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

## 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

(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 (x1, y1) = (-1-1/2, -1+4/2 = (-1, 3/2)

B be the mid-point of QR

B (x2, y2) = (-1+5/2, 4+4/2) = (2, 4)

C be the mid-point of SR

C (x3, y3) = (5+5/2, 4-1/2) = (5, 3/2)

D be the mid-point of SP

D (x4, y4) = (5-1/2, -1-1/2) = (2, -1)

We know that, two point (x1, y1) and (x2, y2)

Are distance = √(x2 – x1)2 + (y2 – y1)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, (x1, y1) = (2, 1), (x2, y2) = (3, -2) and (x3, y3) = (x, y)

We know that the area of a triangle vertex are (x1, y1), (x2, y2) and (x3, y3) is

= 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} Sq. units

Now, 1/2 {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} = 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) = (x2, y2) (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) = (x1, y1) (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) = (x3, y3) (Let)

The area of ∆ABC

= ½ {x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)} 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 (x1, y1) = (-5, 7), B (x2, y2) = (-4, k)

C (x3, y3) = (-1, -6) and D (x4, y4) = (4, 5)

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

= ½ {x1y2 + x2y2 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)} Sq. Units

Now, 1/2 {(x1y2 + x2y3 + x3y4 + x4y1) – (x2y1 + x3y2 + x4y3 + x1y4)} = 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 (x1, y1) = (-2, -1), B (x2, y2) = (4, 0)

c (x3, y3)  = (3, 3) and D (x4, y4) = (-3, 2) Now, Slope of AB (m1) = y2-y1/x2-x1

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

Slope of BC (m2) = y3-y2/x3-x2

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

Slope of CD (m3) = y4-y3/x4-x3 = 2-3/-3-3 = -1/-6

= 1/6

Slope of DA (m4) = y1-y4/x1-x4 = -1-2/-2+3 = -3

Now, slope of AB = slope of CD

m1 = m3

∴AB and CD are parallel

Now, slope of BC = slope of DA

m2 = m4

∴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 – a2 = -6

-a2 + a + 6 = 0

a2 – a – 6 = 0

a2 – (3-2) a – b = 0

a2 – 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 (x1, y1) = (14, 980)

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

Let (x2, y2) = (16, 1220)

Slope of the line = y2 – y1/x2 – x1

= 1220-980/16-14

= 240/2

= 120

∴Equation of the line is

(y – y1) = m (x – x1)

(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 (x1, y1)

x + 3y = 7 —- (i)

This equation slop is (m1) = coefficient of x/coefficient of y

= -1/3

= – 1/3

Let, AB line slope = m2

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

m1× m2 =-1

– 1/3 × m2 = -1

m2 = 3

Throw, the equation of AB is y – y1 = m (x – x1)

(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+x1/2, 8+y1/2) = (1, 2)

3 + x1 = 2

x1 = 2 – 3

x1 = -1

8+y1/2 = 2

8 + y1 = 4

y1 = 4-8

y1 = -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.

Updated: February 3, 2022 — 1:47 pm