**Samacheer Kalvi 10th Maths Solutions Chapter 5 Exercise 5.4 Pdf**

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

Board |
TNSCERT Class 10th Maths |

Class |
10 |

Subject |
Maths |

Chapter |
5 (Exercise 5.4) |

Chapter Name |
Coordinate Geometry |

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

__Exercise – 5.4__

__Exercise – 5.4__

**(1) Find the slope of the following straight lines**

**(i) 5y – 3 = 0 **

**(ii) 7x – 3/17 = 0 **

**Solution: **

(i) 5y – 3 = 0

0.x + 5y – 3 = 0

We know that

Slope (m) = coefficient of x/coefficient of y

= -0/5 [coefficient of x = 0, coefficient of y = 5]

= 0

Thus, the slope (m) = 0

**(ii) 7x – 3/17 = 0**

7x + 0.y – 3/17 = 0

We know that, slope (m) = – coefficient of x/coefficient of y

= -7/0

= Undefined

Thus, The slope (m) = undefined.

** **

**(2) Find the slope of the line which is**

**(i) Parallel to y = 0.7x – 11 **

**(ii) Perpendicular to the line x = -11 **

**Solution: **

(i) Now, y = 0.7x – 11 —- (i)

This equation slope (m_{1}) = 0.7

Now, equation of line parallel to this equation is y = m_{2}x + c —– (ii)

When slope = m_{2}

Now, (i) and (ii) equation parallel

So, m_{1} = m_{2}

∴ m_{2} = 0.7

Thus, the required slope is 0.7

(ii) Given equation

x = -11

0.y = 1.x + 11

∴ Slope (m_{1}) = coefficient of x/coefficient of y

= -1/0

= undefined

Now, line perpendicular slope = m_{2} (let)

Now, m_{1}× m_{2} = -1

-1/0 × m_{2} = -1

-m_{2} = -1×0

m_{2} = 0

Thus, the required slope is 0.

** **

**(3) Check whether the given lines are parallel or perpendicular**

**(i) x/3 + y/4 + 1/7 = 0 and 2x/3 + y/2 + 1/10 = 0 **

**(ii) 5x + 23y + 14 = 0 and 23x – 5y + 9 = 0 **

**Solution: **

(i) x/3 + y/4 + 1/7 = 0

x/3 = y/4 = – 1/4

4x + 3y/12 = – 1/7

28x + 21y = -12

28x + 21y + 12 = 0

Slope of the straight line x/3 + y/4 + 1/7 = 0 is

m_{1} = coefficient of x/coefficient of y

m_{1} = -28/21

m_{1} = – 4/3

Now, 2x/3 + y/2 + 1/10 = 0

2x/3 + y/2 = – 1/10

4x+3y/6 = – 1/10

40x + 30y = -6

40x + 30y + 6 = 0

Slope of the straight line 2x/3 + y/2 + 1/10 = 0 is

m_{2} = – coefficient of x/coefficient of y

m_{2} = -40/30

m_{2} = – 4/3

Now, m_{1} = m_{2}

So, the line is parallel

**(ii) 5x + 23y + 14 = 0 **

Slope of the straight line 5x + 23y + 14 = 0 is

m_{2} = – coefficient of x/coefficient of y

m_{2} = -5/23

m_{2} = – 5/23

Now, 23x – 5y + 9 = 0

Slope of the straight line 23x – 5y + 9 = 0 is

m_{2} = – coefficient of x/coefficient of y

m_{2} = -23/-5

m_{2} = 23/5

Now, m_{1} × m_{2}

= -5/23 × 23/5

= -1

So, the straight line is perpendicular

** **

**(4) If the straight lines 12y = – (p + 3) x + 12, 12x – 7y = 16 are perpendicular then find ‘p’. **

**Solution: **

12y = – (p + 3) x + 12

(p + 3) x + 12y – 12 = 0

Slope of this line is

m_{1} = coefficient of x/coefficient of y

m_{1} = – (p+3)/12

Now, 12x – 7y = 16

12x – 7y – 16 = 0

Slope of this line is

m_{2} = – coefficient of x/coefficient of y

m_{2} = -12/-7

m_{2} = 12/7

Now, two line is perpendicular

So, m_{1 }× m_{2} = -1

-(p+3)/12 × 12/7 = -1

– (p + 3) = -7

p + 3 = 7

p = 7 – 3

p = 4

Thus, the value of p = 4

** **

**(5) Find the equation of a straight line passing through the point p (-5, 2) and parallel to the line joining the points Q (3, -2) and R (-5, 4). **

