RRB NTPC - Co Ordinate Geometry - Problems and Solutions

1.

The vertices of a quadrilateral ABCD are A(0, 0), B(3,3), C(3, 6) and D(0, 3). Then , ABCD is a 

   A.) square
   B.) parallelogram
   C.) rectangle
   D.) rhombus

Answer: Option 'B'

AB2 = (3-0)2 + (3-0)2 = 18 
BC2 = (3-3)2 + (6-3)2 = 9 
CD2 = (0-3)2 + (3-6)2 =18 
AD2 = (0-0)2 + (3-0)2 = 9 
AB = CD = √18 => 3√2, 
BC = AD = √9 
AC2 = (3-0)2 + (6-0)2 = 9 + 36 = 45 
BD2 = (0-3)2 + (3-3)2 = 9 + 0 = 9 
AC ≠ BD 
ABCD is a parallelogram.

2.

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

3.

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.

4.

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)

5.

The co-ordinates of the end points of a diameter AB of a circle are A(-6, 8) and B(-10, 6). Find the  co-ordinates of its centre. 

   A.) -8, 7
   B.) -7, 8
   C.) 7, -8
   D.) -8, -7

Answer: Option 'A'

The center O is the mid point of AB. 
Co-ordinates of O are [(-6-10)/2, (8+6)/2] 
= -16/2, 14/2 
= (-8, 7)

6.

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

7.

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)

8.

If for a line m = tanϑ < 0, then 

   A.) ϑ is acute
   B.) ϑ is obtuse
   C.) ϑ = 90°
   D.) ϑ = 60°

Answer: Option 'B'

m = tanϑ < 0 => is obtuse

9.

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.

10.

The vertices of a ΔABC are A(-6, 18), B(12, 0) and C(9, −21). The centroid of ΔABC is: 

   A.) (5, 1)
   B.) (5, -1)
   C.) (4, -1)
   D.) None of these

Answer: Option 'B'

CENTROID OF A TRIANGLE 
The point of intersection of all the medians of a triangle is called its centroid. 
If A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of ΔABC,
then the co-ordinates of its centroid are [1/3(x1 + x2 + x3), 1/3(y1 + y2 + y3)]
= [1/3(-6+12+9), 1/3(18+0-21)] 
= [15/3, -3/3] 
= (5, -1)

11.

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.

12.

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.

13.

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

14.

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

15.

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.

16.

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.

17.

The end points of a line segment AB are A(-6, 4) and B(12,24). Its midpoint is:

   A.) (14, 4)
   B.) (-4, 14)
   C.) (3, 14)
   D.) (-3, 14)

Answer: Option 'D'

Midpoint is C((-6+12)/2, (4+24)/2) 
(-6/2, 28/2) = (-3, 14)

18.

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

19.

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

20.

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.


Co Ordinate Geometry Download Pdf