**Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.2 Pdf**

TN Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.2: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.2 – Coordinate Geometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 5 Exercise 5.2 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 5.2

**TN Samacheer Kalvi 10th Maths Solutions Chapter 5 – Coordinate Geometry**

Board |
TNSCERT Class 10th Maths |

Class |
10 |

Subject |
Maths |

Chapter |
5 (Exercise 5.2) |

Chapter Name |
Coordinate Geometry |

**TNSCERT Class 10th Maths Pdf | all Exercise Solution**

__Exercise 5.2__

__Exercise 5.2__

**(1) What is the slope of a line whose inclination with positive direction of x -axis is**

**(i) 90****°**

**(ii) 0°**

**Solution: **

(i) Now, Given that θ = 90°

We know that, slope (m) = tanθ

= tan90°

= Undefined

Thus the required slope is undefined

(ii) Given that **θ = 0****°**

We know that

slope (m) = tanθ

= tanθ

= 0

Thus the required slope is 0.

** **

**(2) What is the inclination of a line whose slope is (i) 0 (ii) 1.**

**Solution: **

(i) Now, given that slope (m) = 0

We know that slope (m) = tanθ

∴ tanθ = m

tanθ = 0

tanθ = tan0

θ = 0°

∴ Thus, the required inclination is 0°.

**(ii) Given that slope (m) = 1 **

Now, slope (m) = tanθ

∴ tanθ = m

tanθ = 1

tanθ = tan45°

θ = 45°

Thus the required inclination is 45°

** **

**(3) Find the slope of a line joining the points**

**(i) (5, √5) with the origin **

**(ii) (sin****θ, – cosθ) and (- sinθ, cos) **

**Solution: **

(i) Let, (x_{1}, y_{1}) = (5, √5) and (x_{2}, y_{2}) = (0, 0)

We know that, slope (m) = y_{2} – y_{1}/x_{2} – x_{1}

∴ m = y_{2} – y_{1}/x_{2} – x_{1}

m = 0 – √5/0 – 5 = 1/√5

∴ slope (m) = 1/√5

Therefore, the slope (m) = 1/√5

**(ii) Let, (x _{1}, y_{1}) = (sin**

**θ, – cosθ)**

and (x_{2} y_{2}) = (-sinθ, cosθ)

We know that, Slope (m) = y_{2} – y_{1}/x_{2} – x_{1}

∴ m = y_{2} – y_{1}/x_{2} – x_{1}

m = cosθ + cosθ/-sinθ – sinθ

m = cosθ/sinθ

m = cotθ

Thus, the slope (m) = – cotθ

** **

**(4)What is the slope of a line perpendicular to the line joining A(5, 1) and P where P is the mid-point of the segment (4, 2) and (-6, 4). **

**Solution: **

Now, P is the mid-point of the segment joining (4, 2) and (-6, 4)

∴ P = (4-6/2, 2+4/2) = (-1, 3)

Now, Let A (x_{1}, y_{1}) = (5, 1) and P (x_{2}, y_{2}) = (-1, 5)

We know that, slope (m) = y_{2}-y_{1}/x_{2}-x_{1}

∴ m = 3-1/-1-5

m = 2/-6

m = – 1/3

∴ Thus, the required slope (m) = – 1/3

** **

**(5) Show that the given points are collinear: (-3, -4), (7, 2) and (12, 5) **

**Solution: **

Let A (x_{1}, y_{1}) = (-3, -4), B (x_{2}, y_{2}) = (7, 2) and C (x_{3}, y_{3}) = (12, 5)

We know that,

Slope of AB (m_{1}) = y_{2} – y_{1}/x_{2} – x_{1} = 2 + 4/7 + 3 = 6/10 = 3/5

Slope of BC (m_{2}) = y_{3} – y_{2}/x_{3} – x_{2} = 5 – 2/12 – 7 = 3/5

Now, m_{1} = m_{2}

Therefore, the points A, B, C all lie in a same straight line.

Hence, the points (-3, -4), (7, 2) and (12, 5) are collinear.

** **

**(6) If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a. **

**Solution: **

Let, A (x_{1}, y_{1}) = (3, -1), B (x_{2}, y_{2}) = (a, 3) and c (x_{3}, y_{3}) = (1, -3).

Now, slope of AB (m_{1}) = y_{2} – y_{1}/x_{2} – x_{1} = 3+1/a-3 = 4/a-3

Slope of BC (m_{2}) = y_{3} – y_{2}/x_{3} – x_{2} = -3-3/1-a = -6/1-a

Now, A, B, and C three points are collinear

So, m_{1} = m_{2}

4/a-3 = -6/1-a

-6a + 18 = 4 – 4a

6a – 4a = 18 – 4

2a = 14

a = 7

Thus, the value of a = 7.

