Distance Formula MCQ Quiz in বাংলা - Objective Question with Answer for Distance Formula - বিনামূল্যে ডাউনলোড করুন [PDF]

Concept:

Let A = (x,y,z) and B = (a, b, c) be any two points then distance between them is ,

By distance formula: \(\rm AB = \sqrt {(x-a)^2+(y-b)^2+(z-c)^2}\)

Calculations:

 Consider A = (7, 1, -3)  = (x, y, z)

and B = (4, 5 , \(λ\)) = (a, b, c)

given the distance between the points A = (7, 1, -3) and  B= (4, 5, λ) is 13 units.

By distance formula

\(\rm AB = \sqrt {(x-a)^2+(y-b)^2+(z-c)^2}\)

⇒\(\rm 13 = \sqrt {(7-4)^2+(1-5)^2+(-3-λ)^2}\) 

Squaring on both side 

⇒\(\rm 13^2 = {(7-4)^2+(1-5)^2+(-3-λ)^2}\)

⇒169 = 9 + 16 + 9 + 6\(λ\) + \(λ^2\)

⇒ \(λ^2\)+ 6\(λ\) - 135 = 0

⇒\(\rm (λ+15)(λ-9) = 0\)

⇒ \(\rmλ = -15, 9\)

Concept:

Distance Formula: The distance 'd' between two points (x₁, y₁) and (x₂, y₂) in a point is given by:

d² = (x₁ - x₂)² + (y₁ - y₂)² or d2 = (x2 - x1)2 + (y2 - y1)2

Calculation:

Using the distance formula, we get:

5² = (-1 - 3)² + (a + 1)²

⇒ 25 = 16 + a² + 2a + 1

⇒ a2 + 2a - 8 = 0

⇒ a2 + 4a - 2a - 8 = 0

⇒ a(a + 4) - 2(a + 4) = 0

⇒ (a + 4)(a - 2) = 0

⇒ a + 4 = 0 OR a - 2 = 0

⇒ a = -4 OR a = 2.

Concept:

The distance between the two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is given by:

​\(d=\frac{\left | c_1-c_2 \right |}{\sqrt{a^2+b^2}}\)​

Calculation:

The given equations are y = 2x + 4 and 6x = 3y + 5.

These can also be written as 2x - y + 4 = 0 and 2x - y - 5/3 = 0.

Now, since the lines are parallel.

\(\Rightarrow d=\frac{\left | c_1-c_2 \right |}{\sqrt{a^2+b^2}}\)

\(\Rightarrow d=\frac{\left | 4+\frac{5}{3} \right |}{\sqrt{2^2+(-1)^2}}\)

\(\Rightarrow d=\frac{\left | \frac{17}{3} \right |}{\sqrt{5}}\)

\(\Rightarrow d=\frac{17\sqrt{5}}{15}\)

Hence, the distance between the parallel lines is ​​\(\frac{17\sqrt 5}{15}\) unit.

Concept:

Let  A = (x₁,y₁) and  B = (x₂ ,y₂) be any two points.

Then Distance between A And B is given by distance formula.

AB = \(\rm \sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\) 

Calculation:

Given: Difference of distance from points (3, 0) and (-3, 0) is 4

Consider A = (3, 0) and B = (-3, 0)

Let a point be p(x, y)

As we know distance is always positive

So, |PA - PB| = 4

\(\Rightarrow \rm \sqrt{(x-3)^2+(y-0)^2} - \sqrt{(x+3)^2+(y-0)^2}= 4 \\\Rightarrow \rm \sqrt{(x-3)^2+(y)^2}=4+\sqrt{(x+3)^2+(y)^2}\)

