1.
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
2.
Find the co-ordinates of the centroid of ΔABC whose vertices are A(7, -3), B(5, -4) and C(-3, -5)?
Answer: Option 'B'
x1 = 7, x2 = 5, x3 = -3 and y1 = -3, y2 = -4, y3 = -5
= [(7 + 5 - 3)/3, (-3 - 4 - 5)/3]
= (12-3)/3, -12/3)
= (9/3, -12/3)
= (3, -4)
3.
If for a line m = tanϑ > 0, then
Answer: Option 'A'
m = tanϑ < 0 => is acute
4.
The distance between the points A(5, -7) and B(2, 3) is:
Answer: Option ''
AB2 = (2 - 5)2 + (3 + 7)2
=> (-3)2 + (10)2
=> 9 + 100 => √109
5.
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)
6.
Find the distance of the point A(3, -3) from the origin.
Answer: Option 'A'
OA = √32+(-3)2 = √9+9 = √18 = 3√2
7.
The vertices of a quadrilateral ABCD are A(0, 0), B(3,3), C(3, 6) and D(0, 3). Then , ABCD is a
Answer: Option 'B'
AB2 = (3-0)2 + (3-0)2 = 18
BC2 = (3-3)2 + (6-3)2 = 9
CD2 = (0-3)2 + (3-6)2 =18
AD2 = (0-0)2 + (3-0)2 = 9
AB = CD = √18 => 3√2,
BC = AD = √9
AC2 = (3-0)2 + (6-0)2 = 9 + 36 = 45
BD2 = (0-3)2 + (3-3)2 = 9 + 0 = 9
AC ≠ BD
ABCD is a parallelogram.
8.
The points A(-4, 0), B(1, -4), and C(5, 1) are the vertices of
Answer: Option 'A'
AB2 = (1 + 4)2 + (-4 - 0)2
= 25 + 16 = 41,
BC2 = (5 - 1)2 + (1 + 4)2 = 42 + 52
= 16 + 25 = 41
AC2 = (5 + 4)2 + (1 - 0)2
= 81 + 1 = 82
AB = BC and AB2 = BC2 = AC2
ΔABC is an isosceles right angled triangle
9.
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)
10.
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.
11.
The end points of a line segment AB are A(-6, 4) and B(12,24). Its midpoint is:
Answer: Option 'D'
Midpoint is C((-6+12)/2, (4+24)/2)
(-6/2, 28/2) = (-3, 14)
12.
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.
13.
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.
14.
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)
15.
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
16.
In which quadrant does the point(9, 0) lie?
Answer: Option 'A'
The point (9, 0) lies in y-axis.
17.
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.
18.
In which quadrant does the point(1, 5) lie?
Answer: Option 'A'
The point (1, 5) lies in 1st quadrant.
19.
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.
20.
Find the area of ΔABC whose vertices are A(9, -5), B(3, 7) and (-2, 4).
Answer: Option 'C'
Here, x1 = 9, x2 = 3, x3 = -2 and y1 = -5, y2 = 7, y3 = 4
= 1/2 [9(7-4) + 3(4+5) + (-2)(-5-7)]
= 1/2 [9(3) + 3(9) - 2(-12)]
= 1/2 [27 + 27 + 24]
= 1/2 [78]
= 39 sq.units