** **

**(7)The line through the points (-2, a) and (9, 3) has slope – ½. Find the value of a. **

**Solution: **

Let, A (x_{1}, y_{1}) = (-2, a) and B (x_{2}, y_{2}) = (9, 3)

Given that slope of AB (m) = – 1/2

We know that,

Slope of AB (m) = y_{2} – y_{1}/x_{2} – x_{1}

∴ y_{2} – y_{1}/x_{2} – x_{1} = m

3-a/9+2 = – 1/2

– a-3/11 = – 1/2

– a-3/11 = 1/2

2a – 6 = 11

2a = 17

a = 17/2

Thus, the value of a = 17/2

** **

**(8) The line through the points (-2, 6) and (4, 8) is perpendicular to the line though the points (8, 12) and (x, 24). Find value of z. **

**Solution: **

Let, A (x_{1}, y_{1}) = (-2, 6), B (x_{2}, y_{2}) = (4, 8)

C (x_{3}, y_{3}) = (8, 12) and D (x_{4}, y_{4}) = (x, 24)

Now, Slope of AB (m_{1}) = y_{2} – y_{1}/x_{2} – x_{1} = 8-6/4+2 = 2/6 = 1/3

Slope of CD (m_{2}) = y_{4} – y_{3}/x_{4} – x_{3} = 24-12/x – 8 = 12/x-3

Now, AB is perpendicular to the line CD

So, m_{1}× m_{2} = -1

1/3 × 12/x-8 = -1

-3 (x – 8) = 12

-3x + 24 = 12

-3x = 12 – 24

-3x = -12

x = 4

Thus, the value of x = 4

** **

**(9) Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem**

**(i) A (1, -4), B (2, -3) and C (4, -7) **

**(ii) L (0, 5), M (9, 12) and N (3, 14) **

**Solution:-**

(i) Let A (x_{1}, y_{1}) = (1, -4), B (x_{2}, y_{2}) = (2, -3)

and C (x_{3}, y_{3}) = (4, -7)

Slope of AB = y_{2} – y_{1}/x_{2} – x_{1} = -3+4/2-1 = 1

Slope of BC = y_{3} – y_{2}/x_{3} – x_{2}

= -7+3/4-2

= -4/2 = -2

Slope of CA = y_{1} – y_{3}/x_{1} – x_{3}

= -4+7/1-4

= 3/-3 = -1

Now, Slope of AB × slope CA

= 1 × -1

= -1

∴ AB⊥CA

∴ ∠BAC = 90°

∴ Triangle ABC is a right angled triangle

Now, AB = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

= √(2-1)^{2} + (-3+4) = √1+1 = √2

BC = √(x_{3} – x_{2})^{2} + (y_{3} – y_{2})^{2}

= √(4-2)^{2} + (-7+3)^{2} = √4+16 = √20

CA = √(x_{1} – x_{3})^{2} + (y_{1} – y_{3})^{2}

= √(1 – 4)^{2} + (-4+7)^{2} = √9+9 = √18

Now, (BC)^{2}

= (√20)^{2} = 20

= 18+2

= (√18)^{2} + (√2)

= (CA)^{2} + (AB)^{2}

∴ (BC)^{2} = (CA)^{2} + (AB)^{2}

Thus, Pythagoras theorem is verified.

(ii) Let, L (x_{1}, y_{1}) = (0, 5), M (x_{2}, y_{2}) = (9, 12) and N (x_{3}, y_{3}) = (3, 14)

Slope of LM = y_{2} – y_{1}/x_{2} – x_{1}

= 12-5/9-0

= 7/9

Slope of MN = y_{3} – y_{2}/x_{3} – x_{2}

= 14-12/3-9

= – 2/6 = – 1/3

Slope of NL = y_{1} – y_{3}/x_{1} – x_{3}

= 5 – 14/0-3 = -9/-3 = 3

Now, Slope of MN × Slope NL

= 1/3 × 3

= -1

∴ MN⊥NL

∴ ∠MNL = 90°

Triangle LMN is a right angle triangle Now, LM = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

= √(9 – 0)^{2} + (12 + 5)^{2} = √81 + 49 = √130

MN = √(x_{3} – x_{2})^{2} + (y_{3} – y_{2})^{2}

= √(3 – 9)^{2} + (14 – 12)^{2} = √36+4 = √40

NL = √(x_{1} – x_{3})^{2} + (y_{1} – y_{3})

= √(0 – 3)^{2} + (5 – 14)^{2} = √9 + 81 = √50

Now, (LM)^{2}

= 130

= 90 + 40

= (√90)^{2 }+ (√40)^{2}

= (NL)^{2} + (MN)^{2}

∴ (LM)^{2} = (NL)^{2} + (MN)^{2}

This, Pythagoras theorem is verified.

