RRB NTPC - Co Ordinate Geometry - Problems and Solutions

1.

Find the value of k for which the points A(-2, 5), B(3, k) and C(6, 1) are collinear. 

   A.) 5
   B.) 4
   C.) 7
   D.) 1

Answer: Option 'D'

x>1 = -2, x>2 = 3, x>3 = 6 and y>1 = 5, y>2 = k, y>3 = -1 
Now Δ = 0 <=> -2(k + 1) + 3(-1 + 5) + 6(5 - k) = 0 
      <=> -2 (k+1) + 3(4) + 6(5-k) = 0 
      <=> -2k-2 + 12+30 -6k = 0 
      <=> 40 - 8k = 0 
      <=> -8k = -40 
      <=> k = 5.

2.

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. 

3.

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

4.

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

5.

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.

Co Ordinate Geometry Download Pdf