MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers

Straight Lines Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Straight Lines Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 10 Quiz

Class 11 Maths Chapter 10 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Straight Lines Class 11 Maths MCQ online test

Q1.	In a ΔABC, 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
	A.(1, 4)
	B.(7, -2)
	C.(4, 1)
	D.none of these

Ans: (7, -2)
The equation of median through B is x + y = 5
The point B lies on it.
Let the coordinates of B are (x₁, 5 – x₁)
Now CF is a median through C,
So coordiantes of F i.e. mid-point of AB are
((x₁ + 1)/2, (5 – x₁ + 2)/2)
Now since this lies on x = 4
⇒ (x₁ + 1)/2 = 4
⇒ x₁ + 1 = 8
⇒ x₁ = 7
Hence, the cooridnates of B are (7, -2)

Q2.	The equation of straight line passing through the point (1, 2) and perpendicular to the line x + y + 1 = 0
	A.y – x + 1 = 0
	B.y – x – 1 = 0
	C.y – x + 2 = 0
	D.y – x – 2 = 0

Ans: y – x – 1 = 0
Given straight line is: x + y + 1 = 0
⇒ y = -x – 1
Slope = -1
Now, required line is perpendicular to this line.
So, slope = -1/-1 = 1
Hence, the line is
y – 2 = 1 × (x – 1)
⇒ y – 2 = x – 1
⇒ y – 2 – x + 1 = 0
⇒ y – x – 1 = 0

Q3.	The points (-a, -b), (0, 0), (a, b) and (a², ab) are
	A.vertices of a square
	B.vertices of a parallelogram
	C.collinear
	D.vertices of a rectangle

Ans: collinear
Let the four points are P(-a, -b), O(0, 0), Q(a, b) and R(a², ab)
Now,
m₁ = slope of OP = b/a
m₂ = slope of OQ = b/a
m₃ = slope of OR = b/a
Since m₁ = m₂ = m₃
So, the points O, P, Q, R are collinear.

Q4.	The equation of the line through the points (1, 5) and (2, 3) is
	A.2x – y – 7 = 0
	B.2x + y + 7 = 0
	C.2x + y – 7 = 0
	D.x + 2y – 7 = 0

Ans: 2x + y – 7 = 0
Given, points are: (1, 5) and (2, 3)
Now, equation of line is
y – y₁ = {(y₂ – y₁)/(x₂ – x₁)} × (x – x₁)
⇒ y – 5 = {(3 – 5)/(2 – 1)} × (x – 1)
⇒ y – 5 = (-2) × (x – 1)
⇒ y – 5 = -2x + 2
⇒ 2x + y – 5 – 2 = 0
⇒ 2x + y – 7 = 0

Q5.	The slope of a line which passes through points (3, 2) and (-1, 5) is
	A.3/4
	B.-3/4
	C.4/3
	D.-4/3

Ans: -3/4
Given, points are (3, 2) and (-1, 5)
Now, slope m = (5 – 2)/(-1 – 3)
⇒ m = -3/4
So, the slope of the line is -3/4

Q6.	The 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
	A.15
	B.12
	C.13
	D.14

Ans: 14
Let the A.P. are 1, A₁, A₂, A₃ …… A_m, 31
a = 1, a_n = 31 and n = m + 2
Now, a_n = a + (n – 1)d
⇒ 31 = 1 + (m + 2 – 1)d
⇒ 30 = (m + 1)d
⇒ d = 30/(m + 1)
Again, A₇ = a + 7d = 1 + 7[30/(m + 1)] …………….. 1
and A_m-1 = a + (m – 1)d = 1 + (m – 1)[30/(m + 1)] ………. 2
From equation 1 and 2, we get
A₇/A_m-1 = 5/9
⇒ 1 + 7[30/(m + 1) / 1 + (m – 1)[30/(m + 1)] = 5/9
⇒ [m + 1 + 7(30)] / [m + 1 + 30 m – 30] = 5/9
⇒ [m + 211] / [31 m – 29] = 5/9
⇒ 9[m + 211] = 5[31 m – 29]
⇒ 9 m + 1899 = 155 m – 145
⇒ 146 m = 2044
⇒ m = 2044/146
⇒ m = 14
So, the value of m is 14

