1.
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.
2.
Find the vertices of triangle are A(2, 8), B(-4, 3) and (5, -1). The area of ΔABC is:
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?
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:
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:
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.
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
Answer: Option 'B'
m = tanϑ < 0 => is obtuse
8.
In which quadrant does the point(9, -2) lie?
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
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 = ?
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).
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 = ?
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.
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).
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?
Answer: Option 'B'
Answer: Option 'B'
The point (0, 9) lies in x-axis.