Question

Class 11 Maths MCQ Questions of Straight Lines with Answers?

Answer 1

 Class 11 Maths MCQ Questions of Straight Lines with Answers are furnished here for the students who are preparing for the Class 11 Board Exams and other competitive exams. These MCQ Questions for Class 11 Maths with Solutions cover all topics involved in the latest CBSE Class 11 Maths Syllabus.

Class 11 Maths Straight Lines covers the following important concepts such as:

• General equation of a line
• Various forms of the equation of a line
• The slope of a line
• The distance of a point from a line.

To score good marks in the final examination, Just start your preparation for the Class 11 maths board examination with MCQ Questions available over here which will help you to solve the problems in the annual examination.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The locus of a point, whose abscissa and ordinate are always equal is

(a) x + y + 1 = 0
(b) x – y = 0
(c) x + y = 1
(d) none of these

2. The equation of straight line passing through the point (1, 2) and parallel to the line y = 3x + 1 is

(a) y + 2 = x + 1
(b) y + 2 = 3 × (x + 1)
(c) y – 2 = 3 × (x – 1)
(d) y – 2 = x – 1

3. 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

4. The equation of the line which cuts off equal and positive intercepts from the axes and passes through the point (α, β) is

(a) x + y = α + β
(b) x + y = α
(c) x + y = β
(d) None of these

5. Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are coincedent 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

6. The equation of the line passing through the point (2, 3) with slope 2 is

(a) 2x + y – 1 = 0
(b) 2x – y + 1 = 0
(c) 2x – y – 1 = 0
(d) 2x + y + 1 = 0

7. The slope of the line ax + by + c = 0 is

(a) a/b
(b) -a/b
(c) -c/b
(d) c/b

8. 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

9. 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)

10. 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)

11. The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis. Then the equation of line is

(a) x + y = 14
(b) \(\sqrt3 y\) + x = 14
(c) \(\sqrt 3x\) + y = 14
(d) None of these

12. If two vertices of a triangle are (3, -2) and (-2, 3) and its orthocenter is (-6, 1) then its third vertex is

(a) (5, 3)
(b) (-5, 3)
(c) (5, -3)
(d) (-5, -3)

13. The sum of squares of the distances of a moving point from two fixed points (a, 0) and (-a, 0) is equal to 2c² then the equation of its locus is

(a) x² – y² = c² – a²
(b) x² – y² = c² + a²
(c) x² + y² = c² – a²
(d) x² + y² = c² + a²

14. What can be said regarding if a line if its slope is zero

(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

15. Two lines are perpendicular if the product of their slopes is

(a) 0
(b) 1
(c) -1
(d) None of these

16. y-intercept of the line 4x – 3y + 15 = 0 is

(a) -15/4
(b) 15/4
(c) -5
(d) 5

17. The equation of the locus of a point equidistant from the point A(1, 3) and B(-2, 1) is

(a) 6x – 4y = 5
(b) 6x + 4y = 5
(c) 6x + 4y = 7
(d) 6x – 4y = 7

18. If the midpoint of the section of the straight line intercepted between the axes is (1,1), then the equation of the line is

(a) 2x+y=3
(b) 2x−y=1  
(c) x−y=0
(d) x+y=2

19. The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y - 11 = 0 are

(a) (-6, 5)
(b) (5, 6)
(c) (-5, 6)  
(d) (6, 5)

20. The coordinates of P and Q are (-3, 4) and (2, 1), respectively. If PQ is extended to R such that PR = 2QR, then what are the coordinates of R?

(a) (3, 7)
(b) (2, 4)
(c) (-1/2,5/2)
(d) (7, - 2)

Answer 2

1. Answer: (b) x – y = 0

Explanation: Let the coordinate of the variable point P is (x, y)

Now, the abscissa of this point = x

and its ordinate = y

Given, abscissa = ordinate

⇒ x = y

⇒ x – y = 0

So, the locus of the point is x – y = 0

2. Answer: (c) y – 2 = 3 × (x – 1)

Explanation: Given straight line is: y = 3x + 1

Slope = 3

Now, required line is parallel to this line.

So, slope = 3

Hence, the line is

y – 2 = 3 × (x – 1)

3. Answer: (b) θ is an obtuse angle

Explanation: 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

4. Answer: (a) x + y = α + β

Explanation: Let the equation of the line be x/a + y/b = 1 which cuts off intercepts a and b with the coordinate axes.

It is given that a = b, therefore the equation of the line is

x/a + y/a = 1

⇒ x + y = a …..1

But it is passes through (α, β)

So, α + β = a

Put this value in equation 1, we get

x + y = α + β

5. Answer: (d) a₁/a₂ = b₁/b₂ = c₁/c₂

Explanation: For co-incident lines

\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

6. Answer: (c) 2x – y – 1 = 0

Explanation: Given, the point (2, 3) and slope of the line is 2

By, slope-intercept formula,

y – 3 = 2(x – 2)

⇒ y – 3 = 2x – 4

⇒ 2x – 4 – y + 3 = 0

⇒ 2x – y – 1 = 0

7. Answer: (b) -a/b

Explanation: Give, equation of line is ax + by + c = 0

⇒ by = -ax – c

⇒ y = (-a/b)x – c/b

It is in the form of y = mx + c

Now, slope m = -a/b

8. Answer: (c) y = mx

Explanation: 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

9. Answer: (c) tan^-1 (5/4)

Explanation: 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)

10. Answer: (b) (7, – 2)

Explanation: 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 co-ordiantes 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 co-oridnates of B are (7, -2)

Answer 3

11. Answer: (c) \(\sqrt{3}x+y=14\)

Explanation: Given, The length of the perpendicular from the origin to a line is 7 and the line makes an angle of 150 degrees with the positive direction of the y-axis.

