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, 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

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'**

**x _{1} = 7, x_{2} = 5, x_{3} = -3 and y_{1} = -3, y_{2} = -4, y_{3} = -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 ''**

**AB ^{2} = (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(x**

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'**

**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.

8.

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

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'**

**x _{1} = -2, x_{2} = 5, x_{3} = 8 and y_{1} = 6, y_{2} = k, y_{3} = 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'**

**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.

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, 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

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'**

**AB ^{2} = (13-1)^{2} + (9+3)^{2}**

= 12^{2} + 12^{2} = 288.

BC^{2} = (10-13)^{2} + (12-9)^{2}

= -3^{2} + 3^{2} = 9+9 = 18

CD^{2} = (10+2)^{2} + (12-0)^{2}

= 12^{2} + 12^{2} = 288

AD^{2} = (-2-1)^{2} + (0+3)^{2} = (9+9) =18

AB = CD and BC = AD

AC^{2} = (10-1)^{2} + (12+3)^{2}

= 9^{2} + 15^{2} = 81 + 225 = 306.

BD^{2} = (-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, 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

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