# RRB NTPC - Co Ordinate Geometry - Problems and Solutions

1.

If the A(2, 3), B(5, k), and C(6, 7) are collinear, then k = ?

A.) 11
B.) 12
C.) 18
D.) 6

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.

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

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

3.

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.

4.

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)

5.

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)