16.
Find the area of ΔABC whose vertices are A(2, -5), B(4, 9) and (6, -1).
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?
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 = ?
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.
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?
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.
Answer: Option 'B'
AB = √(b-0)2-(0-a)2
= √b2+a2
= √a2+b2.
22.
If for a line m = tanϑ > 0, then
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.
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 = ?
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.