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

Explanation:

A\(\rm (11,\frac{\pi}{3})\) and B\(\rm (8, \frac{-\pi}{6})\)

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

∠O = \({\pi\over 3}+{\pi\over 6}={\pi\over2}\)

So, distance between the points 

= AB = \(\sqrt{11^2+8^2}=\sqrt{121+64}\) = \(\sqrt{185}\) units.

Option (2) is true.

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:

Shortest Distance Between Skew Lines:

• The shortest distance between two skew lines can be found using the formula:
  • d = |(b - a) × v| / |v|
• Here, d is the shortest distance, a and b are points on the respective lines, and v is the direction vector of the line formed by the cross product of the two direction vectors.
• The formula uses the cross product of the direction vectors of the two lines to determine the perpendicular distance between them.
• If the direction vectors of the two lines are v₁ = (a₁, b₁, c₁) and v₂ = (a₂, b₂, c₂), then the cross product is v₁ × v₂ = (b₁c₂ - b₂c₁, a₂c₁ - a₁c₂, a₁b₂ - a₂b₁).
• Then, the shortest distance is calculated using the above formula.

Calculation:

Given,

The first line is given by:

x - 3 / -1 = y - 5 / 1 = z / 1

So, the direction vector of the first line is v₁ = (-1, 1, 1), and a point on the line is a = (3, 5, 0).

The second line is given by:

1 - x / 1 = 2y - 10 / 2 = z + 1 / 1

So, the direction vector of the second line is v₂ = (1, 2, 1), and a point on the line is b = (1, 5, -1).

Now, we calculate the vector b - a = (1 - 3, 5 - 5, -1 - 0) = (-2, 0, -1).

The cross product of v₁ × v₂ is:

v₁ × v₂ = (1×1 - 2×1, 1×(-1) - (-1)×1, (-1)×2 - 1×1)

v₁ × v₂ = (-1, 0, -3)

The magnitude of the cross product is:

|v₁ × v₂| = √((-1)² + 0² + (-3)²) = √(1 + 9) = √10

The magnitude of the direction vector v₁ is:

|v₁| = √((-1)² + 1² + 1²) = √3

The shortest distance between the lines is:

d = |(-2, 0, -1) × (-1, 0, -3)| / √3 = √10 / √3 = √(10/3)

∴ The shortest distance between the lines is √(14/3).

The correct answer is option (4) √(14/3).

Calculation

\(L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}\)

\(\mathrm{PQ}=\frac{\mathrm{x}-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=λ\)

⇒ \(\mathrm{Q}\left(λ+\frac{15}{7}, 4 λ+\frac{32}{7}, 7 λ+7\right)\)

Since Q lies on line L 

So, \(\frac{λ+\frac{15}{7}+1}{3}=\frac{7 λ+7+5}{7}\)

⇒ 7λ + 22 = 21λ + 36

⇒ λ = -1

∴ Point \(\mathrm{Q}\left(\frac{8}{7}, \frac{4}{7}, 0\right)\)

\(\mathrm{PQ}=\sqrt{\left(\frac{15}{7}-\frac{8}{7}\right)^{2}+\left(\frac{32}{7}-\frac{4}{7}\right)^{2}+(7-0)}\)

\(P Q=\sqrt{66}\)

⇒ (PQ)² = 66

Hence option 3 is correct

Calculation

Let the parallel line is

\(\frac{x-1}{1}=\frac{y-4}{2}=\frac{z-0}{3}\)

so their point of intersection is 

(λ + 1, 2λ + 4 3λ) = (2t + 2, 3t + 6, 4t + 3)

⇒ λ = 2t + 1

⇒  2λ + 4 = 3t + 6 ⇒ t = 0

So POI is (2, 6, 3) 

So distance = \(\sqrt{(2-1)^{2}+(6-4)^{2}+(3-0)^{2}}=\sqrt{14}\)

Hence option 2 is correct

Concept:

Distance between parallel lines:

• The distance between the lines y = mx + c₁ and y = mx + c₂ is \(\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {1{\rm{\;}} + {\rm{\;}}{{\rm{m}}^2}} }}\) 
• The distance between the lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is \(\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} }}\)

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 = \(\frac{{\left| {\frac{{15}}{2}{\rm{\;}} - {\rm{\;}}9} \right|}}{{\sqrt {{3^2} + {4^2}} }} = \frac{{\left( {\frac{3}{2}} \right)}}{5} = \frac{3}{{10}}\)

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 = \(\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}=\frac{|3|}{\sqrt{100}}=\frac{3}{10}\) units

Concept:

Dot product of two perpendicular lines is zero.

