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.

## Straight Lines Class 11 MCQs Questions with Answers

Question 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.

## Answer

Answer: (b) x – y = 0

Hint:

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

Question 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

## Answer

Answer: (c) y – 2 = 3 × (x – 1)

Hint:

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)

Question 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

## Answer

Answer: (b) θ is an obtuse angle

Hint:

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 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

## Answer

Answer: (a) x + y = α + β

Hint:

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

Question 5.

Two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are coincedent if

(a) a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

(b) a_{1}/a_{2} ≠ b_{1}/b_{2} = c_{1}/c_{2}

(c) a_{1}/a_{2} ≠ b_{1}/b_{2} ≠ c_{1}/c_{2}

(d) a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

## Answer

Answer: (d) a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Hint:

Two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are coincedent if

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

Question 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

## Answer

Answer: (c) 2x – y – 1 = 0

Hint:

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

Question 7.

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

(a) a/b

(b) -a/b

(c) -c/b

(d) c/b

## Answer

Answer: (b) -a/b

Hint:

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

Question 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

## Answer

Answer: (c) y = mx

Hint:

Equation of the line passing through (x_{1}, y_{1}) and slope m is

(y – y_{1}) = m(x – x_{1})

Now, required line is

(y – 0 ) = m(x – 0)

⇒ y = mx

Question 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)

## Answer

Answer: (c) tan^{-1} (5/4)

Hint:

Given, lines are:

x – 2y = 5 ………. 1

and y – 2x = 5 ………. 2

From equation 1,

x – 5 = 2y

⇒ y = x/2 – 5/2

Here, m_{1} = 1/2

From equation 2,

y = 2x + 5

Here. m_{2} = 2

Now, tan θ = |(m_{1} + m_{2})/{1 + m_{1} × m_{2}}|

= |(1/2 + 2)/{1 + (1/2) × 2}|

= |(5/2)/(1 + 1)|

= |(5/2)/2|

= 5/4

⇒ θ = tan^{-1} (5/4)

Question 10.

Two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are parallel if

(a) a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

(b) a_{1}/a_{2} ≠ b_{1}/b_{2} = c_{1}/c_{2}

(c) a_{1}/a_{2} ≠ b_{1}/b_{2} ≠ c_{1}/c_{2}

(d) a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

## Answer

Answer: (a) a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Hint:

Two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are parallel if

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Question 11.

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.

## Answer

Answer: (b) x – y = 0

Hint:

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

Question 12.

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)

## Answer

Answer: (b) (7, – 2)

Hint:

The equation of median through B is x + y = 5

The point B lies on it.

Let the coordinates of B are (x_{1}, 5 – x_{1})

Now CF is a median through C,

So co-ordiantes of F i.e. mid-point of AB are

((x_{1}+1)/2, (5 – x_{1}+ 2)/2)

Now since this lies on x = 4

⇒ (x_{1} + 1)/2 = 4

⇒ x_{1} + 1 = 8

⇒ x_{1} = 7

Hence, the co-oridnates of B are (7, -2)

Question 13.

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) √3y + x = 14

(c) √3x + y = 14

(d) None of these

## Answer

Answer: (c) √3x + y = 14

Hint:

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

⇒ √3x/2 + y/2 = 7

⇒ √3x + y = 7×2

⇒ √3x + y = 14

Question 14.

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)

## Answer

Answer: (d) (-5, -3)

Hint:

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 ………… 1

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 ………… 2

Multiply equation 2 by 2, we get

4x – 2y = -14 ……… 3

Subtract equation 1 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)

Question 15.

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

## Answer

Answer: (c) x² + y² = c² – a²

Hint:

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

Question 16.

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

Answer: (c) 2x + y – 7 = 0

Hint:

Given, points are: (1, 5) and (2, 3)

Now, equation of line is

y – y_{1} = {(y_{2} – y_{1})/(x_{2} – x_{1})} × (x – x_{1})

⇒ 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 17.

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

## Answer

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

Hint:

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.

Question 18.

Two lines are perpendicular if the product of their slopes is

(a) 0

(b) 1

(c) -1

(d) None of these

## Answer

Answer: (c) -1

Hint:

Let m_{1} is the slope of first line and m_{2} is the slope of second line.

Now, two lines are perpendicular if m_{1} × m_{2} = -1

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

Question 19.

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

(a) -15/4

(b) 15/4

(c) -5

(d) 5

## Answer

Answer: (d) 5

Hint:

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

Question 20.

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

## Answer

Answer: (b) 6x + 4y = 5

Hint:

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

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