Equation of a Line MCQ Quiz - Objective Question with Answer for Equation of a Line - Download Free PDF

Last updated on Apr 23, 2025

Latest Equation of a Line MCQ Objective Questions

Equation of a Line Question 1:

Comprehension:

Direction: Consider the following for the items that follow:

Let L : x + y + z + 4 = 0 = 2x - y - z + 8 be a line and P : x + 2y + 3z + 1 = 0 be a plane. 

What is the point of intersection of L and P? 

  1. (4, 3, -3)
  2. (4, -3, 3)
  3. (-4, -3, -3)
  4. (-4, -3, 3)

Answer (Detailed Solution Below)

Option 4 : (-4, -3, 3)

Equation of a Line Question 1 Detailed Solution

Explanation:

Given:

L = x + y + z + 4 = 0 = 2x – y – z + 8 and

P = x + 2y + 3z + 1 = 0 is a plane.

To find the point of intersection of L and P,

The point must satisfy both line L and plane P and Option (d) (–4, –3, 3) will satisfy both line L and plane P.

So option (d) is the correct answer.

Equation of a Line Question 2:

Comprehension:

Direction: Consider the following for the items that follow:

Let L : x + y + z + 4 = 0 = 2x - y - z + 8 be a line and P : x + 2y + 3z + 1 = 0 be a plane. 

What are the direction ratios of the line?  

  1. 〈2, 1, -1〉
  2. 〈0, -1, 2〉
  3. 〈0, 1, -1〉
  4. 〈2, 3, -3〉

Answer (Detailed Solution Below)

Option 3 : 〈0, 1, -1〉

Equation of a Line Question 2 Detailed Solution

Explanation:

Given:

L = x + y + z + 4 = 0 = 2x – y – z + 8 and

P = x + 2y + 3z + 1 = 0 is a plane. 

⇒ Let direction ratios of line (a, b, c)

Since line is obtained by the intersection of planes.

⇒ x + y + z + 4 = 0 and 2x – y – z + 8 = 0

So a + b + c = 0....(i)

⇒ 2a – b – c = 0...(ii)

Solve (i) and (ii)

a1+1=b2+1=c12=λ

⇒ a = 0, b = 3λ  and c = –3λ 

⇒ Direction ratios of line is (0, 3, –3) or (0, 1, –1).

∴ Option (c) is correct.

Equation of a Line Question 3:

Let the line passing through the points (–1, 2, 1) and parallel to the line x12=y+13=z4 intersect the line x+23=y32=z41 at the point P. Then the distance of P from the point Q(4, – 5, 1) is :

  1. 5
  2. 10
  3. 5√6
  4. 5√5

Answer (Detailed Solution Below)

Option 4 : 5√5

Equation of a Line Question 3 Detailed Solution

Calculation

Equation of line through point (–1, 2,1) is →qImage67b1d5bf5c75d69925f3a7ef

⇒ x+12=y23=z14(2)=λ

So, [x=2λ1y=3λ+2z=4λ+1

By (1) → x+23=y32=z41=μ( Let )

So, [x=3μ2y=2μ+3z=μ+4

For intersection point ‘P’

x = 2λ – 1 = 3μ – 2

y = 3λ + 2 = 2μ + 3 [λ=1μ=1]

z = 4λ + 1 = µ + 4 

So, point P(x, y, z) = (1, 5, 5)

& Q(4, – 5, 1) 

∴ PQ = 9+100+16

125=55

Hence option 4 is correct

Equation of a Line Question 4:

The equation of the straight line passing through the points O(0, 0) and A(2, 5) is

  1. 2x = 5y
  2. 2y = 5x
  3. 3x = 5y
  4. 3y = 5x

Answer (Detailed Solution Below)

Option 2 : 2y = 5x

Equation of a Line Question 4 Detailed Solution

- pehlivanlokantalari.com

To find the equation of the straight line passing through the points O(0, 0) and A(2, 5), follow these steps:

  1. Find the Slope of the Line:
    The slope m of a line passing through two points (x1, y1) and (x2, y2) is given by:
    m=y2y1x2x1
    Substitute the coordinates of O(0, 0) and A(2, 5):
    m=5020=52
  2. Use the Point-Slope Form:
    The point-slope form of a line is:
    yy1=m(xx1)
    Using the slope m = 52 and the point O(0, 0) :
    y0=52(x0)
    Simplify:
    y=52x
  3. Write the Equation in Standard Form:
    The standard form of a line is Ax + By + C = 0. Rearrange the equation:
    52xy=0
    Multiply through by 2 to eliminate the fraction:
    5x2y=0