Q7.	The ortho centre of the triangle formed by lines xy = 0 and x + y = 1 is :
	A.(0, 0)
	B.( 1/3, 1/3)
	C.( 1/2, 1/2)
	D.none of these

Ans: (0, 0)
Given lines are:
xy = 0 and x + y = 1
⇒ x = 0, y = 0 and x + y = 1
MCQ Questions for Class 11 Maths Chapter 10 Straight Lines with Answers 1

In a triangle OAB, OA and OB are the altitudes which intersect at O.
So, the required orthocentre is (0, 0)

Q8.	Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
	A.a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂
	B.a₁ /a₂ ≠ b₁ /b₂ = c₁ /c₂
	C.a₁ /a₂ ≠ b₁ /b₂ ≠ c₁ /c₂
	D.a₁ /a₂ = b₁ /b₂ = c₁ /c₂

Ans: a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂

Q9.	If the line x/a + y/b = 1 passes through the points (2, -3) and (4, -5), then (a, b) is
	A.a = 1 and b = 1
	B.a = 1 and b = −1
	C.a = −1 and b = 1
	D.a = −1 and b = -1

Ans: a = −1 and b = -1
Given equation of the line is x/a + y/b = 1
⇒ bx + ay = ab
It is given that this line passes through (2, -3)
⇒ b(2) + a(-3) = ab
⇒ 2b – 3a = ab ——– (1)
It also passes through (4, -5)
⇒ 4b – 5a = ab ——– (2)
On solving equation (1) and (2), we get
a = -1 and b = -1

Q10.	The angle between the lines x – 2y = y and y – 2x = 5 is
	A.tan^-1 (1/4)
	B.tan^-1 (3/5)
	C.tan^-1 (5/4)
	D.tan^-1 (2/3)

Ans: tan^-1 (5/4)
Given, lines are:
x – 2y = 5 ………. 1
and y – 2x = 5 ………. 2
From equation 1,
x – 5 = 2y
⇒ y = x/2 – 5/2
Here, m₁ = 1/2
From equation 2,
y = 2x + 5
Here. m₂ = 2
Now, tan θ = |(m₁ + m₂)/{1 + m₁ × m₂}|
= |(1/2 + 2)/{1 + (1/2) × 2}|
= |(5/2)/(1 + 1)|
= |(5/2)/2|
= 5/4
⇒ θ = tan^-1 (5/4)

Q11.	The points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units is
	A.(0, 32/3) and (0, 8/3)
	B.(0, -32/3) and (0, 8/3)
	C.(0, -32/3) and (0, -8/3)
	D.(0, 32/3) and (0, -8/3)

Ans: (0, 32/3) and (0, -8/3)
Given equation of line is (x/3) + (y/4) = 1
⇒ 4x + 3y = 12
⇒ 4x + 3y – 12 = 0 ……………. 1
Let (0, b) is the point of the y-axis whose distance from given line is 4 unit.
When we compare equation 1 with general form of the equation Ax + By + C = 0, we get
A = 4, B = 3, C = -12
Now perpendicular distance of a line Ax + By + C = 0 from a point (x₁, y₁) is
d = |Ax₁ + By₁ + C|/√(A² + B²)
So perpendicular distance of a line 4x + 3y – 12 = 0 from a point (0 ,b) is
4 = |4×0 + 3×b – 12|/√(4² + 3²)
⇒ 4 = |3b – 12|/√(16 + 9)
⇒ 4 = |3b – 12|/√25
⇒ 4 = |3b – 12|/5
⇒ 4 × 5 = |3b – 12|
⇒ |3b – 12| = 20
Now
3b – 12 = 20 and 3b – 12 = -20
⇒ 3b = 20 12 and 3b = -20 + 12
⇒ 3b = 32 and 3b = -8
⇒ b = 32/3 and b = -8/3
So the points are (0, 32/3) and (0, -8/3)

