**Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.5 Pdf**

TN Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.5: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.5 – Coordinate Geometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 5 Exercise 5.5 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 5.5

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

Board |
TNSCERT Class 10th Maths |

Class |
10 |

Subject |
Maths |

Chapter |
5 (Exercise 5.5) |

Chapter Name |
Coordinate Geometry |

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

__Exercise – 5.5__

__Exercise – 5.5__

__Multiple choice questions:__

**(1) The area of triangle formed by the points (−5, 0), (0,−5) and (5,0) is**

**(A) 0 sq. units **

**(B) 25 sq. units **

**(C) 5 sq. units **

**(D) None of these **

**Solution: **

Let, (x_{1}, y_{1}) = (-5, 0), (x_{1}, y_{2}) = (0, -5), (x_{3}, y_{3}) = (5, 0)

The area of the triangle

= 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units

= ½ {-5 (-5 – 0) + 0 (0 – 0) + 5 (0 + 5)} Sq. units

= 1/2 {25 + 25} Sq. units

= 25 sq. units (B)

** **

**(2) A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the Y axis. The path travelled by the man is**

**(A) x = 10 **

**(B) y = 10 **

**(C) x = 0 **

**(D) y = 0 **

**Solution: **

AO = wall

OB = distance the man to wall

OB is x – axis.

So, x = OB = 10 [Given that]

∴ x = 10 (A)

** **

**(3) The straight line given by the equation x = 11 is**

**(A) Parallel to X axis**

**(B) parallel to Y axis**

**(C) Passing through the origin**

**(D) Passing through the point (0,11)**

**Solution: **

Given figures, see that parallel + y-axis (B)

** **

**(4) If (5,7), (3,p) and (6,6) are collinear, then the value of p is**

**(A) 3 **

**(B) 6 **

**(C) 9 **

**(D) 12**

**Solution: **

Let, (x_{1}, y_{1}) = (5, 7), (x_{2}, y_{2}) = (3, p), (x_{3}, y_{3}) = (6, 6)

Now three points are collinear

1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} = 0

x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}) = 0

5 (p – 6) + 3 (6 – 7) + 6 (7 – p) = 0

5p – 30 – 3 + 42 – 6p = 0

-p + 9 = 0

p = 9

Thus, the value of p = 9 (C)

** **

**(5) The point of intersection of 3x – y = 4 and x + y = 8 is **

**(A) (5, 3) **

**(B) (2, 4) **

**(C) (3, 5) **

**(D) (4, 4) **

**Solution: **

Given that, x – y = 4 —- (i)

x + y = 8 —– (ii)

Now, (i) + (ii), we get, 3x + x – y + y = 4 + 8

4x = 12

x = 3

From (ii) x = 3 putting, we get

3x – y = 4

9 – y = 4

y = 9 – 4

y = 5

Thus, the intersection point (3, 5) (C)

** **

**(6) The slope of the line joining (12, 3), (4, a) is 1/8. The value of ‘a’ is **

**(A) 1 **

**(B) 4 **

**(C) -5 **

**(D) 2 **

**Solution:**

Now, the point slope is

= a-3/4-12

= a-3/-8

Now, a-3/-8 = 1/8

8a – 24 = -8

8a = 16

a = 12

∴ Thus, the value of a is 2 (D)

** **

**(7) The slope of the line which is perpendicular to a line joining the points (0, 0 and (-8, 8) is **

**(A) -1 **

**(B) 1 **

**(C) 1/3 **

**(D) -8 **

**Solution: **

(0, 0) and (-8, 8) two points slope is m_{2} = 8-0/-8-0 = -1

∴ Perpendicular line slope is m_{1} (let)

Now, m_{1}× m_{2} = -1

m_{1} × -1 = -1

m_{1} = 1 (B)

** **

**(8) If slope of the line PQ is 1/****√3 then slope of the perpendicular bisector of PQ is **

**(A) √3 **

**(B) – √3 **

**(C) 1/√3 **

**(D) 0 **

**Solution: **

Let, m_{1} = 1/√3 and perpendicular bisector slope = m_{2}

Now, m_{1}× m_{2} = -1

1/√3 × m_{2} = -1

m_{2} = – √3 (B)

