RRB NTPC - Co Ordinate Geometry - Problems and Solutions

  • 1. In which quadrant does the point(-4, -7) lie?
   A.) 1st
   B.) 2nd
   C.) 3rd
   D.) 4th

Answer: Option 'C'

The point (-4, -7) lies in 3rd quadrant. 

  • 2. In which quadrant does the point(1, 5) lie? 
   A.) 1st
   B.) 2nd
   C.) 3rd
   D.) 4th

Answer: Option 'A'

The point (1, 5) lies in 1st quadrant.

  • 3. In which quadrant does the point(9, -2) lie? 
   A.) 1st
   B.) 2nd
   C.) 3rd
   D.) 4th

Answer: Option 'D'

The point (9, -2) lies in 4th quadrant.

  • 4. In which quadrant does the point(-7, 6) lie? 
   A.) 1st
   B.) 2nd
   C.) 3rd
   D.) 4th

Answer: Option 'B'

The point (-7, 6) lies in 2nd quadrant.

  • 5. In which quadrant does the point(0, 9) lie? 
   A.) x-axis
   B.) y-axis
   C.) 3rd
   D.) None of these

Answer: Option 'B'

Answer: Option 'B'
The point (0, 9) lies in x-axis.

  • 6. In which quadrant does the point(9, 0) lie? 
   A.) x-axis
   B.) y-axis
   C.) 4th
   D.) None of these

Answer: Option 'A'

The point (9, 0) lies in y-axis.

  • 7. Find the distance of the point A(4, -4) from the origin. 
   A.) 3√2
   B.) 2√8
   C.) 6√2
   D.) 8√2

Answer: Option 'D'

OA = √42+(-4)2 = √16+16 = √32 = 8√2 

  • 8. Find the distance of the point A(3, -3) from the origin. 
   A.) 3√2
   B.) 3√6
   C.) 6√2
   D.) 7√2

Answer: Option 'A'

OA = √32+(-3)2 = √9+9 = √18 = 3√2

  • 9. P is a point on x-axis at a distance of 4 units from y-axis to its right. The co-ordinates of P are: 
   A.) (4, 0)
   B.) (0, 4)
   C.) (4, 4)
   D.) (-4, 4)

Answer: Option 'A'

The co-ordinates of P are A(4, 0)

  • 10. A is a point on y-axis at a distance of 5 units from x-axis lying below x-axis. The co-ordinates of A are:
   A.) (5, 0)
   B.) (-5, 0)
   C.) (0, 5)
   D.) (0, -5)

Answer: Option 'D'

The co-ordinates of A are A(0, -5)

  • 11. Find the distance of the point A(4, -2) from the origin. 
   A.) 4√5 units
   B.) 2√5 units
   C.) 5√2 units
   D.) 7√2 units

Answer: Option 'B'

OA = √4 - 02+(-2 - 0)2 = √16+4 = √20 = √4 × 5 = 2√5 units

  • 12. Find the distance between the points A(-4, 7) and B(2, -5).
   A.) 8√5 Units
   B.) 6√5 Units
   C.) 6√4 Units
   D.) None of these

Answer: Option 'B'

AB = √(2+4)2 + (-5-7)2
= √62 + (-12)2
= √36+144 = √180
=√36 × 5 = 6√5 units.

  • 13. The distance between the points A(b, 0) and B(0, a) is. 
   A.) √a2-b2.
   B.) √a2+b2.
   C.) √a+b
   D.) a-b

Answer: Option 'B'

AB = √(b-0)2-(0-a)2 
= √b2+a2 
= √a2+b2.

  • 14. If the distance of the point P(x, y) from A(a, 0) is a + x, then y2 = ?
   A.) 8ax
   B.) 6ax
   C.) 4ax
   D.) 2ax

Answer: Option 'C'

√(x-a)2+(y-0)2 = a + x 
= (x-a)2+y2 
= (a+x)2 => y2 = (x-a)2-(x-a)2-4ax => y2 = 4ax

  • 15. The distance between the points A(5, -7) and B(2, 3) is:
   A.) 109
   B.) 5√7
   C.) √109
   D.) None of these

Answer: Option ''

AB2 = (2 - 5)2 + (3 + 7)2 
=> (-3)2 + (10)2 
=> 9 + 100 => √109

  • 16. Find the area of ΔABC whose vertices are A(9, -5), B(3, 7) and (-2, 4). 
   A.) 29 units
   B.) 35.9 sq.units
   C.) 39 sq.units
   D.) 39.5 sq.units

Answer: Option 'C'

Here, x1 = 9, x2 = 3, x3 = -2 and y1 = -5, y2 = 7, y3 = 4
= 1/2 [9(7-4) + 3(4+5) + (-2)(-5-7)] 
= 1/2 [9(3) + 3(9) - 2(-12)] 
= 1/2 [27 + 27 + 24] 
= 1/2 [78] 
= 39 sq.units 

  • 17. Find the area of ΔABC whose vertices are A(2, -5), B(4, 9) and (6, -1). 
   A.) 9 units
   B.) 5 sq.units
   C.) 7 sq.units
   D.) 6 sq.units

Answer: Option 'D'

