Distance Formula MCQ Quiz - Objective Question with Answer for Distance Formula - Download Free PDF

Calculation:

Given points A(3, −1) and B(1, 1). Let P be the midpoint of AB, and Q be a point on the perpendicular bisector of AB that lies √2 units from P.

Compute the midpoint P of AB:

Compute the slope of AB:

Therefore, the equation of line AB is:

The perpendicular bisector of AB must pass through P(2, 0) and have slope perpendicular to −1 (i.e. slope +1):

Any point Q on this bisector satisfies y = x - 2. Write Q = (x, x − 2).

We require the distance PQ = √2. Since P(2, 0),

⇒ 

Thus,

⇒ 

If x = 3, then y = 3 - 2 = 1. So one solution is Q(3, 1).

If x = 1, then y = 1 - 2 = -1. So the other solution is Q(1, −1).

Hence, the correct answer is Option 2.

Calculation:

Given points A(0,0), B(p,1), and C(1,q), which form an equilateral triangle. We are told (0

Compute the squared lengths of the sides:

Because the triangle is equilateral, all three squared lengths are equal:

⇒ 

⇒ 

both p and q are positive in (0,1)

Let p = q = t Then

⇒ 

⇒ 

Equate AB² and BC²:

⇒ 

Simplify:

⇒ 

So t satisfies:

⇒ 

Since 0 1) is not allowed. Hence,

⇒ 

Therefore:

Hence, the correct answer is Option 4.

Explanation:

A and B

Here we can see that both point make an right angle triangle.

∠O = 

So, distance between the points 

= AB =  =  units.

Option (2) is true.

b

The distance between two points in a 3D space is calculated using the formula:

Distance =  

Calculation:

Substituting the given points (x₁, y₁, z₁) = (7, 1, -3) and (x₂, y₂, z₂) = (4, 5, λ):

Distance = √((4 - 7)² + (5 - 1)² + (λ + 3)²)

Given that Distance = 13, we get:

13 = √((4 - 7)² + (5 - 1)² + (λ + 3)²)

Squaring both sides:

13² = (4 - 7)² + (5 - 1)² + (λ + 3)²

169 = (-3)² + (4)² + (λ + 3)²

⇒ 169 = 9 + 16 + (λ + 3)²

⇒ 169 = 25 + (λ + 3)²

⇒ (λ + 3)² = 169 - 25

⇒ (λ + 3)² = 144

Taking the square root of both sides:

λ + 3 = ±√144

⇒ λ + 3 = ±12

Solving for λ:

Case 1: λ + 3 = 12 ⇒ λ = 12 - 3 = 9

Case 2: λ + 3 = -12 ⇒ λ = -12 - 3 = -15

Conclusion:

The possible values of λ are 9 and -15.

However, as per the given options, the correct answer is:

∴ λ = 9

Calculation

Given L₁ : 

L₂ : 

Let line L passing from A(7, –2, 11) and parallel to L₂

⇒ L : 

B lies on line L

Point B lies on

⇒ 

⇒ -3λ - 6 = 0 

⇒ λ = -2 

B ⇒ (3, 4, -1) 

Hence option 2 is correct

Concept:

Distance between parallel lines:

• The distance between the lines y = mx + c₁ and y = mx + c₂ is  
• The distance between the lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is

Calculation:

Given lines are 6x + 8y + 15 = 0 and 3x + 4y + 9 = 0

⇒ 6x + 8y + 15 = 0

Take 2 common from above equation, we get

⇒ 3x + 4y + 15/2 = 0       ---(1)

And 3x + 4y + 9 = 0       ---(2)

Equation 1 and 2 are parallel to each other.

∴ The distance between the lines =

Alternate MethodParallel lines have the same slope and will never intersect.

Two lines y = m₁x + c₁

and y = m₂x + c₂

are said to be parallel if:

m₁ = m₂

Example : 

Line 1: 3x + 4y = 1

Line 2: 3x + 4y = 5

Application:

We have,

Line 1: 6x + 8y + 15 = 0

And, Line 2: 3x + 4y + 9 = 0

Line 2 can also be written as,

6x + 8y + 18 = 0

Since, both the line are parallel, hence its graph will be similar to:

We have,

c₂ - c₁ = 18 - 15 = 3

a² + b² = 6² + 8² = 100

Using formula of distance between straight lines,

D =  units

Concept:

Let A (x₁, y₁) and B (x2, y2) be the end points of the line AB. C be the mid point of the line AB.

