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.

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

Answer: Option 'C'

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 [5] 
= 2 1/2 sq.units

3.

In which quadrant does the point(1, 5) lie? 

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

Answer: Option 'A'

The point (1, 5) lies in 1st quadrant.

4.

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

Answer: Option 'B'

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)

5.

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)

Answer: Option 'A'

The co-ordinates of P are A(4, 0)

6.

Find the value of k for which the points A(-2, 6), B(5, k) and C(8, 3) are collinear. 

   A.) 5
   B.) 4
   C.) 7
   D.) 1

Answer: Option 'A'

x1 = -2, x2 = 5, x3 = 8 and y1 = 6, y2 = k, y3 = 3 
Now Δ = 0 <=> -2(k - 3) + 5(3 - 6) + 8(6 - k) = 0 
      <=> -2 (k-3) + 5(-3) + 8(6-k) = 0 
      <=> -2k + 6 - 15 + 48 - 8k = 0 
      <=> 39 - 10k = 0 
      <=> k = -(39/10) 
      <=> k = 3.9.

7.

If for a line m = tanϑ < 0, then 

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

Answer: Option 'B'

m = tanϑ < 0 => is obtuse

8.

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.

9.

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

Answer: Option 'D'

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.

10.

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

11.

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

Answer: Option 'B'

AB = √(2+4)2 + (-5-7)2
= √62 + (-12)2
= √36+144 = √180
=√36 × 5 = 6√5 units.

12.

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.

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

Answer: Option 'B'

AB = √(b-0)2-(0-a)2 
= √b2+a2 
= √a2+b2.

14.

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

15.

In which quadrant does the point(0, 9) lie? 

   A.) x-axis
   B.) y-axis
   C.) 3rd
   D.) None of these

Answer: Option 'B'

Answer: Option 'B'
The point (0, 9) lies in x-axis.

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