Final Answer:

The equation of the straight line passing through O(0, 0) and A(2, 5) is:

5x2y=0

Equation of a Line Question 5:

The equation of straight line passing through origin making angle 60° with positive direction of X-axis is

  1. x - √3y = 0
  2. √3x - y = 0
  3. y = 3x
  4. x = 3y

Answer (Detailed Solution Below)

Option 2 : √3x - y = 0

Equation of a Line Question 5 Detailed Solution

- pehlivanlokantalari.com

To find the equation of a straight line passing through the origin and making an angle of 600 with the positive direction of the  X-axis, follow these steps:

  1. Understand the Slope of the Line:
    The slope m of a line is related to the angle θ it makes with the positive X-axis by the formula:
    m=tan(θ)
    Given θ=60 , the slope is:
    m=tan(60)=3
  2. Use the Point-Slope Form:
    The point-slope form of a line passing through the origin (0, 0) with slope m is:
    y=mx
    Substitute m =3}:
    y=3x
  3. Write the Equation in Standard Form:
    The standard form of a line is Ax + By + C = 0. Rearrange the equation:
    3xy=0

Final Answer:

The equation of the straight line passing through the origin and making an angle of 60with the positive X-axis is:

3xy=0

Top Equation of a Line MCQ Objective Questions

P is a point on the line segment joining the points (3, 2, -1) and (6, -4, -2). If x coordinate of P is 5, then its y coordinate is:

  1. 2
  2. 1
  3. -1
  4. -2

Answer (Detailed Solution Below)

Option 4 : -2

Equation of a Line Question 6 Detailed Solution

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Concept:

Equation of line joining two points (x1,y1,z1) and (x2,y2,z2) is given as

xx1x2x1=yy1y2y1=zz1z2z1

A point lies on the line only when its coordinates satisfy the equation of the line.

Calculation:

Given:

P = (5,y,z)

The equation of line joining (3,2,-1) and (6,-4,-2) is 

x363=y242=z+12+1=x33=y26=z+11

so if point P lies on the line then it must satisfy the above equation

533=y26=z+11

23=y26

⇒ - 4 = y - 2

⇒ y =  - 2

Hence y coordinate of P is -2

What are the direction ratios of the line of intersection of the planes x = 3z + 4 and y = 2z - 3?

  1. 〈1, 2, 3〉  
  2. 〈2, 1, 3〉  
  3. 〈3, 2, 1〉  
  4. 〈1, 3, 2〉  

Answer (Detailed Solution Below)

Option 3 : 〈3, 2, 1〉  

Equation of a Line Question 7 Detailed Solution

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Concept:

 The equation of a line with direction ratio (a, b, c) that passes through the point (x1, y1, z1) is given by the formula: xx1a=yy1b=zz1c

If two planes intersect each other, the intersection will always be a line.

 

Calculation:

Given:

x = 3z + 4 and y = 2z - 3

⇒ x - 4 = 3z and y + 3 = 2z

x43=zand y+32=z

x43=y+32=z1

If two planes intersect each other, the intersection will always be a line.

Now,  direction ratios of the line are 〈3, 2, 1〉

Determine the vector equation for the line, given the cartesian equation of a line is x+53=y72=z+32?

  1. r=(3i^+5j^6k^)+λ(2i^+4j^+2k^)
  2. r=(5i^+7j^3k^)+λ(2i^+4j^+2k^)
  3. r=(5i^+7j^3k^)+λ(3i^+2j^+2k^)
  4. r=(3i^+5j^6k^)+λ(3i^+2j^+2k^)

Answer (Detailed Solution Below)

Option 3 : r=(5i^+7j^3k^)+λ(3i^+2j^+2k^)

Equation of a Line Question 8 Detailed Solution

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Concept:

Equation of a line in cartesian form:- xx1a=yy1b=zz1c

Equation of a line in vector form:- r=a+λb 

where, a=x1i^+y1j^+z1k^ and b=ai^+bj^+ck^

Calculation:

We have,

Cartesian equation:- x+53=y72=z+32

⇒ x(5)3=y72=z(3)2      ------- equation (1)

⇒ xx1a=yy1b=zz1c           -------- equation (2)

Comparing (1) and (2)

⇒ x1 = -5, y= 7, z1 = -3

⇒ a = 3, b = 2, c = 2

⇒ a=5i^+7j^3k^

⇒ b=3i^+2j^+2k^

Equation of line in vector form:- r=a+λb

⇒ r=(5i^+7j^3k^)+λ(3i^+2j^+2k^)

