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

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

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

Board |
TNSCERT Class 10th Maths |

Class |
10 |

Subject |
Maths |

Chapter |
5 (Unit Exercise 5) |

Chapter Name |
Coordinate Geometry |

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

__UNIT Exercise – 5__

__UNIT Exercise – 5__

**(1) PQRS is a rectangle formed by joining the points P (-1, -1), Q (-1, 4), R (5, 4) and S (5, -1). A, B, C and D are the mid-points of PQ, QR, RS and SP respectively. Is the quadrilateral ABCD a square, a rectangle or a rhombus? Justify your answer. **

**Solution: **

Now, A be the mid-point of PQ

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

B be the mid-point of QR

B (x_{2}, y_{2}) = (-1+5/2, 4+4/2) = (2, 4)

C be the mid-point of SR

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

D be the mid-point of SP

D (x_{4}, y_{4}) = (5-1/2, -1-1/2) = (2, -1)

We know that, two point (x_{1}, y_{1}) and (x_{2}, y_{2})

Are distance = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

Now, The length of AB is

= √(2+1)^{2} + (4 – 3/2)

= √(3)^{2} + (3/2)^{2}

= √9+25/4

= √36+25/4

= √61/4

= √61/2

The length of BC is

= √(5-2)^{2} + (3/2 – 4)^{2}

= √(7)^{2} + (- 5/2)^{2}

= √9 + 25/4

= √36+25/4

= √61/2

The length of CD is

= √(2-5)^{2} + (-1-3/2)^{2}

= √(-3)^{2} + (-5/2)^{2}

= √9+25/4

= √36+25/4

= √61/4

The length of DA is

= √(-1-2)^{2} + (3/2+1)^{2}

= √(-3)^{2} + (5/2)^{2}

= √9+25/4

= √36+25/2

= √61/2

Now, AB = BC = CD = DA = √61/2

Now, four sides are equal.

So, ABCD is rhombus

Thus, the ABCD is a rhombus.

** **

**(2) The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, -2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex. **

**Solution: **

Let, (x_{1}, y_{1}) = (2, 1), (x_{2}, y_{2}) = (3, -2) and (x_{3}, y_{3}) = (x, y)

We know that the area of a triangle vertex are (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) is

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

Now, 1/2 {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} = 5

2 (-2-y) + 3 (y – 1) + x (1 + 2) = 10

-4-2y+3y – 3 + 3x = 10

3x + y – 7 = 10

3x + y = 17 —– (i)

Given that, y = x + 3

x – y = -3 —– (ii)

(i) now, (i) + (ii), we get

3x + y = 17

x – y = -3

__________

4x = 14

x = 14/4 = 7/2

From (i), x = 7/2 putting, we get

3x + y = 17

3 × 7/2 + y = 17

y = 17 – 21/2 = 34-21/2

y = 13/2

Thus, the third vertex is (7/2, 13/2)

** **

**(3) Find the area of a triangle formed by the lines 3x + y – 2 = 0, 5x + 2y – 3 = 0 and 2x – y – 3 = 0 **

**Solution: **

Now, Equation of AB = 3x + y – 2 = 0 —- (i)

Equation of BC = 5x + 2y – 3 = 0 —- (ii)

Equation of CA = 2x – y – 3 = 0 —- (iii)

Now, (i) × 2 – (ii) × 1; We get

**6x + 2y – 4 = 0 **

**5x + 2y – 3 = 0 **

** (-) (-) (+)**

**_____________**

**x – 1 = 0**

x = 1

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

3x + y – 2 = 0

3.2 + y – 2 = 0

3 – 2 + y = 0

1 + y = 0

y = -1

∴ Now, D = (1, -1) = (x_{2}, y_{2}) (let)

Now, (i) + (ii), we get

**3x + y – 2 = 0**

**2x – y – 3 = 0**

**_____________**

**5x – 5 = 0**

x = 1

From (iii), x = 1 Putting, we get

2x – y – 3 = 0

2 – y – 3 = 0

y = -1

∴ A = (1, 1) = (x_{1}, y_{1}) (Let)

Now, (ii) × 1 + (iii) × 2, we get

**5x + 2y – 3 = 0 **

**4x – 2y – 6 = 0 **

**______________**

**9x – 9 = 0**

9x = 9

x = 1

From (ii), x = 1, Putting, we get

5x + 2y – 3 = 0

5 + 2y – 3 = 0

2 + 2y = 0

y = -1

∴ Now, C = (1, -1) = (x_{3}, y_{3}) (Let)

