation 1,
\nx – 5 = 2y
\n\u21d2 y = x\/2 – 5\/2
\nHere, m1<\/sub> = 1\/2
\nFrom equation 2,
\ny = 2x + 5
\nHere. m2<\/sub> = 2
\nNow, tan \u03b8 = |(m1<\/sub> + m2<\/sub>)\/{1 + m1<\/sub> \u00d7 m2<\/sub>}|
\n= |(1\/2 + 2)\/{1 + (1\/2) \u00d7 2}|
\n= |(5\/2)\/(1 + 1)|
\n= |(5\/2)\/2|
\n= 5\/4
\n\u21d2 \u03b8 = tan-1<\/sup> (5\/4)<\/p>\n<\/details>\n

\n

Question 11.
\nThe points on the y-axis whose distance from the line x\/3 + y\/4 = 1 is 4 units is
\n(a) (0, 32\/3) and (0, 8\/3)
\n(b) (0, -32\/3) and (0, 8\/3)
\n(c) (0, -32\/3) and (0, -8\/3)
\n(d) (0, 32\/3) and (0, -8\/3)<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) (0, 32\/3) and (0, -8\/3)
\nGiven equation of line is (x\/3) + (y\/4) = 1
\n\u21d2 4x + 3y = 12
\n\u21d2 4x + 3y – 12 = 0 ……………. 1
\nLet (0, b) is the point of the y-axis whose distance from given line is 4 unit.
\nWhen we compare equation 1 with general form of the equation Ax + By + C = 0, we get
\nA = 4, B = 3, C = -12
\nNow perpendicular distance of a line Ax + By + C = 0 from a point (x1<\/sub>, y1<\/sub>) is
\nd = |Ax1<\/sub> + By1<\/sub> + C|\/\u221a(A\u00b2 + B\u00b2)
\nSo perpendicular distance of a line 4x + 3y – 12 = 0 from a point (0 ,b) is
\n4 = |4\u00d70 + 3\u00d7b – 12|\/\u221a(4\u00b2 + 3\u00b2)
\n\u21d2 4 = |3b – 12|\/\u221a(16 + 9)
\n\u21d2 4 = |3b – 12|\/\u221a25
\n\u21d2 4 = |3b – 12|\/5
\n\u21d2 4 \u00d7 5 = |3b – 12|
\n\u21d2 |3b – 12| = 20
\nNow
\n3b – 12 = 20 and 3b – 12 = -20
\n\u21d2 3b = 20 12 and 3b = -20 + 12
\n\u21d2 3b = 32 and 3b = -8
\n\u21d2 b = 32\/3 and b = -8\/3
\nSo the points are (0, 32\/3) and (0, -8\/3)<\/p>\n<\/details>\n

\n

Question 12.
\nEquation of the line passing through (0, 0) and slope m is
\n(a) y = mx + c
\n(b) x = my + c
\n(c) y = mx
\n(d) x = my<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) y = mx
\nEquation of the line passing through (x1<\/sub>, y1<\/sub>) and slope m is
\n(y – y1<\/sub>) = m(x – x1<\/sub>)
\nNow, required line is
\n(y – 0 ) = m(x – 0)
\n\u21d2 y = mx<\/p>\n<\/details>\n

\n

Question 13.
\nThe distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is
\n(a) 3\/10
\n(b) 2\/3
\n(c) 3\/2
\n(d) 7\/10<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) 3\/10
\nGiven equations are:
\n3x + 4y = 9
\n\u21d2 3x + 4y – 9 = 0 and
\n6x + 8y = 15
\n\u21d2 6x + 8y – 15 = 0
\n\u21d2 3x + 4y – 15\/2 = 0
\nNow, compare these lines with a1<\/sub> x + b1<\/sub> y + c1<\/sub> = 0 and a2<\/sub> x + b2<\/sub> y + c2<\/sub> = 0, we get
\na1<\/sub> = 3, b1<\/sub> = 4, c1<\/sub> = -9 and
\na2<\/sub> = 3, b2<\/sub> = 4, c2<\/sub> = -15\/2
\nNow, distance between two parallel line = |c1<\/sub> – c2<\/sub>|\/\u221a(a1<\/sub>\u00b2 + b1<\/sub>\u00b2)
\n= |-9 + 15\/2|\/\u221a(3\u00b2 + 4\u00b2)
\n= |(-18 + 15)\/2|\/\u221a25
\n= |(-3\/2)|\/5
\n= (3\/2)\/5
\n= 3\/10<\/p>\n<\/details>\n

\n

Question 14.
\nWhat can be said regarding if a line if its slope is negative
\n(a) \u03b8 is an acute angle
\n(b) \u03b8 is an obtuse angle
\n(c) Either the line is x-axis or it is parallel to the x-axis.
\n(d) None of these<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) \u03b8 is an obtuse angle
\nLet \u03b8 be the angle of inclination of the given line with the positive direction of x-axis in the anticlockwise sense.
\nThen its slope is given by m = tan \u03b8
\nGiven, slope is positive
\n\u21d2 tan \u03b8 < 0
\n\u21d2 \u03b8 lies between 0 and 180 degree
\n\u21d2 \u03b8 is an obtuse angle<\/p>\n<\/details>\n

