# 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

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

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

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

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

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

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

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

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)

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)

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

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

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

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

√(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

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

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 
= 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

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 
= 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

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

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

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 
= 2 1/2 sq.units