Here, x1 = 2, x2 = 4, x3 = 6 and y1 = -5, y2 = 9, y3 = -1
= 1/2 [2(9+1) + 4(-1+5) + 6(5-9)] 
= 1/2 [2(10) + 4(4) + 6(-4)] 
= 1/2 [20 + 16 - 24] 
= 1/2 [12] 
= 6 sq.units

  • 18. The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is ? 
   A.) equilateral
   B.) right angled
   C.) isosceles
   D.) scalene

Answer: Option 'C'

AB2= (-5 - 0)2 + (-3 - 0)2 = 16 + 9 = 25 
BC2 = (3 + 5)2 + (1-3)2 = 82 + (-2)2 = 64 + 4 = 68 
AC2 = (3 - 0)2 + (1 - 6)2 = 9 + 25 = 34. 
AB = AC. ==> ΔABC is isosceles.

  • 19. The points A(-4, 0), B(1, -4), and C(5, 1) are the vertices of 
   A.) An isosceles right angled triangle
   B.) An equilateraltriangle
   C.) A scalene triangle
   D.) None of these

Answer: Option 'A'

AB2 = (1 + 4)2 + (-4 - 0)2 
= 25 + 16 = 41, 
BC2 = (5 - 1)2 + (1 + 4)2 = 42 + 52
= 16 + 25 = 41 
AC2 = (5 + 4)2 + (1 - 0)2 
= 81 + 1 = 82 
AB = BC and AB2 = BC2 = AC2 
ΔABC is an isosceles right angled triangle

  • 20. Find the vertices of triangle are A(2, 8), B(-4, 3) and (5, -1). The area of ΔABC is: 
   A.) 5 1/2 sq.units
   B.) 2 1/3 sq.units
   C.) 2 1/2 sq.units
   D.) None of these

Answer: Option 'C'

Here, x1 = 2, x2 = -4, x3 = 5 and y1 = 8, y2 = 2, y3 = -1
= 1/2 [2(3+1) - 4(-1-8) + 5(8-3)] 
= 1/2 [2(4) + 4(-9) + 5(5)] 
= 1/2 [16 - 36 + 25] 
= 1/2 [5] 
= 2 1/2 sq.units

  • 21. Find the value of k for which the points A(-2, 5), B(3, k) and C(6, 1) are collinear. 
   A.) 5
   B.) 4
   C.) 7
   D.) 1

Answer: Option 'D'

x>1 = -2, x>2 = 3, x>3 = 6 and y>1 = 5, y>2 = k, y>3 = -1 
Now Δ = 0 <=> -2(k + 1) + 3(-1 + 5) + 6(5 - k) = 0 
      <=> -2 (k+1) + 3(4) + 6(5-k) = 0 
      <=> -2k-2 + 12+30 -6k = 0 
      <=> 40 - 8k = 0 
      <=> -8k = -40 
      <=> k = 5.

  • 22. Find the value of k for which the points A(-2, 6), B(5, k) and C(8, 3) are collinear. 
   A.) 5
   B.) 4
   C.) 7
   D.) 1

Answer: Option 'A'

x1 = -2, x2 = 5, x3 = 8 and y1 = 6, y2 = k, y3 = 3 
Now Δ = 0 <=> -2(k - 3) + 5(3 - 6) + 8(6 - k) = 0 
      <=> -2 (k-3) + 5(-3) + 8(6-k) = 0 
      <=> -2k + 6 - 15 + 48 - 8k = 0 
      <=> 39 - 10k = 0 
      <=> k = -(39/10) 
      <=> k = 3.9.

  • 23. If the A(2, 3), B(5, k), and C(6, 7) are collinear, then k = ? 
   A.) 11
   B.) 12
   C.) 18
   D.) 6

Answer: Option 'D'

x1 = 2, x2 = 5, x3 = 6 and y1 = 3, y2 = k, y3 = 7 
Now Δ = 0 <=> 2(k - 7) + 5(7 - 3) + 6(3 - k) = 0 
      <=> 1/2 [2 (k-7) + 5(4) + 6(3-k)] = 0 
      <=> 2k - 14 + 20 + 18 - 6k = 0 
      <=> 24 - 4k = 0 
      <=> 4k = 24 
      <=> k = 6.

  • 24. If the points A(1, 2), B(2, 4), and C(k, 6) are collinear, then k = ?
   A.) 3
   B.) -3
   C.) 4
   D.) -4

Answer: Option 'B'

x1 = 1, x2 = 2, x3 = k and y1 = 2, y2 = 4, y3 = 6 
= 1(4 - 6) + 2 (6 - 2) + k(2 - 4) 
= -2 + 12 - 4 + 2k - 4k = 0 
= 6 - 2k = 0 
= -2k = 6 
= k = -3.

  • 25. Find the co-ordinates of the centroid of ΔABC whose vertices are A(7, -3), B(5, -4) and C(-3, -5)?
   A.) 3, -3
   B.) 3, -4
   C.) 4, -3
   D.) None of these

Answer: Option 'B'

x1 = 7, x2 = 5, x3 = -3 and y1 = -3, y2 = -4, y3 = -5 
= [(7 + 5 - 3)/3, (-3 - 4 - 5)/3] 
= (12-3)/3, -12/3) 
= (9/3, -12/3) 
= (3, -4)

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