\n

Question 15.
\nTwo lines a1<\/sub> x + b1<\/sub> y + c1<\/sub> = 0 and a2<\/sub> x + b2<\/sub> y + c2<\/sub> = 0 are parallel if
\n(a) a1<\/sub> \/a2<\/sub> = b1<\/sub> \/b2<\/sub> \u2260 c1<\/sub> \/c2<\/sub>
\n(b) a1<\/sub> \/a2<\/sub> \u2260 b1<\/sub> \/b2<\/sub> = c1<\/sub> \/c2<\/sub>
\n(c) a1<\/sub> \/a2<\/sub> \u2260 b1<\/sub> \/b2<\/sub> \u2260 c1<\/sub> \/c2<\/sub>
\n(d) a1<\/sub> \/a2<\/sub> = b1<\/sub> \/b2<\/sub> = c1<\/sub> \/c2<\/sub><\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (a) a1<\/sub> \/a2<\/sub> = b1<\/sub> \/b2<\/sub> \u2260 c1<\/sub> \/c2<\/sub>
\nTwo lines a1<\/sub> x + b1<\/sub> y + c1<\/sub> = 0 and a2<\/sub> x + b2<\/sub> y + c2<\/sub> = 0 are parallel if
\na1<\/sub>\/a2<\/sub> = b1<\/sub>\/b2<\/sub> \u2260 c1<\/sub>\/c2<\/sub><\/p>\n<\/details>\n

\n

Question 16.
\nThe slope of a line making inclination of 30\u00b0 with the positive direction of x-axis is
\n(a) 1\/2
\n(b) \u221a3
\n(c) \u221a3\/2
\n(d) 1\/\u221a3<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (d) 1\/\u221a3
\nHere inclination of the line is 30\u00b0
\nSo, slope of the line m = tan 30\u00b0 = 1\/\u221a3<\/p>\n<\/details>\n

\n

Question 17.
\nThe perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2) is
\n(a) 5
\n(b) 4
\n(c) 2
\n(d) 1<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) 2
\nThe perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2)
\nd = |4 \u00d7 (-1) + 3 \u00d7 3 + 5|\/\u221a(4\u00b2 + 3\u00b2)
\n\u21d2 d = |-4 + 9 + 5|\/\u221a(16 + 9)
\n\u21d2 d = 10\/\u221a(25)
\n\u21d2 d = 10\/5
\n\u21d2 d = 2<\/p>\n<\/details>\n

\n

Question 18.
\nThe inclination of the line 5x – 5y + 8 = 0 is
\n(a) 30\u00b0
\n(b) 45\u00b0
\n(c) 60\u00b0
\n(d) 90\u00b0<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (b) 45\u00b0
\nGiven line is: 5x – 5y + 8 = 0
\n\u21d2 5y = 5x + 8
\n\u21d2 y = (5\/5)x + 8\/5
\n\u21d2 y = x + 8\/5
\nNow tan \u03b8 = 1
\n\u21d2 tan \u03b8 = tan 45\u00b0
\n\u21d2 \u03b8 = 45\u00b0
\nSo, the inclination of the line is 45\u00b0<\/p>\n<\/details>\n

\n

Question 19.
\nThe points (-a, -b), (0 , 0), (a, b) and (a\u00b2, ab) are
\n(a) vertices of a square
\n(b) vertices of a parallelogram
\n(c) collinear
\n(d) vertices of a rectangle<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) collinear
\nLet the four points are P(-a, -b), O(0, 0), Q(a, b) and R(a\u00b2, ab)
\nNow,
\nm1<\/sub> = slope of OP = b\/a
\nm2<\/sub> = slope of OQ = b\/a
\nm3<\/sub> = slope of OR = b\/a
\nSince m1<\/sub> = m2<\/sub> = m3<\/sub>
\nSo, the points O, P, Q, R are collinear.<\/p>\n<\/details>\n

\n

Question 20.
\nGiven the three straight lines with equations 5x + 4y = 0, x + 2y – 10 = 0 and 2x + y + 5 = 0, then these lines are
\n(a) none of these
\n(b) the sides of a right angled triangle
\n(c) concurrent
\n(d) the sides of an equilateral triangle<\/p>\n

\nAnswer<\/span><\/summary>\n

Answer: (c) concurrent
\nSince the determinant of these lines is equal to zero
\ni.e.
\n|5 4 0|
\n|1 2 -10| = 0
\n|2 1 -5|
\nSo, these three lines are concurrent.<\/p>\n<\/details>\n

\n

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use\u00a0MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download …<\/p>\n

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