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

Answer: (b) (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)

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

Answer: (c) 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.

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

Answer: (b) -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

Answer: (d) 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

Answer: (a) (0, 0)
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)

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

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

Answer: (c) 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)

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)

Answer: (c) 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

Answer: (a) 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

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

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

Answer: (d) 1/√3
Here inclination of the line is 30°
So, slope of the line m = tan 30° = 1/√3

Answer: (c) 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

Answer: (b) 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°

Answer: (c) 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.

Answer: (c) concurrent
Since the determinant of these lines is equal to zero
i.e.
|5 4 0|
|1 2 -10| = 0
|2 1 -5|
So, these three lines are concurrent.