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

**Answer: Option 'C'**

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

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

**Answer: Option 'A'**

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

- 3. In which quadrant does the point(9, -2) lie?

**Answer: Option 'D'**

**The point (9, -2) lies in 4th quadrant.**

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

**Answer: Option 'B'**

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

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

**Answer: Option 'B'**

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

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

**Answer: Option 'A'**

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

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

**Answer: Option 'D'**

**OA = √42+(-4)2 = √16+16 = √32 = 8√2 **

- 8. Find the distance of the point A(3, -3) from the origin.

**Answer: Option 'A'**

**OA = √32+(-3)2 = √9+9 = √18 = 3√2**

- 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. A is a point on y-axis at a distance of 5 units from x-axis lying below x-axis. The co-ordinates of A are:

**Answer: Option 'D'**

**The co-ordinates of A are A(0, -5)**

- 11. 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**

- 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. 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}.

- 14. 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

- 15. 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

- 16. 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

- 17. 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

- 18. 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.

- 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 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

- 21. 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.

- 22. 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.

- 23. 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.

- 24. 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.

- 25. 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)

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