# 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

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

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

√(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

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

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

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°

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

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