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


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