Distance of a point from a Line MCQ Quiz in தமிழ் - Objective Question with Answer for Distance of a point from a Line - இலவச PDF ஐப் பதிவிறக்கவும்

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:

The perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by 

CALCULATION:

Here, we have to find the points which are on the y-axis such that the perpendicular distance between the points and the line  is 3 units.

The given equation of line can be re-written as: 4x - 3y - 12 = 0

Let P = (0, y)

Here a = 4, b = - 3 and d = 3

Now substitute x1 = 0 and y1 = y in the equation 4x - 3y - 12 = 0

⇒ |4⋅ x1 - 3 ⋅ y1 - 12| = |0 - 3y - 12|

⇒ 

As we know that, the perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by 

⇒ 

⇒ |- 3y - 12| = 15

⇒ y = 1 or - 9

So, the points are: (0, 1) and (0, - 9)

Hence, option D is the correct answer.

Concept : 

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

Calculations : 

Given let (1, 1) = (x1, y1 )

Let d1 is the distance of the line 3x + 4y = 1 from the point (1, 1).

Let d2 is the distance of the line 4x + 3y + 2k = 0 from the point (1, 1).

The two straight lines 3x + 4y = 1 and 4x + 3y + 2k = 0 are equidistant from the point (1, 1).

Therefore, d1 = d2.

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

Hence,  the distance of the line 3x + 4y - 1  = 0 from the point (1, 1) is d1 = 

d1 = 

Similarly,  the distance of the line 4x + 3y + 2k = 0 from the point (1, 1) is d2 = 

d2 = 

Since, d1 = d2.

6 = 7+2k

k = 

Concept:

Perpendicular Distance of a Point from a Line

Let us consider a plane given by the Cartesian equation, Ax + By + C = 0 and a point whose coordinate is, (x₁, y₁)

Now, distance = 

Calculation:

Given:

Perpendicular Distance of a Point (a, b) from a Line the line 8x + 6y + 1 = 0 is 1

⇒ 8a + 6b + 1 = ± 10

⇒ 8a + 6b + 1 = 10 and 8a + 6b + 1 = -10

∴ 8a + 6b -9 = 0 and 8a + 6b + 11 = 0

So statement 2 and 3 are correct.

Concept:

Perpendicular Distance of a Point from a Line:

Let us consider a line Ax + By + C = 0 and a point whose coordinate is (x₁, y₁)

Calculation:

Given: equation of line is 3x + 4y = 7 and a point is (3, 2)

We know the distance of a line from is given by, 

So, distance of a point (3, 2) from a line 3x + 4y - 7 = 0 is given by,

= 10/5

= 2 units 

Hence, option (2) is correct.

Calculation

Writing P in terms of parametric co-ordinates (2 + r cos θ, 3 + r sin θ)

As tan θ = √3 

⇒ 

P must satisfy 2x - 3y + 28 = 0

So,

⇒ r = 4 + 6√3

Hence option (4) is correct

CONCEPT:

The perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by 

CALCULATION:

Here, we have to find the distance of the line 5x + 3y = 6 from the origin

Let P = (0, 0)

⇒ x1 = 0 and y1 = 0

Here, a = 5 and b = 3

Now substitute  x1 = 0 and y1 = 0 in the equation 5x + 3y - 6 = 0, we get

⇒ |5⋅ x1 + 3 ⋅ y1 - 6| = |0 + 0 - 6| = 6

⇒ 

As we know that, the perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by 

⇒ 

Hence, option B is the correct answer.

Concept Used:

Perpendicular distance from a point (x₁, y₁) to the line ax + by + c = 0 is given by:

d =

Calculation

Given line:

Multiply by 12: 4x + 3y = 12

4x + 3y - 12 = 0

Points on x-axis are of the form (x₁, 0)

Perpendicular distance = 4

4 =

⇒ 4 =

⇒ 4 =

⇒ 4 =

⇒ 20 = |4x₁ - 12|

Case 1: 4x₁ - 12 = 20

⇒ 4x₁ = 32

⇒ x₁ = 8

Point: (8, 0)

Case 2: 4x₁ - 12 = -20

⇒ 4x₁ = -8

⇒ x₁ = -2

Point: (-2, 0)

∴ The points are (8, 0) and (-2, 0).

Hence option 1 is correct

Calculation

Given

⇒ λ + 2 = 4k - 3, - λ = 3k - 2, 8λ + 7 = k - 2

⇒ k = 1, λ = -1

∴ P = (1,1, -1)

Projection of  is

⇒ l = 

∴ 

⇒ 14l² = 108

Concept:

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

Calculation:

Point on the X-axis (x, 0)

So the distance from (x, 0) to the point (x, y)

Distance should be positive so, distance = |y|

Hence, option (4) is correct.