The area of ∆ABC

= ½ {x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})} Sq. units

= ½ {1 (-1+1) + 1 (-1 + 1) + 1 (-1 + 1)} Sq. units

= ½ × 0 Sq. units

= 0 Sq. units

** **

**(4) If vertices of a quadrilateral are at A (-5, 7), B (-4, k), C (-1, -6) and D (4, 5) and its area is 72 Sq. units. Find the value of k. **

**Solution: **

Let, A (x_{1}, y_{1}) = (-5, 7), B (x_{2}, y_{2}) = (-4, k)

C (x_{3}, y_{3}) = (-1, -6) and D (x_{4}, y_{4}) = (4, 5)

We know that, the area of the quadrilateral of ABCD is

= ½ {x_{1}y_{2} + x_{2}y_{2} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})} Sq. Units

Now, 1/2 {(x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{4} + x_{4}y_{1}) – (x_{2}y_{1} + x_{3}y_{2} + x_{4}y_{3} + x_{1}y_{4})} = 72

{(-5k + 24 – 5 + 28) – (-28 – k – 24 – 25)} = 72×2

(-5k + 47) – (-5 – 77) = 72×2

-5k + 47 + k + 77 = 72×2

-4k + 124 = 72×2

– 4k = 144 – 124

-4k = 20

k = -5

Thus, the value of k = -5.

** **

**(5) Without using distance formula, show that the points (-2, – 1), (4, 0), (3, 3) and (-3, 2) are vertices of a parallelogram. **

**Solution: **

Now, Let A (x_{1}, y_{1}) = (-2, -1), B (x_{2}, y_{2}) = (4, 0)

c (x_{3}, y_{3}) = (3, 3) and D (x_{4}, y_{4}) = (-3, 2)

Now, Slope of AB (m_{1}) = y_{2}-y_{1}/x_{2}-x_{1}

= 0+1/4+2 = 1/6

Slope of BC (m_{2}) = y_{3}-y_{2}/x_{3}-x_{2}

= 3+0/3-4 = 3/-1 = -3

Slope of CD (m_{3}) = y_{4}-y_{3}/x_{4}-x_{3} = 2-3/-3-3 = -1/-6

= 1/6

Slope of DA (m_{4}) = y_{1}-y_{4}/x_{1}-x_{4} = -1-2/-2+3 = -3

Now, slope of AB = slope of CD

m_{1} = m_{3}

∴AB and CD are parallel

Now, slope of BC = slope of DA

m_{2} = m_{4}

∴BC and DA are parallel

Thus, the ABCD be a parallelogram.

** **

**(6) Find the equations of the lines, whose sum and product of intercepts are 1 and –6 respectively.**

**Solution: **

Let x intercept = a

∴y intercept = 1 – a

∴Now, x intercept x y intercept = -6

a× (1-a) = -6

a – a^{2} = -6

-a^{2} + a + 6 = 0

a^{2} – a – 6 = 0

a^{2} – (3-2) a – b = 0

a^{2} – 3a + 2a – 6 = 0

a (a – 3) + 2 (a – 3) = 0

(a – 3) (a + 2) = 0

a – 3 = 0

a = 3

a + 2 = 0

a = -2

When a = 3

y intercept = 1 – a = 1 – 3 = -2

∴Now, equation of a line is

x/a + y/b = 1

x/3 + y/-2 = 1

x/3 – y/2 = 1

2x-3y/6 = 1

2x – 3y = 6

2x – 3y – 6 = 0

When, a = -2

y intercept = 1+2 = 3

∴Equation of line is

x/-2 + y/3 = 1

-3x + 2y/6

-3x + 2y = 6

3x – 2y = -6

3x – 2y + 6 = 0

Thus, the required equation

2x – 3y – 6 = 0

3x – 2y + 6 = 0

** **

**(7) The owner of a milk store finds that, he can sell 980 litres of milk each week at****₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could be sell weekly at ₹17/litre? **

**Solution: **

Given that, sell 980 litres of milk each week at ₹14/litre.

Let selling price = x and demand = y let (x_{1}, y_{1}) = (14, 980)

Given that, sell 1220 litres of mill each week at ₹16/litre.