** **

**(9) If A is a point on the Y axis whose ordinate is 8 and B is a point on the X axis whose abscissae is 5 then the equation of the line AB is**

**(A) 8x + 5y = 40 **

**(B) 8x – 5y = 40 **

**(C) x = 8 **

**(D) y = 5 **

**Solution: **

Now, x – intercepts = 5

y – intercepts = 8

Now, the given equation is x/5 + y/8 = 1

8x + 5y/40 = 1

8x + 5y = 40 (A)

** **

**(10) The equation of a line passing through the origin and perpendicular to the line7x – 3y + 4 = 0 is **

**(A) 7x – 3y + 4 = 0 **

**(B) 3x – 7y + 4 = 0 **

**(C) 3x + 7y = 0 **

**(D) 7x – 3y = 0 **

**Solution: **

7x – 3y + 4 = 0 is perpendicular line equation is -3x – 7y + k = 0

k = 0

Now, the required line equation is

-3x – 7y + 0 = 0

3x + 7y = 0 (C)

** **

**(11) Consider four straight lines**

**(i) l _{2}; 3y = 4x + 5 **

**(ii) l _{2}; 4y = 3x – 1 **

**(iii) l _{3}; 4y + 3x – 1 **

**(iv) l _{4}; 4x + 3y = 2 **

**Which of the following statement is true?**

**(A) l _{1} and l_{2} are perpendicular **

**(B) l _{1} and l_{4} are parallel **

**(C) l _{2} and l_{4} are perpendicular **

**(D) l _{2} and l_{3} are parallel **

**Solution: **

Now, slope of l_{1} = – (-4)/2 = 4/3

Slope of l_{2} = -(-3)/4 = 3/4

Slope of l_{3} = -3/4 = – 3/4

Slope of l_{11} = -4/3 = – 4/3

Now, slope of l_{2 }× slope of l_{4}

= ¾ × – 4/3

= -1

∴ l_{2} an l_{4} are perpendicular (C)

** **

**(12) A straight line has equation 8y = 4x + 21. Which of the following is true**

**(A) The slope is 0.5 and the y intercept is 2.6 **

**(B) The slope is 5 and the y intercept is 1.6 **

**(C) The slope is 0.5 and the y intercept is 1.6 **

**(D) The slope is 5 and the y intercept is 2.6 **

**Solution: **

8y = 4x + 21

y = 4/8 x + 21/8

Slope = 4/8 = 1/2

= 0.5

and y intercepts = 21/8

= 2.6

(A)

** **

**(13) When proving that a quadrilateral is a trapezium, it is necessary to show**

**(A) Two sides are parallel. **

**(B) Two parallel and two non-parallel sides.**

**(C) Opposite sides are parallel. **

**(D) All sides are of equal length.**

**Solution: **

(B) Two parallel and two non-parallel sides.

** **

** **

**(14) When proving that a quadrilateral is a parallelogram by using slopes you must find**

**(A) The slopes of two sides **

**(B) The slopes of two pair of opposite sides**

**(C) The lengths of all sides **

**(D) Both the lengths and slopes of two sides**

**Solution: **

**Ans: **

(B) The slopes of two pair of opposite sides.

** **

**(15) (2, 1) is the point of intersection of two lines.**

**(A) x – y – 3 = 0; 3x – y – 7 = 0 **

**(B) x + y = 3; 3x + y = 7 **

**(C) 3x + y = 3; x + y = 7 **

**(D) x + 3y – 3 = 0; x – y – 7 = 0 **

**Solution: **

Now, option (A) x – y – 3 = 0

Putting the point (2, 1)

2 – 1 – 3 = 0

1 – 3 = 0

-2 ≠ 0

Option (B) x + y = 3

Putting the point (2, 1)

2 + 1 = 3

3 = 3

and 3x + y = 7

3.2 + 2 = 7

6 + 1 = 7

7 = 7

Option (C) 3x + y = 3

Putting the point (2, 1)

3.2 + 1 = 3

6+1 = 3

7≠3

Option (D), x + 3y – 3 = 0

Putting the point (2, 1)

2 + 3.1 – 3 = 0

2+3 – 3 = 0

2 ≠ 0

∴ Ans (B)

**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 5.5 Coordinate Geometry Solved by Expert Teacher.**