# RRB NTPC - Co Ordinate Geometry - Problems and Solutions

21.

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

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)

22.

Find the co-ordinates of a point P which divides the join of A(5, -4) and B(10, 8) in the ratio 3 : 2.

A.) 9, 5
B.) 7, 8
C.) 8, 7
D.) 9, -7

Required point is P = (mx2 + nx1)/m+n, (my2 + ny1)/m+n
P((3(10) + 2(5))/5, (3(15) + 2(-5))/5)
=(30 + 10)/5, (45-10)/5
= P(8, 7)

23.

A point C divides the join of A(2,5) and B(3,9) in the ratio 3 : 4. The co-ordinates of C are:

A.) (15/7, 46/7)
B.) (15/7, 47/7)
C.) (15/7, 40/7)
D.) None of these

x1=2, x2=3 and y1=3, y2=7
= [(3 × 3 + 4 × 2)/7, (3 × 9 + 4 × 5)/7]
= 9+6/7, 27+20/7
= (15/7, 47/7)

24.

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)

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

25.

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

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)

26.

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

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,
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.

27.

The points A(1, -3), B(13, 9), C(10, 12) and D(-2, 0) taken in order are the vertices of

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

AB2 = (13-1)2 + (9+3)2
= 122 + 122 = 288.
BC2 = (10-13)2 + (12-9)2
= -32 + 32 = 9+9 = 18
CD2 = (10+2)2 + (12-0)2
= 122 + 122 = 288
AD2 = (-2-1)2 + (0+3)2 = (9+9) =18
AB = CD and BC = AD
AC2 = (10-1)2 + (12+3)2
= 92 + 152 = 81 + 225 = 306.
BD2 = (-2-13)2 + (0-9)2
= 225 + 81 = 306
AC = BD
ABCD is a rectangle.

28.

If for a line m = tanϑ < 0, then

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

m = tanϑ < 0 => is obtuse

29.

If for a line m = tanϑ > 0, then

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