1.

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(x**

2.

A point C divides the join of A(2,5) and B(3,9) in the ratio 3 : 4. The co-ordinates of C are:

**Answer: Option 'B'**

**x _{1}=2, x_{2}=3 and y_{1}=3, y_{2}=7**

= [(3 × 3 + 4 × 2)/7, (3 × 9 + 4 × 5)/7]

= 9+6/7, 27+20/7

= (15/7, 47/7)

3.

In which quadrant does the point(-4, -7) lie?

**Answer: Option 'C'**

**The point (-4, -7) lies in 3rd quadrant. **

4.

Find the area of ΔABC whose vertices are A(9, -5), B(3, 7) and (-2, 4).

**Answer: Option 'C'**

**Here, x _{1} = 9, x_{2} = 3, x_{3} = -2 and y_{1} = -5, y_{2} = 7, y_{3} = 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

5.

If the points A(1, 2), B(2, 4), and C(k, 6) are collinear, then k = ?

**Answer: Option 'B'**

**x _{1} = 1, x_{2} = 2, x_{3} = k and y_{1} = 2, y_{2} = 4, y_{3} = 6 **

= 1(4 - 6) + 2 (6 - 2) + k(2 - 4)

= -2 + 12 - 4 + 2k - 4k = 0

= 6 - 2k = 0

= -2k = 6

= k = -3.

6.

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.

7.

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

**Answer: Option 'B'**

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

8.

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)**

9.

In which quadrant does the point(-7, 6) lie?

**Answer: Option 'B'**

**The point (-7, 6) lies in 2nd quadrant.**

10.

Find the distance of the point A(4, -2) from the origin.

**Answer: Option 'B'**

**OA = √4 - 0 ^{2}+(-2 - 0)^{2} = √16+4 = √20 = √4 × 5 = 2√5 units**

11.

If for a line m = tanϑ < 0, then

**Answer: Option 'B'**

**m = tanϑ < 0 => is obtuse**

12.

If for a line m = tanϑ > 0, then

**Answer: Option 'A'**

**m = tanϑ < 0 => is acute**

13.

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

**Answer: Option 'A'**

**The point (9, 0) lies in y-axis.**

14.

The distance between the points A(b, 0) and B(0, a) is.

**Answer: Option 'B'**

**AB = √(b-0) ^{2}-(0-a)^{2} **

= √b^{2}+a^{2}

= √a^{2}+b^{2}.

15.

If the A(2, 3), B(5, k), and C(6, 7) are collinear, then k = ?

**Answer: Option 'D'**

**x _{1} = 2, x_{2} = 5, x_{3} = 6 and y_{1} = 3, y_{2} = k, y_{3} = 7 **

Now Δ = 0 <=> 2(k - 7) + 5(7 - 3) + 6(3 - k) = 0

<=> 1/2 [2 (k-7) + 5(4) + 6(3-k)] = 0

<=> 2k - 14 + 20 + 18 - 6k = 0

<=> 24 - 4k = 0

<=> 4k = 24

<=> k = 6.

16.

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)**

17.

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)**

18.

If the distance of the point P(x, y) from A(a, 0) is a + x, then y^{2} = ?

**Answer: Option 'C'**

**√(x-a) ^{2}+(y-0)^{2} = a + x **

= (x-a)^{2}+y^{2}

= (a+x)^{2} => y^{2} = (x-a)^{2}-(x-a)^{2}-4ax => y^{2} = 4ax

19.

The points A(-4, 0), B(1, -4), and C(5, 1) are the vertices of

**Answer: Option 'A'**

**AB ^{2} = (1 + 4)^{2} + (-4 - 0)^{2} **

= 25 + 16 = 41,

BC^{2} = (5 - 1)^{2} + (1 + 4)^{2} = 4^{2} + 5^{2}

= 16 + 25 = 41

AC^{2} = (5 + 4)^{2} + (1 - 0)^{2}

= 81 + 1 = 82

AB = BC and AB^{2} = BC^{2} = AC^{2}

ΔABC is an isosceles right angled triangle

20.

Find the area of ΔABC whose vertices are A(2, -5), B(4, 9) and (6, -1).

**Answer: Option 'D'**

**Here, x _{1} = 2, x_{2} = 4, x_{3} = 6 and y_{1} = -5, y_{2} = 9, y_{3} = -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

21.

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

**Answer: Option 'A'**

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

22.

The distance between the points A(5, -7) and B(2, 3) is:

**Answer: Option ''**

**AB ^{2} = (2 - 5)^{2} + (3 + 7)^{2} **

=> (-3)^{2} + (10)^{2}

=> 9 + 100 => √109

23.

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'**

**AB ^{2} = (3-0)^{2} + (3-0)^{2} = 18 **

BC^{2} = (3-3)^{2} + (6-3)^{2} = 9

CD^{2} = (0-3)^{2} + (3-6)^{2} =18

AD^{2} = (0-0)^{2} + (3-0)^{2} = 9

AB = CD = √18 => 3√2,

BC = AD = √9

AC^{2} = (3-0)^{2} + (6-0)^{2} = 9 + 36 = 45

BD^{2} = (0-3)^{2} + (3-3)^{2} = 9 + 0 = 9

AC ≠ BD

ABCD is a parallelogram.

24.

Find the vertices of triangle are A(2, 8), B(-4, 3) and (5, -1). The area of ΔABC is:

**Answer: Option 'C'**

**Here, x _{1} = 2, x_{2} = -4, x_{3} = 5 and y_{1} = 8, y_{2} = 2, y_{3} = -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

25.

The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is ?

**Answer: Option 'C'**

**AB ^{2}= (-5 - 0)^{2} + (-3 - 0)^{2} = 16 + 9 = 25 **

BC^{2} = (3 + 5)^{2} + (1-3)^{2} = 8^{2} + (-2)^{2} = 64 + 4 = 68

AC^{2} = (3 - 0)^{2} + (1 - 6)^{2} = 9 + 25 = 34.

AB = AC. ==> ΔABC is isosceles.

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