Samacheer Kalvi 10th Maths Book Solutions Chapter 5 – Coordinate Geometry is available here. The Samacheer Kalvi 10th Maths book answers of Chapter 5, available at BYJU’S, contain step by step explanations designed by our Mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on Samacheer Kalvi 10th Maths solutions. Practising the Samacheer Kalvi Class 10 Maths Book Chapter 5 Questions and Solutions is the best way to ace the exams.

Chapter 5 of the Samacheer Kalvi 10th Maths guide will help the students to solve problems related to the area of a triangle, collinearity of three points, area of a quadrilateral, the inclination of a line, the slope of different lines, the concept of a straight line.

### Samacheer Kalvi 10th Maths Chapter 5: Coordinate Geometry Book Exercise 5.1 Questions and Solutions

**Question 1: Find the area of the triangle whose vertices are (1,–1), (–4, 6) and (–3, –5).**

**Solution: **

Area of △ABC

= (1 / 2) [(6 + 20 + 3) – (4 – 18 – 5)]

= (1 / 2) [29 – (-19)]

= (1 / 2) [29 + 19]

= (1 / 2) 48

= 24 square units

**Question 2: Determine whether the sets of points are collinear: (a, b + c), (b, c + a) and (c, a + b)**

**Solution:**

= (1 / 2) [a (c + a) + b (a + b) + c (b + c) – b (b + c) – c (c + a) – a (a + b)]

= (1 / 2) [ac + a^{2 }+ ba + b^{2 }+ cb + c^{2} – b^{2} – bc – c^{2} – ac – a^{2} – ab]

= 0

**Question 3: A triangular-shaped glass with vertices at A (-5, -4), B (1, 6) and C (7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied?**

**Solution:**

Area of △ABC

= (1 / 2) [(-30 – 4 – 28) – (-4 + 42 + 20)]

= (1 / 2) [-62 – (58)]

= (1 / 2) [-62 – 58]

= (1 / 2) (120)

= 60 square feet

Area covered by one bucket of paint = 6 square feets

Required number of bucket = 60 / 6

= 10 buckets

**Question 4: In the figure, find the area of **

**(i) △AGF **

**(ii) △FED **

**(iii) quadrilateral BCEG**

**Solution:**

(i) △AGF

= (1 / 2) [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]

= (1 / 2) [(-22) – (-29.5)]

= (1 / 2) (-22+29.5)

= (1 / 2) [7.5]

= 3.75 square units.

(ii) △FED

F (-2, 3) E (1.5, 1) and D (1, 3)

= (1 / 2) [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]

= (1 / 2) [(5.5) – (-0.5)]

= (1 / 2) * 6

Area of △FED = 3 square units.

(iii) B (-4, -2) C (2, -1) E (1.5, 1) G (-4.5, 0.5)

= (1 / 2) [(4 + 2 + 0.75 + 9) – (-4 – 1.5 – 4.5 – 2)]

= (1 / 2) [(15.75) – (-12)]

= (1 / 2) (15.75 + 12)

= (1 / 2) (27.75)

= 13.875 square units.

### Samacheer Kalvi 10th Maths Chapter 5: Coordinate Geometry Book Exercise 5.2 Questions and Solutions

**Question 1: Show that the given vertices form a right-angled triangle and check whether it satisfies Pythagoras theorem: A (1, -4), B (2, -3) and C (4, -7)**

**Solution:**

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

= (-3 – (-4)) / (2 – 1)

= (-3 + 4) / 1

= 1

Slope of BC = (-7 – (-3)) / (4 – 2)

= (-7 + 3) / 2

= 4 / 2

= 2

Slope of CA = (-7 – (-4)) / (4 – 1)

= (-7 + 4) / 3

= -3 / 3

= -1

Slope of AB * Slope of CA = -1

1 (-1) = -1

-1 = -1

Hence the given points are vertices of the right triangle.