Distance between two points (x1, y1, z1) and (x2, y2, z2) is given by, \(\rm \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

Calculation:

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

Let, \(\rm \dfrac{x-0}{1}=\dfrac{y-0}{0}=\dfrac{z-0}{0} =k\)

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 =

 \(\rm \sqrt {(2-2)^2+(0-3)^2+(0-4)^2}\\ =\sqrt{9+16}\\ =5\)

Hence, option (2) is correct. 

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 = \(\rm (\dfrac {x_1+x_2}{2},\dfrac{y_1+y_2}{2})\)

By distance formula AB = \(\rm\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}\)

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 = \(\rm (\dfrac {8+6}{2}, \dfrac{4+6}{2})\)

D = \(\rm (7 ,5)\)

Here, then length of median from A = AD = \(\rm \sqrt {(10-7)^2+ (5-5)^2}\)

⇒AD = \(\rm \sqrt {9+0}\)

⇒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:

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

D = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

Calculation:

Given line 3y = 4x + 5

⇒ 4x - 3y + 5 = 0

(x1, y1) = (2, 1)

∴ D = \(\rm \left|ax_1+by_1+c\over\sqrt{a^2+b^2}\right|\)

⇒ D = \(\rm \left|4\times 2 + (-3)\times 1+5\over\sqrt{4^2+(-3)^2}\right|\)

⇒ D = \(\rm \left|10\over5\right|\) = 2

Given;

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

Concept:

The formula when two points are equidistant is-

\(\sqrt {(x-x_{1})^2 + (y-y_{1})^2} = \sqrt {(x-x_{2})^2 + (y-y_{2})^2} \)

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-

 \(⇒ \sqrt {(x+1)^2 + (y-1)^2} = \sqrt {(x-3)^2 + (y+2)^2} \)

⇒ (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:\(\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

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 \(\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

⇒ \(\sqrt {{{\left( {{a} - {3}} \right)}^2} + {{\left( {{2} - {4}} \right)}^2}} = 8\)

By squaring both the sides we get

⇒ (a - 3)² + 4 = 64

⇒ a² + 9 - 6a - 60 = 0

⇒ a² - 6a - 51 = 0

⇒ \(a = \frac{{6 \pm \sqrt {240} }}{2} = 3 \pm 2\sqrt {15} \)

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:\(\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

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:\(\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

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 \(\rm \frac{y_2-y_1}{x_2-x_1}\)

Calculation:

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

 \(=\rm \frac{1-k}{3-0} \\=\frac{1-k}{3}\)

3x - 4y - 5 = 0 

⇒4y = 3x - 5

⇒ y = \(\frac{3}{4}\rm x-\frac{5}{4}\) 

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

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

\(\rm \frac{1-k}{3}\times \frac{3}{4}=-1.....(\text {Product of slopes of perpendicular lines} )\\ \Rightarrow 1-k=-4\\ \Rightarrow k =5 \)

Hence, option (3) is correct. 

Concept:

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

D = \(\rm \left|c_1-c_2\over\sqrt{a^2+b^2}\right|\)

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 = \(\rm \left|c_1-c_2\over\sqrt{a^2+b^2}\right|\)

⇒ D = \(\rm \left|-12-(-7)\over\sqrt{4^2+3^2}\right|\)

⇒ D = \(\rm \left|-5\over5\right|\) = 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:

 \(\left( {\frac{{{{\rm{x}}_1} + {{\rm{x}}_2}}}{2},\frac{{{{\rm{y}}_1} + {{\rm{y}}_2}}}{2}} \right)\).

Calculation:

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

\(\left( {{\rm{a}},2{\rm{b}}} \right) = \left( {\frac{{10 + {\rm{k}}}}{2},\frac{{ - 6 + 4}}{2}} \right) = \left( {5 + \frac{{\rm{k}}}{2},-1} \right)\)

Therefore,

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

2b = - 1

Given, a – 2b = 7

\(\left( {\frac{{\rm{k}}}{2} + 5{\rm{\;}}} \right) + 1 = 7\)

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

\(\frac{{\rm{k}}}{2} = 1\)

k = 2