Q12.	Equation of the line passing through (0, 0) and slope m is
	A.y = mx + c
	B.x = my + c
	C.y = mx
	D.x = my

Ans: y = mx
Equation of the line passing through (x₁, y₁) and slope m is
(y – y₁) = m(x – x₁)
Now, required line is
(y – 0 ) = m(x – 0)
⇒ y = mx

Q13.	The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is
	A.3/10
	B.2/3
	C.3/2
	D.7/10

Ans: 3/10
Given equations are:
3x + 4y = 9
⇒ 3x + 4y – 9 = 0 and
6x + 8y = 15
⇒ 6x + 8y – 15 = 0
⇒ 3x + 4y – 15/2 = 0
Now, compare these lines with a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0, we get
a₁ = 3, b₁ = 4, c₁ = -9 and
a₂ = 3, b₂ = 4, c₂ = -15/2
Now, distance between two parallel line = |c₁ – c₂|/√(a₁² + b₁²)
= |-9 + 15/2|/√(3² + 4²)
= |(-18 + 15)/2|/√25
= |(-3/2)|/5
= (3/2)/5
= 3/10

Q14.	What can be said regarding if a line if its slope is negative
	A.θ is an acute angle
	B.θ is an obtuse angle
	C.Either the line is x-axis or it is parallel to the x-axis.
	D.None of these

Ans: θ is an obtuse angle
Let θ be the angle of inclination of the given line with the positive direction of x-axis in the anticlockwise sense.
Then its slope is given by m = tan θ
Given, slope is positive
⇒ tan θ < 0
⇒ θ lies between 0 and 180 degree
⇒ θ is an obtuse angle

Q15.	Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
	A.a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂
	B.a₁ /a₂ ≠ b₁ /b₂ = c₁ /c₂
	C.a₁ /a₂ ≠ b₁ /b₂ ≠ c₁ /c₂
	D.a₁ /a₂ = b₁ /b₂ = c₁ /c₂

Ans: a₁ /a₂ = b₁ /b₂ ≠ c₁ /c₂
Two lines a₁ x + b₁ y + c₁ = 0 and a₂ x + b₂ y + c₂ = 0 are parallel if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Q16.	The slope of a line making inclination of 30° with the positive direction of x-axis is
	A.1/2
	B.√3
	C.√3/2
	D.1/√3

Ans: 1/√3
Here inclination of the line is 30°
So, slope of the line m = tan 30° = 1/√3

Q17.	The perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2) is
	A.5
	B.4
	C.2
	D.1

Ans: 2
The perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2)
d = |4 × (-1) + 3 × 3 + 5|/√(4² + 3²)
⇒ d = |-4 + 9 + 5|/√(16 + 9)
⇒ d = 10/√(25)
⇒ d = 10/5
⇒ d = 2

Q18.	The inclination of the line 5x – 5y + 8 = 0 is
	A.30°
	B.45°
	C.60°
	D.90°

Ans: 45°
Given line is: 5x – 5y + 8 = 0
⇒ 5y = 5x + 8
⇒ y = (5/5)x + 8/5
⇒ y = x + 8/5
Now tan θ = 1
⇒ tan θ = tan 45°
⇒ θ = 45°
So, the inclination of the line is 45°

Q19.	The points (-a, -b), (0 , 0), (a, b) and (a², ab) are
	A.vertices of a square
	B.vertices of a parallelogram
	C.collinear
	D.vertices of a rectangle

Q20.	Given the three straight lines with equations 5x + 4y = 0, x + 2y – 10 = 0 and 2x + y + 5 = 0, then these lines are
	A.none of these
	B.the sides of a right angled triangle
	C.concurrent
	D.the sides of an equilateral triangle