** **

**(10) Show that the given points form a parallelogram: **

**A (2.5, 3.5), B (10, -4), C (2.5, – 2.5) and D (-5, 5) **

**Solution: **

Let, A (x_{1}, y_{1}) = (2.5, 3.5) B (x_{2}, y_{2}) = (10, -4)

C (x_{3}, y_{3}) = (2.5, -2.5) and D (x_{3}, y_{4}) = (-5, 5)

Slope of AB (m_{1}) = y_{2} – y_{1}/x_{2} – x_{1} = -4-3.5/10-2.5 = -7.5/7.5 = -1

Slope of DC (m_{2}) = y_{4}-y_{3}/x_{4} – x_{3} = 5+2.5/-5-2.5 = 7.5/7.5 = -2

Now, m_{1} = m_{2}

So, AB is parallel to DC —- (i)

Slope of BC (m_{3}) = y_{3} – y_{2}/x_{3} – x_{2} = -2.5+4/2.5-10 = – 1.5/7.5 = – 1/5

Slope of AD (m_{4}) = y_{4} – y_{1}/x_{4} – x_{1} = 5-3.5/-5-2.5 = 1.5/7.5 = – 1/5

Now, m_{4} = m_{4}

So, BC is parallel to AD —- (ii)

From (i), (ii), we get ABCD is a parallelogram.

Thus the A, B, C, D are points from a parallelogram.

** **

**(11) If the points A (2, 2), B (-2, -3), C (1, -3) and D (x, y) form a parallelogram then find the value of x and y. **

**Solution: **

Let, A (x_{1}, y_{1}) = (2, 2), B (x_{2}, y_{2}) = (-2, -3)

C (x_{3}, y_{3}) = (1, -3) and D (x_{4}, y_{4}) = (x, y)

Now, slope of AB (m_{1}) = y_{2} – y_{1}/x_{2} – x_{1} = -3-2/-2-2 = 5/4

Slope of CD (m_{2}) = y_{4} – y_{3}/x_{4} – x_{3} = y + 3/x –1

Now, AB is parallel to DC

So, Slope of AB (m_{1}) = Slope of CD (m_{2})

∴ m_{1} = m_{2}

5/4 = y+3/x-1

5x – 5 = 4y + 12

5x – 4y = +12+5

5x – 4 = 17 —- (i)

Now, Slope of BC (m_{3}) = y_{3} – y_{2}/y.x_{3} – x_{2} = -3+3/1+2 = 0/3 = 0

Slope of AD (m_{4}) = y_{4} – y_{1}/x_{4} – x_{1} = y-1/x-2

Now, BC is parallel to AD

∴ Slope of BC (m_{3}) = slope of AD (m_{4})

∴ m_{3} = m_{4}

= y-2/x-2

y-2 = 0

y = 2

From (i) putting y = 2, we get

5x – 4y = -7

5x – 4×2 = 17

5x – 8 = 17

5x = 17+8

5x = 25

x = 5

Thus, the value of x = 5 and y = 2

** **

**(12) Let, A (3, -4), B (9, -4), C (5, -7) and D (7, -7). Show that ABCD is a trapezium. **

**Solution: **

Let, A (x_{1}, y_{1}) = (3, -4), B (x_{2}, y_{2}) = (9, -4)

C (x_{3}, y_{3}) = (5, -7) and D (x_{4}, y_{4}) = (7, -7)

Now, Slope of AB (m_{1}) = y_{2}+y_{1}/x_{2}-x_{1} = -4+4/9-3 = 0/6 = 0

Slope of CD (m_{2}) = y_{4}-y_{3}/x_{4}-x_{3} = -7+7/7-5 = 0/2 = 0

Now, m_{1} = m_{2}

∴ Slope of AB = slope of CD

So, AB is parallel to CD —– (i)

Now, slope of BC (m_{3}) = y_{3}-y_{2}/x_{3}-x_{2} = -7+7/5-9 = -3/-4 = 3/4

Slope of AD (m_{4}) = y_{4}-y_{1}/x_{4}-x_{1} = -7+4/7-3 = -3/4 = – 3/4

Now, m_{3}≠ m_{4}

Slope of BC ≠ Slope of AD

So, BC is not parallel to AD —- (ii)

From (i) and (ii), we get

ABCD is a trapezium.

Thus, ABCD is a trapezium [Proved]

** **

**(13) A quadrilateral has vertices at A (-4, -2), B (5, -1), C (6, 5) and D (-7, 6). Show that the mid-points of its sides form a parallelogram. **

**Solution: **

Let, P is the mid-point of AB.

∴ P (x_{1}, y_{1}) = (-4+5/2, -2-1/2) = (1/2, – 3/2)

Let, Q is the mid-point of BC

∴ Q (x_{2}, y_{2}) = (5+6/2, -1+5/2) = (11/2, 2)

Let, R is the mid-point of CD

R (x_{3}, y_{3}) = (6-7/2, 5+6/2) = (- 1/2, 11/2)

Let, S is the mid-point of DA

S (x_{4}, y_{4}) = (-7-4/2, 6-2/2) = (-11/2, 2)

Now, m_{1} = m_{2}

Slope of PQ = slope of RS

∴ PQ is parallel to RS —- (i)

Now, m_{3} = m_{4}

Slope of QR = slope of AD

∴ QR is parallel to AD —- (ii)

From (i) and (ii), we get

PQRS is a parallelogram.

Thus, the quadrilateral ABCD mid-points of its sides form a parallelogram. [Proved]

**Here we posted the solution of TN Tamilnadu Board Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry. TNSCERT Samacheer Kalvi 10th Maths Chapter 5 Ex 5.2 Coordinate Geometry Solved by Expert Teacher.**