**1. Which type of triangle is formed using the vertices A (1, 2), B (−3, 12) and C (−5, −6)?**

a) Isosceles triangle

b) Equilateral triangle

c) Scalene triangle

d) Right-angled triangle

**Answer:** c) Scalene triangle

**Explanation: **We know that:

→ An equilateral triangle has all sides equal.

→ An isosceles triangle has two sides equal.

→ A scalene triangle has no sides equal.

→ A right-angled triangle satisfies the Pythagoras Theorem.

We can find the lengths of the sides of a triangle using the distance formula.

Distance between any two points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by:

**d = √[(x**_{2}** − x**_{1}**)**^{2}** + (y**_{2}** − y**_{1}**)**^{2}**]**

We are given **△**ABC with vertices A (1, 2), B (−3, 12) and C (−5, −6).

Now, using the distance formula, find the length of AB, BC and CA.

→ AB = √[((− 3) − 1)^{2} + (12 − 2)^{2}]

= √[(− 4)^{2} + (10)^{2}]

= √[16 + 100]

= √116

= 2√29 units

→ BC = √[((− 5) − (− 3))^{2} + ((− 6) − 12)^{2}]

= √[(− 5 + 3)^{2} + (− 6 − 12)^{2}]

= √[(− 2)^{2} + (− 18)^{2}]

= √[4 + 324]

= √328

= 2√82 units

→ CA = √[((− 5) − 1)^{2} + ((− 6) − 2)^{2}]

= √[(− 6)^{2} + (− 8)^{2}]

= √[36 + 64]

= √100

= 10 units

Now, AB^{2} = (√116)^{2}

= 116

BC^{2} = (√328)^{2}

= 328

CA^{2} = (10)^{2}

= 100

Thus, AB^{2} + CA^{2} ≠ BC^{2}

Thus, it does not satisfy the Pythagoras theorem.

Since the length of all three sides is different and it doesn’t satisfy the Pythagoras theorem, thus **△**ABC is a scalene triangle.

**2. The point which divides the line segment joining the points (5, –9) and (–2, 5) in ratio 2 : 3 internally lies in which quadrant?**

a) First quadrant

b) Second quadrant

c) Third quadrant

d) Fourth quadrant

**Answer: **d) Fourth quadrant

**Explanation:**Here,

x_{1} = 5

x_{2} = − 2

y_{1} = − 9

y_{2} = 5

m = 2

n = 3

Let P(x, y) divides the line segment joining the points (5, – 2) and (− 9, 5) in the ratio 2 : 3 internally. Thus the coordinates of P are

**3. If the point P(5, 4) lies on the line segment joining points A(3, 2) and B(9, 8), then which of the following is true?**

a) BP = AB/3

b) AP = BP/2

c) AP = BP/3

d) BP = AB/2

**Answer: **b) AP = BP/2

**Explanation: **We are given that point P(5, 4) lies on the line segment joining points A(3, 2) and B(9, 8).

Thus,

AP = √[(x_{2} − x_{1})^{2} + (y_{2} − y_{1})^{2}]

= √[(5 − 3)^{2} + (4 − 2)^{2}]

= √[(2)^{2} + (2)^{2}]

= √[4 + 4]

= √8

= 2√2

BP = √[(x_{2} − x_{1})^{2} + (y_{2} − y_{1})^{2}]

= √[(5 − 9)^{2} + (4 − 8)^{2}]

= √[(− 4)^{2} + (− 4)^{2}]

= √[16 + 16]

= √32

= 4√2

AB = √[(x_{2} − x_{1})^{2} + (y_{2} − y_{1})^{2}]

= √[(9 − 3)^{2} + (8 − 2)^{2}]

= √[(6)^{2} + (6)^{2}]

= √[36 + 36]

= √72

= 6√2

We know that: AP = 2√2 and BP = 4√2

∴ AP = BP/2

**4. Out of the following points, which three points satisfy the condition of collinearity?**

a) (5, −3), (4, 1) and (−2, 7)

b) (1, 2), (3, −2) and (0, 5)

c) (−4, −1), (2, 3) and (−3, 6)

d) (−6, 10), (−4, 6) and (3, −8)

**Answer: **d) (−6,10), (−4, 6) and (3, −8)

**Explanation: **We know that the area of a triangle formed by three collinear points is zero.

Also, If A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of a △ABC, then its area is given by:

