Co Ordinate Geometry - Problems and Solutions

16.

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

17.

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. 

18.

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

19.

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)

20.

In which quadrant does the point(9, -2) lie? 

   A.) 1st
   B.) 2nd
   C.) 3rd
   D.) 4th

Answer: Option 'D'

The point (9, -2) lies in 4th quadrant.

21.

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.

22.

If for a line m = tanϑ > 0, then 

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

Answer: Option 'A'

m = tanϑ < 0 => is acute

23.

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.

24.

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.

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