Let (x_{2}, y_{2}) = (16, 1220)

Slope of the line = y_{2} – y_{1}/x_{2} – x_{1}

= 1220-980/16-14

= 240/2

= 120

∴Equation of the line is

(y – y_{1}) = m (x – x_{1})

(y – 980) = 120 (x – 14)

y – 980 = 120x – 1680

120x – y = 1680 – 980

120x – y = 700

Now, given that x = 17

120×17 – y = 700

y = 2040 – 700

y = 1340

Thus, the demand is 1340 litres

** **

**(8) Find the image of the point (3, 8) with respect to the line z + 3y = 7 assuming the line to be a plane mirror. **

**Solution: **

Let, B the mirror point of A is (x_{1}, y_{1})

x + 3y = 7 —- (i)

This equation slop is (m_{1}) = coefficient of x/coefficient of y

= -1/3

= – 1/3

Let, AB line slope =** m _{2}**

Now, x + 3y = 7 and AB is perpendicular.

m_{1}× m_{2} =-1

– 1/3 × m_{2} = -1

m_{2} = 3

Throw, the equation of AB is y – y_{1} = m (x – x_{1})

(y – 8) = 3 (x – 3)

3x – 9 = y – 8

3x – y = -8+9

3x – y = 1 — (ii)

Now, (ii) × 1 + (i) × 3, we get

x + 3y = 7

9x – 3y = 3

__________

10x = 10

x = 1

From (ii) x = 1 putting, we get

3x – y = 1

3 – y = 1

y = 3 – 1 = 2

y = 2

∴Point 0 = (1, 2)

Now, 0 be the mid-point of AB.

Now, (3+x_{1}/2, 8+y_{1}/2) = (1, 2)

3 + x_{1} = 2

x_{1} = 2 – 3

x_{1} = -1

8+y_{1}/2 = 2

8 + y_{1} = 4

y_{1} = 4-8

y_{1} = -4

Thus, the mirror point B is (-1, -4)

** **

**(9) Find the equation of a line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes. **

**Solution: **

Now, 4x + 7y = 3 —- (i)

2x – 3y = -1 —– (ii)

Now, (i) × 3 + (ii) × 7, we get

**12x + 21y = 9**

**14x – 21y = -7 **

**_____________**

**26x = 2**

x = 2/26

x = 1/13

From (i) putting x = 1/13, we get

4x + 7y = 3

4 × 1/13 + 7y = 3

7y = 3 – 4/13

7y = 39 – 4/13

7y = 35/13

y = 5/13

The intersection point is = (1/13, 5/13)

Let, x intercepts = y intercepts = a

∴Equation of line is x/a + y/a = 1

(1/13, 5/13) point passing through this line

1/13a + 5/13a = 1

1+5/13a = 1

13a = 6

a = 6/13

∴The required equation the line is

x/(6/13) + y/(6/13) = 1

13x/6 + 13y/6 = 1

13x + 13y = 6

13x + 13y – 6 = 0

Thus, the required equation is 13x + 13y – 6 = 0

** **

**(10) A person standing at a junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 seek to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find the equation of the path that he should follow. **

**Solution: **

Given that 2x – 3y = -4 —- (i)

3x + 4y = 5 —- (ii)

Now, (i) × 4 + (ii) × 3, we get

8x – 12y = -16

9x + 12y = 15

____________

17x = -1

x = – 1/17

From (i) putting x = – 1/17, we get

2x – 3y = -4

2 × – 1/17 – 3y = -4

3y = 4 – 2/17 = 68-2/17 = 66/17

y = 22/17

Now, two straight paths intersection point is (- 1/17, 22/17)

Now, 6x – 7y + 8 = 0 is perpendicular equation is – 7x – 6y + k = 0 —- (i)

Now, (- 1/17, 22/17) point passing through the equation (i)

– 7 × – 1/17 – 6 × 22/7 + k = 0

7/17 – 132/17 + k = 0

7-132/17 + k = 0

k = 125/17

The required equation is – 7k – 6y + 135/17 = 0

-119x – 102y + 125 = 0

119x + 102y – 125 = 0

Thus, the required equation is 119x + 102y – 125 = 0.

**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 Coordinate Geometry Solved by Expert Teacher.**