Now, equation of line is

x × cos 30 + y × sin 30 = 7

⇒ \(\frac{\sqrt3}{2}\) + y/2 = 7

⇒ \(\sqrt 3x\) + y = 7 × 2

⇒ \(\sqrt 3 x\) + y = 14

12. Answer: (d) (-5, -3)

Explanation: Let the third vertex of the triangle is C(x, y)

Given, two vertices of a triangle are A(3,-2) and B(-2,3)

Now given orthocentre of the circle = H(-6, 1)

So, AH ⊥ BC and BH ⊥ AC

Since the product of the slope of perpendicular lines equal to -1

Now, AH ⊥ BC

⇒ {(-2 – 1)/(3 + 6)} × {(y + 2)/(x – 3)} = -1

⇒ (-3/9) × {(y + 2)/(x – 3)} = -1

⇒ (-1/3)×{(y – 3)/(x + 2)} = -1

⇒ (y – 3)/{3×(x + 2)} = 1

⇒ (y – 3) = 3×(x + 2)

⇒ y – 3 = 3x + 6

⇒ 3x + 6 – y = -3

⇒ 3x – y = -3 – 6

⇒ 3x – 2y = -9 ………… (i)

Again, BH ⊥ AC

⇒ {(3 – 1)/(-2 + 6)} × {(y – 3)/(x + 2)} = -1

⇒ (2/4) × {(y – 3)/(x + 2)} = -1

⇒ (1/2)×{(y – 3)/(x + 2)} = -1

⇒ (y – 3)/{2×(x + 2)} = 1

⇒ (y – 3) = 2×(x + 2)

⇒ y – 3 = 2x + 4

⇒ 2x + 4 – y = -3

⇒ 2x – y = -3 – 4

⇒ 2x – y = -7 ………… (ii)

Multiply equation 2 by 2, we get

4x – 2y = -14 ……… (iii)

Subtract equation i and we get

-x = 5

⇒ x = -5

From equation 2, we get

2×(-5) – y = -7

⇒ -10 – y = -7

⇒ y = -10 + 7

⇒ y = -3

So, the third vertex of the triangle is (-5, -3)

13. Answer: (c) x² + y² = c² – a²

Explanation: Let P(h, k) be any position of the moving point and let A(a, 0) and B(-a, 0) be the given points. Then

PA² + PB² = 2c²

⇒ (h – a)² + (k – 0)² + (h + a)² + (k – 0)² = 2c²

⇒ h² – 2ah + a² + k² + h² + 2ah + a² + k² = 2c²

⇒ 2h² + 2k² + 2a² = 2c²

⇒ h² + k² + a² = c²

⇒ h² + k² = c² – a²

Hence, the locus of (h, k) is x² + y² = c²– a²

14. Answer: (c) Either the line is x-axis or it is parallel to the x-axis.

Explanation: 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 zero

⇒ tan θ = 0

⇒ θ = 0°

⇒ Either the line is x-axis or it is parallel to the x-axis.

15. Answer: (c) -1

Explanation: Let m₁ is the slope of first line and m₂ is the slope of second line.

Now, two lines are perpendicular if m₁ × m₂ = -1

i.e. the product of their slopes is equals to -1

16. Answer: (d) 5

Explanation: Given, equation of line is 4x – 3y + 15 = 0

⇒ 4x – 3y = -15

⇒ 4x/(-15) + (-3)y/(-15) = 1

⇒ x/(-15/4) + 3y/15 = 1

⇒ x/(-15/4) + y/(15/3) = 1

⇒ x/(-15/4) + y/5 = 1

Now, compare with x/a + y/b = 1, we get y-intercept b = 5

17. Answer: (b) 6x + 4y = 5

Explanation: Let P(h, k) be any point on the locus. Then

Given, PA = PB

⇒ PA² = PB²

⇒ (h – 1)² + (k – 3)² = (h + 2)² + (k – 1)²

⇒ h² – 2h + 1 + k² – 6k + 9 = h²+ 4h + 4 + k² – 2k + 1

⇒ -2h – 6k + 10 = 4h – 2k + 5

⇒ 6h + 4k = 5

Hence, the locus of (h, k) is 6x + 4y = 5

18. Answer: (d) x+y=2

Explanation: It is given that (1, 1) is the midpoint of line AB.

1 = a+0/2,1=0+b/2 

⇒a =2 and b=2

Equation of line AB is

x₂ + y₂ =1

⇒x + y =2

19. Answer: (b) (5, 6)

Explanation: Let (h, k) be the coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y - 11 = 0. Then, the slope of the perpendicular line is k−3h−2 .

Again the slope of the given line x + y - 11 = 0 is - 1

Using the condition of perpendicularity of lines, we have

\((\frac{k-3}{h-2})(-1)=-1\)

or k−h =1 ....(1) 

Since (h, k) lies on the given line, we have, h +k−11=0 or h + k =11 ...(2)

Solving (1) and (2), we get h = 5 and k = 6.

Thus (5, 6) are the required coordinates of the foot of the perpendicular.

20. Answer: (d) (7, - 2)

Explanation: As given :

Coordinates of P and Q are (- 3, 4) and (2, 1) respectively.

Let coordinates of R be (x, y).

As given : PR = 2 QR

⇒ PR - QR = QR

⇒ PQ = QR.

So, Q is the mid point of P and R

⇒2 = −3+x/22 and = 4+y/2

⇒x = 7 and y = −2

∴ Coordinates of R = (7,−2)

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