Students can access Class 11 Maths Chapter 10 Straight Lines MCQs online, which will help them to score good marks in the examination. The multiple-choice questions are based on the CBSE syllabus (2022-2023) and NCERT guidelines. These multiple-choice questions have detailed answers and explanations. To obtain all chapter-by-chapter MCQs for Class 11 Maths, click here.
MCQs on Class 11 Maths Chapter 10 Straight Lines
Solve the MCQs for Class 11 Maths Chapter 10 Straight Lines. There are four options for each MCQ, but only one is accurate.
Download PDF – Chapter 10 Straight lines MCQs
Students should select the appropriate choice and compare their answers to those on our website. Also, look over the important questions for class 11 Maths.
1) Two lines are said to be perpendicular if the product of their slope is equal to:
- -1
- 0
- 1
- ½
Answer: (a) -1
Explanation:
When two lines are perpendicular, then the product of their slope is equal to -1. If two lines are perpendicular with slope m1 and m2, then m1.m2 = -1.
2) What is the distance of (5, 12) from the origin?
- 5 units
- 8 units
- 12 units
- 13 units
Answer: (d) 13 units
Explanation: Let the points be A(0, 0) and B(5, 12).
A (0, 0) = (x1, y1)
B(5, 12) = (x2, y2)
The distance between two points, AB = √[(x2-x1)2 + (y2-y1)2]
AB = √[(5-0)2 + (12-0)2]
AB = √(25+144)
AB = √(169)
AB = 13
Hence, the distance of (5, 12) from the origin is 13 units.
3) Two lines are said to be parallel if the difference of their slope is
- -1
- 0
- 1
- None of these
Answer: (b) 0
Explanation:
We know that two lines are said to be parallel if their slope is equal. If m1 and m2 are the slopes of two parallel lines, then it is represented as m1 = m2.
Hence, the difference of their slope should be m1-m2 = 0.
So, option (b) 0 is the correct answer.
4) The equation of a straight line that passes through the point (3, 4) and perpendicular to the line 3x+2y+5=0 is
- 2x-3y+6 = 0
- 2x+3y+6 = 0
- 2x-3y-6 = 0
- 2x+3y-6 = 0
Answer: (a) 2x-3y+6 = 0
Explanation:
The equation of a straight line perpendicular to 3x+2y+5 = 0 is 2x-3y+λ =0 …(1)
This passes through the point (3, 4).
Now, substitute in equation (1), we get
2(2) – 3(4) +λ = 0
4-12+λ =0
-6+λ =0
λ=6
Substituting λ=6 in (1), we get 2x-3y+6 = 0, which is the required equation.
Hence, option (a) 2x-3y+6 =0 is the correct answer.
5) The slope of a line ax+by+c =0 is
- a/b
- -a/b
- c/b
- -c/b
Answer: (b) -a/b
Explanation:
We know that the general equation of a line is ax+by+c = 0.
Rearranging the equation, we get
⟹ by = -ax -c
⟹y = (-a/b)x -(c/b) …(1)
This is of the form, y= mx+c …(2)
By comparing (1) and (2), we get
Slope, m = -a/b.
6) The equation of a line that passes through the points (1, 5) and (2, 3) is:
- 2x + y – 7 = 0
- 2x – y – 7 = 0
- x + 2y – 7 = 0
- 2x + y + 7 = 0
Answer: (a) 2x + y – 7 = 0
Explanation:
We know that the equation of a line passes through two points (x1, y1) and (x2 y2) is
(y-y1)/(x-x1) = (y2-y1)/(x2-x1)
(x1, y1) = (1, 5)
(x2, y2) = (2, 3)
Now, substitute the values in the formula, we get
(y-5)/(x-1) = (3-5)/(2-1)
(y-5)/(x-1) = (-2)/(1)
y-5 = -2(x-1)
y-5 = -2x+2
2x+y-5-2 =0
2x+y-7 =0
Therefore, the equation of a line that passes through the points (1, 5) and (2, 3) is 2x+y-7=0.
7) The distance between the lines 3x+4y=9 and 6x+8y=15 is
- 2/3
- 3/2
- 3/10
- 7/10
Answer: (c) 3/10
Explanation:
Given equations:
3x + 4y = 9 …(i)
6x + 8y =15 …(ii)
⟹3x + 4y = 15/2
The slope of line (i) is -3/4.
The slope of line (ii) is -3/4.
Since slopes are equal, the lines are parallel.
Hence, the distance between two parallel lines = |(c1 – c2)|/√(a2 + b2)
= |(9 – (15/2))|/√(32 + 42)
= 3/10
Therefore, the distance between the lines 3x+4y=9 and 6x+8y=15 is 3/10.
8) The locus of a point, whose abscissa and ordinate are always equal is
- x-y = 0
- x+y =1
- x+y+1 = 0
- None of the above
Answer: (a) x-y =0
Explanation:
Let the abscissa and ordinate of a point “P” be(x, y)
Given condition: Abscissa = Ordinate
(i.e) x =y
Hence, the locus of a point is x-y = 0.
Therefore, option (a) x-y =0 is the correct answer.
9) If A(6, 4) and B(2, 12) are the two points, then the slope of a line perpendicular to line AB is
- -2
- 2
- ½
- -½
Answer: (c) ½
Explanation:
Given points: A(6, 4) = (x1, y1)
B(2, 12) = (x2, y2)
We know that the slope of a line passing through two points (x1, y1) and (x2, y2) is (y2-y1)/(x2-x1).
m = (12-4)/(2-6) = 8/-4 = -2.
We know that the slope of two perpendicular lines m1.m2 = -1.
So, m2 = -1/m1
Hence, the slope of a line perpendicular to line AB is -1/m = -1/-2 = ½.
10) What can be said regarding a line if its slope is negative?
- θ is an acute angle
- θ is an obtuse angle
- Either the line is x-axis or it is parallel to the x-axis.
- None of these
Answer: (b) θ is an obtuse angle
Explanation:
The line with a negative slope makes an obtuse angle with a positive x-axis when measured in the anti-clockwise direction.
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