Distance of a point from a Line MCQ Quiz in বাংলা - Objective Question with Answer for Distance of a point from a Line - বিনামূল্যে ডাউনলোড করুন [PDF]

Calculation

Given L₁ : \( \frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3} \\\)

L₂ : \( \frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6} \)

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

⇒ L : \( \frac{x-7}{2}=\frac{y+3}{-3}=\frac{z-11}{6} \)

B lies on line L

\(\mathrm{B}=(2 λ+7,-3 λ-2,6 λ+11)\)

Point B lies on \(\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}\)

⇒ \(\frac{2 λ+7-6}{1}=\frac{-3 λ-2-4}{0}=\frac{6 λ+11-8}{3}\)

⇒ -3λ - 6 = 0 

⇒ λ = -2 

B ⇒ (3, 4, -1) 

\(\mathrm{AB} =\sqrt{(7-3)^2+(4+2)^2+(11+1)^2}\)

\(⇒\sqrt{16+36+144}\)

\(⇒\sqrt{196}=14 \)

Hence option 2 is correct

CONCEPT:

The perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by \(d = \left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\)

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 \(\frac{x}{3} - \frac{y}{4} = 1\) 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|

⇒ \(\sqrt{a^2+b^2} = \sqrt{4^2 + (- 3)^2} = 5\)

As we know that, the perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by \(d = \left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\)

⇒ \(d = \left| {\frac{{4\cdot x_1- 3 \cdot y_1 - 12 }}{{\sqrt {{4^2} + {(- 3)^2}} }}} \right| = 3\)

⇒ |- 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,\(\rm d =\frac{\left | ax_{1}+by{_{1}}+c \right |}{\sqrt{a^{2}+b^{2}}}\)

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,\(\rm d =\frac{\left | ax_{1}+by{_{1}}+c \right |}{\sqrt{a^{2}+b^{2}}}\)

Hence,  the distance of the line 3x + 4y - 1  = 0 from the point (1, 1) is d1 = \(\rm \frac{\left | (3)(1)+(4)(1)-1\right |}{\sqrt{3^{2}+4^{2}}}\)

d1 = \(\rm\frac {6}{5}\)

Similarly,  the distance of the line 4x + 3y + 2k = 0 from the point (1, 1) is d2 = \(\rm \frac{\left | (4)(1)+(3)(1)+2k\right |}{\sqrt{4^{2}+3^{2}}}\)

d2 = \(\rm\frac{7+2k}{5}\)

Since, d1 = d2.

\(\rm\frac{6}{5} = \frac{7+2k}{5}\)

6 = 7+2k

k = \(\rm\frac{-1}{2}\)

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 = \(\left| {\frac{{{\rm{A}}{{\rm{x}}_1}{\rm{\;}} + {\rm{\;B}}{{\rm{y}}_1}{\rm{\;}} + {\rm{\;c}}}}{{\sqrt {{{\rm{A}}^2}{\rm{\;}} + {\rm{\;\;}}{{\rm{B}}^2}{\rm{\;}}} }}} \right|\)

Calculation:

Given:

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

 \( \Rightarrow \left| {\frac{{8{\rm{a\;}} + {\rm{\;}}6{\rm{b\;}} + {\rm{\;}}1{\rm{\;}}}}{{\sqrt {{8^2}{\rm{\;}} + {\rm{\;\;}}{6^2}{\rm{\;}}} }}} \right| = 1\)

 \( \Rightarrow \frac{{8{\rm{a\;}} + {\rm{\;}}6{\rm{b\;}} + {\rm{\;}}1{\rm{\;}}}}{{10}} = \pm 1\)

⇒ 8a + 6b + 1 = ± 10

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

∴ 8a + 6b -9 = 0 and 8a + 6b + 11 = 0

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)

\(\rm d=|\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}|\)

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, \(\rm d=|\frac{Ax_1+By_1+C}{\sqrt{A^2+B^2}}|\)

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

\(\rm d = |\frac{3(3)+4(2)-7}{\sqrt{3^2+4^2}}|\)

\(\rm = |\frac{9+8-7}{\sqrt{9+16}}|\)

\(\rm = |\frac{10}{\sqrt{25}}|\)

= 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 

⇒ \(\mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{√{3} \mathrm{r}}{2}\right)\)

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

So, \(2\left(2+\frac{r}{2}\right)-3\left(3+\frac{√{3} \mathrm{r}}{2}\right)+28=0\)

⇒ 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 \(d = \left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\)

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

⇒ \(\sqrt{a^2+b^2} = \sqrt{5^2 + 3^2} = \sqrt{34}\)

As we know that, the perpendicular distance d from P (x1, y1) to the line ax + by + c = 0 is given by \(d = \left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2}} }}} \right|\)

⇒ \(d = \left| {\frac{{5\cdot x_1+ 3 \cdot y_1 - 6}}{{\sqrt {{5^2} + {3^2}} }}} \right| = \frac{6}{\sqrt{34}}\)

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 = \(\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\)

Calculation

Given line: \(\frac{x}{3} + \frac{y}{4} = 1\)

Multiply by 12: 4x + 3y = 12

4x + 3y - 12 = 0

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

Perpendicular distance = 4

4 = \(\frac{|4x_1 + 3(0) - 12|}{\sqrt{4^2 + 3^2}}\)

⇒ 4 = \(\frac{|4x_1 - 12|}{\sqrt{16 + 9}}\)

⇒ 4 = \(\frac{|4x_1 - 12|}{\sqrt{25}}\)

⇒ 4 = \(\frac{|4x_1 - 12|}{5}\)

⇒ 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

\(\frac{x-2}{1}=\frac{y}{-1}=\frac{z-7}{8}=λ\)

\(\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}=k\)

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

⇒ k = 1, λ = -1

∴ P = (1,1, -1)

Projection of \(2 \hat{\mathrm{i}}-2 \hat{\mathrm{k}} \text { on } 2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) is

⇒ l = \(\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}}\)

∴ \(l^2=8-\frac{4}{14}=\frac{108}{14}\)

⇒ 14l² = 108

Concept:

Distance between two points (x₁, y₁) and (x₂, y₂) is given by,\(\rm\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Calculation:

Point on the X-axis (x, 0)

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

 \(\rm\sqrt{(x-x)^2+(y-0)^2}\\ =\sqrt{y^2} \\=y\)

Distance should be positive so, distance = |y|

Hence, option (4) is correct.