**Solution: **

Let, Q (x_{1}, y_{1}) = (3, -2) and R (x_{2}, y_{2}) = (-5, 4)

Now, QR line equation is

y – y_{1}/y_{2} – y_{1} = x – x_{1}/x_{2} – x_{1}

y + 2/4+2 = x-3/-5-3

(y+2)/6 = (x-3)/-8

4y + 8 = -3x + 9

3x + 4y – 1 = 0

This equation parallel to the equation of line is

3x + 4y + k = 0, —- (i)

Where k is constant

Now, p (-5, 2) point passing throughout the equation (I

3x + 4y + k = 0

3 × (-5) + 4 × (2) ≠ k = 0

-15 + 8 + k = 0

– 7 + k = 0

k = 7

Thus, the required equation is

3x + 4y + 7 = 0

** **

**(6) Find the equation of a line passing through (6, -2) and perpendicular to the line joining the points (6, 7) and (2, -3). **

**Solution: **

Now, (6, 7) and (2, -3) points joining line equation is

y – 7/-3-7 = x – 6/2-6

y-7/-1 = x-6/-4

2y – 14 = 5x – 30

5x – 2y – 16 = 0

Now, This equation perpendicular to the line equation is

-2x – 5y + k = 0

2x + 5y – k = 0 —- (i)

Now, (6, -2) point paring thought the equation (i)

2x + 5y = k

2 × 6 + 5x (-1) = k

12 – 10 = k

k = 2

From (i), putting x = 1, we get

2x + 5y – 2 = 0

Thus, the required equation is 2x + 5y – 2 = 0

** **

**(7) A (-3, 0) B (10, -2) and C (12, 3) are the vertices of ****∆ABC. Find the equation of the altitude through A and B. **

**Solution: **

Now, slope of BC (m_{1}) = y_{2} – y_{1}/x_{2} – x_{1}

= 3+2/12-10 = 5/2

Now, AD⊥BC

So, slope of AD = m_{2} (let)

Now, m_{1}× m_{2} = -1

5/2 × m_{2} = -1

m_{2} = – 2/5

Now, Equation of the line AD is,

(y – y_{3}) = m_{2} (x – x_{3}) [Let, A (x_{3}, y_{3}) = (-3, 0)]

(y – 0) = – 2/5 (x + 3)

5y = -2x – 6

2x + 5y + 6 = 0

Thus, the equation of the altitude trough A is 2x + 5y + 6 = 0.

Let, A (x_{1}, y_{1}) = (-3, 0) and BC (x_{2}, y_{2}) = (12, 3), B (x_{3}, y_{3}) = (10, -2).

Now, slope of AC (m_{3}) = y_{2}-y_{1}/x_{2}-x_{1} = 3-0/12+3

= 3/15 = 1/5

Let, slope of BD = m_{4}

Now, BD⊥AC

Then, m_{3}× m_{4} = -1

1/3 × m_{4} = -1

m_{4} = -5

Equation of the line BD is

(y – y_{3}) = m_{4} (x – x_{3})

(y + 2) = -5 (x – 10)

y + 2 = – 5x + 50

5x + y – 48 = 0

Thus, the equation of the altitude through B is

5x + y – 48 = 0

** **

**(8) Find the equation of the perpendicular bisector of the line joining the points A (-4, 2) and B (6, -4). **

**Solution: **

D is the mid-point of AB

∴ D (x_{1}, y_{1}) = (-4+6/2, 2-4/2) = (1, -1)

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

Let, slope of CD = m_{2}

Now, AB⊥CD

Then, m_{1}× m_{2} = -1

-3/5 × m_{2} = -1

m_{2} = 5/3

Now, the equation of CD

(y – y_{1}) = m_{2} (x – x_{1})

(y + 1) = 5/3 (x – 1)

3y + 3 = 5x – 5

5x – 3y – 5 – 3 = 0

5x – 3y – 8 = 0

Thus, the required equation is 5x – 3y – 8 = 0.