∴ The vector equation for the line is r=(5i^+7j^3k^)+λ(3i^+2j^+2k^) 

The value of k for which straight line x + y + 3z - 2 = 0 = 2x + y - z - 3 is parallel to the plane 3x + 2y + kz - 4 = 0 is:

  1. -1
  2. 2
  3. 1
  4. 3

Answer (Detailed Solution Below)

Option 2 : 2

Equation of a Line Question 9 Detailed Solution

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Given:

 The straight line x + y + 3z - 2 = 0 = 2x + y - z - 3 is parallel to the plane 3x + 2y + kz - 4 = 0

Concept:

If a line parallel to the plane then dot product of direction ratio of line and normal to plane must be 0 

Calculation:

 The straight line x + y + 3z - 2 = 0 = 2x + y - z - 3 is parallel to the plane 3x + 2y + kz - 4 = 0

So, the direction ratios of normals to the planes are ( 1, 1, 3 ) and ( 2, 1, -1 ) repectively .

So direction ratio of the required line is the cross product of these normals

i.e. |ijk113211| 

= i(-1 - 3) - j(-1 - 6) + k(1 - 2)

= - 4i + 7j - k

Therefore, direction ratio of line is ( -4, 7, -1)

Now line is parallel to 3x + 2y + kz - 4 = 0, So dot product of direction ratio of line and normal to plane must be 0 

∴ 3(-4) + 2(7) + k(-1) = 0

⇒ 2 - k = 0

⇒ k = 2

Hence the option (2) is correct.

If the foot of the perpendicular from the origin to a straight line is at the point (3, -4), then the equation of the line is:

  1. 4x - 3y + 25 = 0
  2. 4x + 3y - 25 = 0
  3. 3x - 4y = 25
  4. 3x - 4y + 25 = 0

Answer (Detailed Solution Below)

Option 3 : 3x - 4y = 25

Equation of a Line Question 10 Detailed Solution

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Concept: 

1. Equation of line of slope m passing through (x1, y1) is given by

(y - y1) = m(x - x1)

2. If two line with slope m1 and m2 are perpendicular to each other, then m1m2 = -1

Calculation:

Slope of line passes through (0, 0) and (3, -4) is 

m1 = 4  03  0 = 43

Let slope of perpendicular at (3, -4) is m2, then

m1m2 = -1 ⇒ m2 = 3/4

We know that, 

Equation of line of slope m passing through (x1, y1) is given by

(y - y1) = m(x - x1)

Hence, equation of line passing through (3, -4) and slope 3/4 is given by

(y + 4) = 34(x  3)

⇒ 3x - 4y = 25

Hence, option 3 is correct.

Determine the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 4x - 2y + 3z - 6 = 0

  1. 1245,1845,2445
  2. 2445,1845,1245
  3. 2429,1229,1829
  4. 1829,1229,2429

Answer (Detailed Solution Below)

Option 3 : 2429,1229,1829

Equation of a Line Question 11 Detailed Solution

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Given:

The plane : 4x - 2y + 3z - 6 = 0

Concept:

For a plane having equation ax + by + cz = d(a, b, c) are direction ratios of the normal to the plane.

Equation of a line passing through (x1, y1, z1) and having direction ratios a, b, c in the cartesian form :

xx1a=yy1b=zz1c

Calculation:

Direction ratios of normal to the plane are 4, -2, and 3. 

Equation of a line passing through the origin (0, 0, 0) and having direction ratios 4, -2, 3 is:

x04=y02=z03=λ

⇒ x = 4λ , y = -2λ and z = 3λ

Satisfying equation of the plane with above coordinates :

⇒ 4(4λ) - 2(-2λ) + 3(3λ) - 6 = 0

⇒ λ = 6/29

∴ Foot of the perpendicular,

⇒ x = 24/29, y = -12/29 and z = 18/29

The equation of line equally inclined to co-ordinate axes and passing through (-3, 2, -5) is

  1. x+31=y21=z+51
  2. x+31=y31=5+z1
  3. x+31=y21=z+51
  4. x+31=2y1=z+51

Answer (Detailed Solution Below)

Option 1 : x+31=y21=z+51

Equation of a Line Question 12 Detailed Solution

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Concept:

Equation of the line in 3D:

If a line passes through a point (x1,y1,z1) and having direction cosines (a,b,c) then the equation of the line is given by:

xx1a=yy1b=zz1c

Direction cosines of the line equally inclined to co-ordinate axes:

If a line is equally inclined to the co-ordinate axes then the direction cosines of the line are (13,13,13)

 

Calculation:

It is given that the line is equally inclined to the co-ordinate axes.

