Collinearity of points MCQ Quiz in తెలుగు - Objective Question with Answer for Collinearity of points - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Last updated on Apr 15, 2025

పొందండి Collinearity of points సమాధానాలు మరియు వివరణాత్మక పరిష్కారాలతో బహుళ ఎంపిక ప్రశ్నలు (MCQ క్విజ్). వీటిని ఉచితంగా డౌన్‌లోడ్ చేసుకోండి Collinearity of points MCQ క్విజ్ Pdf మరియు బ్యాంకింగ్, SSC, రైల్వే, UPSC, స్టేట్ PSC వంటి మీ రాబోయే పరీక్షల కోసం సిద్ధం చేయండి.

Latest Collinearity of points MCQ Objective Questions

Top Collinearity of points MCQ Objective Questions

Collinearity of points Question 1:

If the points (2, - 1, 2), (1, 2, - 3) and (3, k, 7) are collinear, then find the value of k.

  1. 3
  2. - 3
  3. 4
  4. - 4

Answer (Detailed Solution Below)

Option 4 : - 4

Collinearity of points Question 1 Detailed Solution

CONCEPT:

If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then |x1y1z1x2y2z2x3y3z3|=0

CALCULATION:
 
Given: The points (2, - 1, 2), (1, 2, - 3) and (3, k, 7) are collinear
As we know that, if the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then |x1y1z1x2y2z2x3y3z3|=0
Here, x1 = 2, y1 = - 1, z1 = 2, x2 = 1, y2 = 2, z2 = - 3, x3 = 3, y3 = k and z3 = 7
 
⇒ |2121233k7|=0
 
⇒ 2 × (14 + 3k) + 1 × (7 + 9) + 2 × (k - 6) = 0
 
⇒ 28 + 6k + 16 + 2k - 12 = 0
 
⇒ 32 + 8k = 0
 
⇒k = - 4
 
Hence, option D is the correct answer.

Collinearity of points Question 2:

If the points (1, 3, 1), (2, - 1, k) and (0, 7, 3) are collinear, then find the value of k.

  1. 0
  2. 1
  3. - 1
  4. None of these

Answer (Detailed Solution Below)

Option 3 : - 1

Collinearity of points Question 2 Detailed Solution

CONCEPT:

If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then |x1y1z1x2y2z2x3y3z3|=0

CALCULATION:
 
Given: The points (1, 3, 1), (2, - 1, k) and (0, 7, 3) are collinear
As we know that, if the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then |x1y1z1x2y2z2x3y3z3|=0
Here, x1 = 1, y1 = 3, z1 = k, x2 = 2, y2 = - 1, z2 = k, x3 = 0, y3 = 7 and z3 = 3
 
⇒ |13121k073|=0
 
⇒ 1 × (- 3 - 7k) - 3 × (6 - 0) + 1 × (14 - 0) = 0
 
⇒ - 3 - 7k - 18 + 14 = 0
 
⇒ k = - 1
 
Hence, option C is the correct answer.

Collinearity of points Question 3:

The points (-5, 1), (1, k) and (4, -2) are collinear if the value of k is

  1. -1
  2. 2
  3. 3
  4. 1

Answer (Detailed Solution Below)

Option 1 : -1

Collinearity of points Question 3 Detailed Solution

Given:

Points: (-5, 1), (1, k), and (4, -2)

Formula used:

For three points to be collinear, the area of the triangle formed by them must be zero.

Area of a triangle using coordinates:

Area = (1/2) × [x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)]

Calculation:

Since the points are collinear:

(1/2) × [-5(k + 2) + 1(-2 - 1) + 4(1 - k)] = 0

-5(k + 2) + 1(-3) + 4(1 - k) = 0

-5k - 10 - 3 + 4 - 4k = 0

-9k - 9 = 0

-9k = 9

k = -1

∴ The value of k is -1.

Collinearity of points Question 4:

If points P(4,5,x),Q(3,y,4) and R(5,8,0) are colinear, then the value of x+y is

  1. 4
  2. 3
  3. 5
  4. 4

Answer (Detailed Solution Below)

Option 4 : 4

Collinearity of points Question 4 Detailed Solution

PR=(54)i^+(85)j^+(0x)k^

PR=i^+3j^xk^

Again, QR=(53)i^+(8y)j^+(04)k^

QR=2i^+(8y)j^4k^

Since, P, Q and R are co-linear points, hence

12=38y=x4

On Solving, we get:-

x=2,y=2

x+y=4

Hence, answer is option-(D).

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