**Area = ½ [x**_{1}**(y**_{2}** − y**_{3}**) + x**_{2}**(y**_{3}** − y**_{1}**) + x**_{3}**(y**_{1}** − y**_{2}**)]**

**Option a)**

We are given three points (5, −3), (4, 1) and (−2, 7)

Here,

x_{1} = 5

x_{2} = 4

x_{3} = −2

y_{1} = −3

y_{2} = 1

y_{3} = 7

Thus, Area of triangle = ½[x_{1}(y_{2} − y_{3}) + x_{2}(y_{3} − y_{1}) + x_{3}(y_{1} − y_{2})]

= ½[5(1 − 7) + 4(7 − (− 3)) + (− 2)((− 3) − 1)]

= ½[(5)(−6) + 4(10) + (−2)(−4)]

= ½[−30 + 40 + 8]

= ½[18]

= 9

≠ 0

Thus, the points are not collinear.

**Option b)**

We are given three points (1, 2), (3, −2) and (0, 5)

Here,

x_{1} = 1

x_{2} = 3

x_{3} = 0

y_{1} = 2

y_{2} = −2

y_{3} = 5

Thus, Area of triangle = ½[x_{1}(y_{2} − y_{3}) + x_{1}(y_{3} − y_{1}) + x_{3}(y_{1} − y_{2})]

= ½[1(− 2 − 5) + 3(5 − 2) + 0(2 − (− 2))]

= ½[1(− 7) + 3(3) + 0(2 + 2)]

= ½[−7 + 9 + 0]

= ½[2]

= 1

≠ 0

Thus, the points are not collinear.

**Option c)**

We are given three points (−4, −1), (2, 3) and (−3, 6)

Here,

x_{1} = −4

x_{2} = 2

x_{3} = −3

y_{1} = −1

y_{2} = 3

y_{3} = 6

Thus, Area of triangle = ½[x_{1}(y_{2} − y_{3}) + x_{2}(y_{3} − y_{1}) + x_{3}(y_{1} − y_{2})]

= ½[(−4)(3 − 6) + 2(6 − (−1)) + (−3)((−1) − 3)]

= ½[(− 4)(− 3) + 2(7) + (− 3)(− 4)]

= ½[12 + 14 + 12]

= ½[38]

= 19

≠ 0

Thus, the points are not collinear.

**Option d)**

We are given three points (− 6, 10), (−4, 6) and (3, −8)

Here,

x_{1} = −6

x_{2} = −4

x_{3} = 3

y_{1} = 10

y_{2} = 6

y_{3} = −8

Thus, Area of triangle = ½[x_{1}(y_{2} − y_{3}) + x_{2}(y_{3} − y_{1}) + x_{3}(y_{1} − y_{2})]

= ½[(−6)(6 − (−8)) + (−4)(−8 − 10) + 3(10 − 6)]

= ½[(−6)(14) + (−4)(−18) + 3(4)]

= ½[−84 + 72 + 12]

= ½[−84 + 84]

= ½[0]

= 0

So, the area of the triangle is zero. Thus, the points are collinear.

**5. In what ratio does the point (–2, 3) divide the line segment joining the points A (–8, 6) and B (5, –9)?**

a) 7 : 5

b) 7 : 6

c) 5 : 7

d) 6 : 7

**Answer:** d) 6 : 7

**Explanation: **Let point (–2, 3) divide the line segment joining the points A(–8, 6) and B(5, –9) in k : 1 ratio.

We know that if the point M (x, y) divides the line segment joining P (x1, y1) and Q (x2, y2) internally in the ratio m : n, then the coordinates of M are given by the section formula as

(–2, 3) = (**5k - 8/****k + 1****, ****- 9k + 6/****k + 1**)

We know that if (x, y) = (a, b) then x = a and y = b

–2 = **(5k - 8/k + 1)** and 3 = **(-9k + 6/k + 1)**

Using –2 = 5k - 8k + 1 to find the ratio.

**⇒ **–2 = **(5k - 8/k + 1)⇒** –2 (k + 1) = 5k – 8

Ratio = k : 1 = 6/7 : 1 = 6 : 7

Thus, point (– 2, 3) divide the line segment joining the points A(– 8, 6) and B(5, – 9) in the ratio 6 : 7.

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