Distance Formula MCQ Quiz in తెలుగు - Objective Question with Answer for Distance Formula - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept:

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

By distance formula: 

Calculations:

 Consider A = (7, 1, -3)  = (x, y, z)

and B = (4, 5 , ) = (a, b, c)

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

By distance formula

⇒ 

Squaring on both side 

⇒

⇒169 = 9 + 16 + 9 + 6 + 

⇒ + 6 - 135 = 0

⇒

⇒ 

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:

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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.

Hence, the distance between the parallel lines is ​​ 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 =  

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

Squaring both sides, we get

Squaring both sides, we get

Concept:

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

Calculation:

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

∴ 

= 

= 

= .

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 ⇒       ------(i)

bx + ay + c = 0 ⇒ ay = -bx - c ⇒       ------(ii)

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

⇒ 

⇒ 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

Calculation:

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

AP = 3√3

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)

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.

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.