Therefore, the direction cosines of the line are (13,13,13)

It is given that the line passes through (-3, 2, -5).

Therefore, we have (x1,y1,z1)=(3,2,5) and (a,b,c)=(13,13,13).

Thus, the equation of the line is given as follows:

x+313=y213=z+513=k

x+31=y21=z+51

What is the number of possible values of k for which the line joining the points (k, 1, 3) and (1, -2, k + 1) also passes through the point (15, 2, -4)?

  1. Zero
  2. One
  3. Two
  4. Infinite

Answer (Detailed Solution Below)

Option 3 : Two

Equation of a Line Question 13 Detailed Solution

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CONCEPT:

  • The direction ratios of the line joining the points (x1, y1, z1) and (x2, y2, z2) is given by: a = x2 - x1, b = y2 - y1 and c = z2 - z1
  • If a, b, c are the direction ration ratios of a line passing through the point (x1, y1, z1), then the equation of line is given by: xx1a=yy1b=zz1c

CALCULATION:

Given: The line joining the points (k, 1, 3) and (1, -2, k + 1) also passes through the point (15, 2, -4)

As we know that, the direction ratios of the line joining the points (x1, y1, z1) and (x2, y2, z2) is given by: a = x2 - x1, b = y2 - y1 and c = z2 - z1

So, the direction ratios of the line joining the points (k, 1, 3) and (1, -2, k + 1) is: a = 1 - k, b = - 3 and c = k - 2

As we know, if a, b, and c are the direction ratio ratios of a line passing through the point (x1, y1, z1), then the equation of a line is given by: xx1a=yy1b=zz1c

So, the equation of the line with direction ratios a, b, c and passing through the point (k, 1, 3) is given by: xk1k=y13=z3k2

∵ It is given that, the line represented by xk1k=y13=z3k2 also passes through the point (15, 2, -4)

So, substitute x = 15, y = 2 and z = - 4 in the equation xk1k=y13=z3k2

⇒ 15k1k=213=43k2

⇒ 15k1k=13

⇒ - 45 + 3k = 1 - k

⇒ k = 23/2

⇒ 13=7k2

⇒ k - 2 = 21

⇒ k = 23

So, there are two possible values of k as shown above.

Hence, correct option is 3

The foot of the perpendicular from (2, 3) upon the line 4x - 5y + 8 = 0 is

  1. (0 ,0)
  2. (1, 1)
  3. (4178,12875)
  4. (7841,12841)

Answer (Detailed Solution Below)

Option 4 : (7841,12841)

Equation of a Line Question 14 Detailed Solution

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Calculation:

Let (h, k) be the coordinates of the foot of the perpendicular from the point (2, 3) on the line 4x - 5y + 8 = 0.

The slope of the perpendicular line will be k3h2

The slope of the line 4x - 5y + 8 = 0

⇒ 4x - 5y + 8 = 0

⇒ -5y = -4x - 8

⇒ y = 4/5x + 8/5

On comparing the above equation with y = mx + c we get m = 4/5

∴ the slope of the line is 4/5

Using the condition of perpendicularity of lines,

k3h2×45=1

4k - 12 = - 5h + 10

4k + 5h = 22       ----(i)

Since (h, k) lies on the given line 4x - 5y + 8 = 0,

⇒ 4h - 5k + 8 = 0

⇒ 4h - 5k = - 8       ----(ii)

on solving the equations (i) and (ii) using elimination method, we get h = 78/41 and k = 128/41

Therefore, (7841,12841)are the coordinates of the foot of the perpendicular from the point (2, 3) on the line 4x - 5y +8 = 0.

Find the equation of line passing through the points A (2, - 3, 1) and B (3, - 4, - 5) ?

  1. x21=y+31=z16
  2. x21=y+31=z16
  3. x21=y+31=z16
  4. x21=y+31=z16

Answer (Detailed Solution Below)

Option 1 : x21=y+31=z16

Equation of a Line Question 15 Detailed Solution

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Concept:

The cartesian equation of a line passing through A (x1, y1, z1) and B (x2, y2, z2) is given by: xx1x2x1=yy1y2y1=zz1z2z1

Calculation:

Given: A (2, - 3, 1) and B (3, - 4, - 5) are two points

Let x1 = 2, y1 = - 3, z1 = 1, x2 = 3, y2 = - 4 and z2 = - 5.

As we know that, equation of a line passing through A (x1, y1, z1) and B (x2, y2, z2) is given by: xx1x2x1=yy1y2y1=zz1z2z1

So, the required equation of line is:x21=y+31=z16

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