** **

**(9) Find the equation of a straight line through the intersection of lines 7x + 3y = 10, 5x – 4y = 1 and parallel to the line 13x + 5y + 12 = 0 **

**Solution: **

Given that, 7x + 3y = 10 —- (i)

5x – 4y = 1 —- (ii)

Now, (i) × 4 + (ii) × 3, we get

28x + 12y = 40

15x – 12y = 3

_______________

43x = 43

x = 1

From (i), x = 1 putting, we get

7x + 3y = 10

7.1 + 3y = 10

3y = 10 – 7

3y = 3

y = 1

Now, 13x + 5y + 12 = 0 is parallel to the equation is 13x + 5y + k = 0

Now, (1, 1) point passing through this line 13 + 5 + k = 0

k = -18

13x + 5y + 12 = 0 is parallel to the equation is 13x + 5y – 18 = 0

Thus, the required equation is 13x + 5y – 18 = 0

** **

**(10) Find the equation of a straight line through the intersection of lines 5x – 6y = 2, 3x + 2y = 10 and perpendicular to the line 4x – 7y + 13 = 0 **

**Solution: **

Given that, 5x – 6y = 2 —- (i)

3x + 2y = 10 —— (ii)

Now, (i) × 2 + (ii) × 6, we get

10x – 12y = 4

18x + 12y = 60

______________

28x = 64

x = 64/28

x = 16/7

From (i) x = 16/7 putting, we get

5x – 6y = 2

5 × 16/7 – 6y = 2

6y = 80/7 – 2

6y = 80-14/7

y = 66/7×6

y = 11/7

∴ Let, A (x, y) = (16/7, 11/7)

Now, 4x – 7y + 13 = 0 is perpendicular to the line equation is,

-7x – 4y + k = 0 —– (iii)

Now, A (16/7, 11/7) point passing through this equation

– 7 × 16/7 – 4 × 11/7 + k = 0

-112-44/7 ≠ k = 0

k = 156/7

k = 156/7

From (iii) Putting k = 156/7, we get

– 7x – 4y + 156/7 = 0

49x + 28y – 156 = 0

Thus, the 4x – 7y + 13 = 0 is perpendicular to the line equation is

49x + 28y – 156 = 0

** **

**(11) Find the equation of a straight line joining the point of intersection of 3x + y + 2 = 0 and x – 2y – 4 = 0 to the point of intersection of 7x – 3y = -12 and 2y = x + 3 **

**Solution: **

Given that,

3x + y = -2 —– (i)

x – 2y = 4 —– (ii)

7x – 3y = -12 —– (iii)

2y – x + 2y = 3 —– (iv)

Now, (i) × 2 + (ii) × 1, we get

6x + 2y = -y

x – 2y = 4

__________

7x = 0

x = 0

From (i) putting x = 0, we get

3x + y = -2

3.0 + y = -2

y = -2

Let, A (x, y) = (0, -2)

Now, (i) × 2 + (ii) × 3, we get

14x – 6y = -24

-3x + 6y = 9

_____________

11x = -15

x = – 15/11

From (iv), x = – 15/11 putting, we get

– x + 2y = 3

15/11 + 2y = 3

2y = 3 – 15/11

2y = 33-15/11

2y = 18/11

y = 9/11

Let B = (-15/11, 9/11)

The equation of AB,

y+2/(9/11 + 2) = x-0/(-15/11 – 0)

y+2/(31/11) = 11x/-15

– 15 × 11 (y + 2) = 31 × 11 × x

31x = -15y – 30

31x + 15y + 30 = 0

Thus, the required equation is 31x + 15y + 30 = 0.

** **

**(12) Find the equation of a straight line through the point of intersection of the lines 8x + 3y = 18, 4x + 5y = 9 and bisecting the line segment joining the points (5, -4) and (-7, 6). **

**Solution: **

Given that, 8x + 3y = 18 —– (i)

4x + 5y = 9 —- (ii)

Now, (i) × 5 – (ii) × 3, we get

40x + 15y = 90

12x + 15y = 27

(-) (-) (-)

________________

28x = 63

x = 63/28

x = 9/4

From (i), x = 9/4 putting, we get

8x + 3y = 18

28 × 9/4 + 3y = 18

18 + 3y = 18

y = 0

Let A = (9/4, 0)

Now, (5, -4) and (-7, 6) points two mid-points = (5-7/2, -4+6/2)

= (-1, 1)

∴ Let B = (-1, 1)

Now, Equation of AB is

y – 0/1-0 = (x – 9/4)/(-1-9/4)

y/1 = (4x-9/4)/(-4-9/4)

y = 4x-9/-13

4x – 9 = -13y

4x + 13y – 9 = 0

This, the required of the line is 4x + 3y – 9 = 0.

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