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

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.

Explore numerous MCQ Questions of Straight Lines Class 11 with answers provided with detailed solutions by looking below.

Question 1.
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) none of these
(d) (4, 1)

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)

Question 2.
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

Answer: (b) 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

Question 3.
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

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.

Question 4.
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

Answer: (c) 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

Question 5.
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

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

Question 6.
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

Let the A.P. are 1, A1, A2, A3 …… Am, 31
a = 1, an = 31 and n = m + 2
Now, an = 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 Am-1 = a + (m – 1)d = 1 + (m – 1)[30/(m + 1)] ………. 2
From equation 1 and 2, we get
A7/Am-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

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

Given lines are:
xy = 0 and x + y = 1
⇒ x = 0, y = 0 and x + y = 1

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

Question 8.
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

Answer: (a) 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

Question 9.
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

Answer: (d) 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

Question 10.
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)

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)

Question 11.
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)

Answer: (d) (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)

Question 12.
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

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

Question 13.
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

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

Question 14.
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

Answer: (b) 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

Question 15.
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

Answer: (a) 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

Question 16.
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

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

Question 17.
The perpendicular distance of a line 4x + 3y + 5 = 0 from the point (-1, 2) is
(a) 5
(b) 4
(c) 2
(d) 1

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

Question 18.
The inclination of the line 5x – 5y + 8 = 0 is
(a) 30°
(b) 45°
(c) 60°
(d) 90°

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°

Question 19.
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

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.

Question 20.
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