The coordinate of C = 

By distance formula AB = 

Calculations:

Given, Point A(10, 5), B(8, 4) and C(6, 6) are vertices of a triangle, 

Let D be the mid point on the Line BC.

Its coordinates is given by

D = 

D = 

Here, then length of median from A = AD = 

⇒AD = 

⇒AD = 3

Point A(10, 5), B(8, 4) and C(6, 6) are vertices of a triangle, then length of median from A is 3

Concept:

Dot product of two perpendicular lines is zero.

Distance between two points (x1, y1, z1) and (x2, y2, z2) is given by, 

Calculation:

Let M be the foot of perpendicular drawn from the point P(2, 3, 4)

Let, 

x = k, y  = 0, z = 0

So M = (k, 0, 0)

Now direction ratios of PM = (2 - k, 3 - 0, 4 - 0) = (2- k, 3, 4) and direction ratios of given line are 1, 0, 0

PM is perpedicular to the given line so,

(2 - k) (1) + 3(0) + 4 (0) = 0

∴ k = 2

M = (2, 0, 0)

Perpendicular distance PM =

Hence, option (2) is correct. 

Concept:

The distance of a point (x₁, y₁) from a line ax + by + c = 0 

D = 

Calculation:

Given line 3y = 4x + 5

⇒ 4x - 3y + 5 = 0

(x1, y1) = (2, 1)

∴ D = 

⇒ D = 

⇒ D =  = 2

Given;

Coordinates that are (-1, 1) and (3, -2)

Concept:

The formula when two points are equidistant is-

Calculation:

Let the locus point be (x, y),

As the locus of the points that are equidistant from two points (-1, 1) and (3, -2), therefore the equation would be-

⇒ (x + 1)² + (y - 1)² = (x - 3)² + (y + 2)²

∵ (a + b)² = a² + 2ab + b² & 

(a - b)2 = a2 - 2ab + b2 

⇒ x² + 2x + 1 + y² - 2y + 1 = x² - 6x + 9 + y² + 4y + 4

⇒ 2x - 2y + 2 = - 6x + 4y + 13

⇒ 8x - 6y - 11 = 0

Hence, the equation is 8x - 6y - 11 = 0

CONCEPT:

Let A (x₁, y₁) and B (x₂, y₂) be any two points in the XY – plane, then the distance between A and B is given by:

CALCULATION:

Given: The distance between the points (3, 4) and (a, 2) is 8 units

Here, we have to find the value of a.

As we know that, the distance between two points A (x1, y1) and B (x2, y2) is given by 

⇒ 

By squaring both the sides we get

⇒ (a - 3)² + 4 = 64

⇒ a² + 9 - 6a - 60 = 0

⇒ a² - 6a - 51 = 0

⇒ 

Hence, option A is the correct answer.

CONCEPT:

Let A (x₁, y₁) and B (x₂, y₂) be any two points in the XY – plane, then the distance between A and B is given by:

CALCULATION:

Given: The distance between the points (5, - 2) and (1, a) is 5.

Let A = (5, - 2) and B = (1, a)

As we know that, the distance between the points A and B is given by:

Here, x₁ = 5, y₁ = - 2, x₂ = 1 and y₂ = a

Concept:

If two nonvertical lines are perpendicular, then the product of their slopes is −1.

The slope of a line passing through the distinct points (x₁, y₁) and (x₂, y₂) is 

Calculation:

Slope of line passing through points (0, k) and (3, 1)

3x - 4y - 5 = 0 

⇒4y = 3x - 5

⇒ y =  

So, the slope of line 3x - 4y - 5 = 0 is 3/4

Now since line OP and 3x - 4y - 5 = 0 are perepndicular 

Hence, option (3) is correct. 

Concept:

The distance between the parallel lines ax + by + c₁ and ax + by + c₂ is:

D = 

Calculation;

The 2 given lines are:

3y + 4x - 12 = 0

3y + 4x - 7 = 0

a = 4, b = 3, c₁ = -12 and c₂ = -7

∴ The distance between the lines

D = 

⇒ D = 

⇒ D =  = 1

Concept:

Let A(x₁, y₁) and B(x₂, y₂) be any two points on the X-Y plane. Suppose point C is the mid-point of the line segment AB, then the coordinates of point C is:

 .

Calculation:

Mid-point of (10, -6) and (k, 4) :

Therefore,

2b = - 1

Given, a – 2b = 7

k = 2