Squaring both sides, we get

\(\Rightarrow \rm (\sqrt{(x-3)^2+(y)^2})^2=(4+\sqrt{(x+3)^2+(y)^2})^2\\\rm \Rightarrow (x-3)^2+y^2=16+(x+3)^2+y^2+8\sqrt{(x+3)^2+(y)^2}\\\rm \Rightarrow x^2-6x+9 = 16+x^2+6x+9+8\sqrt{(x+3)^2+(y)^2}\\\rm \Rightarrow-12x-16=8\sqrt{(x+3)^2+(y)^2}\\\rm \Rightarrow-3x-4=2\sqrt{(x+3)^2+(y)^2}\)

Squaring both sides, we get

\(\rm \Rightarrow(-3x-4)^2=(2\sqrt{(x+3)^2+(y)^2})^2\\\rm \Rightarrow(3x+4)^2=4 ((x+3)^2+(y)^2)\\\rm \Rightarrow 9x^2+16+24x = 4(x^2+6x+9+y^2)\\\rm \Rightarrow 5x^2-4y^2 = 20\\\rm\therefore \frac{x^2}{4} - \frac{y^2}{5}=1\)

Concept:

1. Standard Equation of a line y = mx + c, where m is the slope of the line.
2. Slopes of parallel lines are equal. The parallel lines are equally inclined with the positive x-axis and hence the slope of parallel lines are equal. If the slopes of two parallel lines are represented as m₁, m₂ then we have m₁ = m₂.

Calculation:

Given:

Lines ax + by + c = 0 and bx + ay + c = 0

Express the two lines in standard format that is y = mx + c

ax + by + c = 0 ⇒ by = -ax - c ⇒ \(\displaystyle y=(\frac{-a}{b})x-\frac{c}{b}\)      ------(i)

bx + ay + c = 0 ⇒ ay = -bx - c ⇒ \(\displaystyle y=(\frac{-b}{a})x-\frac{c}{a}\)      ------(ii)

Given both lines are parallel implies their slopes are equal, that is

⇒ \(\displaystyle \frac{-a}{b}=\frac{-b}{a}\)

⇒ b² = a²

⇒ a2 - b2 = 0

∴ a2 - b2 = 0 is correct. 

Concept:

The distance between the two points P(x₁, x₂, x₃) and Q(y1, y2, y3) is given by

\(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\)

Calculation:

Let the point be A(0, y, 0). Then,

AP = 3√3

\(⇒ \sqrt{(0-1)^2+(y+2)^2+(0-1)^2}=3\sqrt3\)

On squaring both sides, we get

1 + (y + 2)² + 1 = 27

⇒ (y + 2)² = 25

⇒ y + 2 = ±5

⇒ y = 3, -7

Hence, the coordinates of the required point are (0, 3, 0) and (0, -7, 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.

Concept:

The distance between two parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is given by the formula: \(\rm d=\dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}\).

Calculation:

Using the concept above, we have a = 5, b = -12, c₁ = 2 and c₂ = -3.

∴ \(\rm d=\dfrac{|c_1-c_2|}{\sqrt{a^2+b^2}}\)

= \(\rm \dfrac{|2-(-3)|}{\sqrt{5^2+(-12)^2}}\)

= \(\rm \dfrac{|5|}{\sqrt{169}}\)

= \(\dfrac{5}{13}\).

Explanation:

Let m1 be slope of x + y - 2 = 0 and m2 be slope of line perpendicular to it.

Equation is x + y - 2 = 0 

rewritten as  y = -x + 2 = 0  --- [as y = mx + c]

So m1 = -1

We know: m1 × m2 = -1

⇒ m2 = 1

Equation of line passing through (1, 3) and with slope m2 = -1

(y – 3) = 1 × (x – 1)

⇒  x - 1 = y - 3

⇒ x - y + 2 = 0        --- (i)

& x + y - 2 = 0          --- (ii)

Solving (i) and (ii) we get

→ x = 0, y = 2

Concept Used:

To find the distance between a point and the x-axis, you can use the formula, i.e., |y|

where y is the y-coordinate of the point.

Calculation:

In the case of the point P (2, 3), the distance from the x-axis would be: distance = |3| = 3

So the distance of the point P (2, 3) from the x-axis is 3.