**Question 2: Let A (3, -4), B (9, -4) , C (5, -7) and D (7, -7). Show that ABCD is a trapezium.**

**Solution:**

A trapezium will always contain two parallel sides and two non-parallel sides.

Slope of AB :

m = (-4 + 4) / (9 – 3)

m = 0 / 6 = 0

Slope of BC :

m = (-7 + 4) / (5 – 9)

m = -3 / (-4) = 3 / 4

Slope of CD :

m = (-7 + 7) / (7 – 5)

m = 0 / 2 = 0

Slope of DA :

m = (-7 + 4) / (7 – 3)

m = -¾

In the given trapezium, sides AB and CD are parallel.

**Example 3: A quadrilateral has vertices 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:**

The midpoint of the side AB = P

= (-4 + 5) / 2, (-2 – 1) / 2

= P (1 / 2, -3 / 2)

The midpoint of the side BC = Q

= (5 + 6) / 2, (-1 + 5) / 2

= Q (11 / 2, 4 / 2)

= Q (11 / 2, 2)

The midpoint of the side CD = R

= (-7 + 6) / 2, (6 + 5) / 2

= R (-1 / 2, 11 / 2)

The midpoint of the side DA = S

= (-7 – 4) / 2, (-2 + 6) / 2

= S (-11 / 2, 4 / 2)

= S (-11 / 2, 2)

Slopes of Opposite Sides :

Slope of PQ = (7 / 2) / 5 => 7 / 10

Slope of RS = (-7 / 2) / (-5) = 7 / 10

Slope of PQ = (7 / 2) / (-5) => -7 / 10

Slope of RS = (7 / 2) / (-5) = -7 / 10

**Question 4: PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.**

**Solution:**

Distance between SM = √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}

= √(2 – 1)^{2} + (-1-1)^{2}

SM = √5

SQ = 2√5

SQ = 2 PR

2√5 = 2PR

PR = √5

PM = √5 / 2

S (1, 1) and M (2, -1)

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

m = (-1 -1) / (2 – 1)

m = -2 / 1

m = -2

The slope of PR = -1 / Slope of SQ

= -1 / (- 2)

The slope of PR = 1 / 2

Equation of PR :

(y – y_{1}) = m(x – x_{1})

(y + 1) = (1 / 2) (x – 2)

2y + 2 = x – 2

2y = x – 2 – 2

2y = x – 4

y = (x – 4) / 2

Let P(x, y) => P(x, (x – 4) / 2)

Distance between PM = √(x – 2)^{2} + ((x – 4) / 2 + 1)^{2}

√5 / 2 = √(x – 2)^{2} + ((x – 4) / 2 + 1)^{2}

Taking squares on both sides,

5 / 4 = x^{2} – 4x + 4 + [(x – 4 + 2) / 2]^{2}

5 / 4 = x^{2} – 4x + 4 + [(x – 2) / 2]^{2}

5 / 4 = x^{2} – 4x + 4 + [(x^{2} – 4x + 4) / 4]

5 / 4 = (4x^{2} – 16x + 16 + x^{2} – 4x + 4) / 4

5 = 5x^{2} – 20x + 20

5x^{2} – 20x + 15 = 0

x^{2} – 4x + 3 = 0

(x – 1) (x – 3) = 0

x = 1 and x = 3

If x = 1 then y = (1 – 4) / 2 => y = -3 / 2

If x = 3 then y = (3 – 4) / 2 => y = -1 / 2

So, the required points are (1, -3 / 2) and (3, -1 / 2).

### Samacheer Kalvi 10th Maths Chapter 5: Coordinate Geometry Book Exercise 5.3 Questions and Solutions

**Question 1: The equation of a straight line is 2 (x − y) + 5 = 0 . Find its slope, inclination and intercept on the y-axis.**

**Solution:**

2(x – y) + 5 = 0

2x – 2y + 5 = 0

2y = 2x + 5

y = (2 / 2)x + (5 / 2)

y = x + (5 / 2)

Slope (m) = 1

The angle of inclination :

m = 1

tan θ = 1

θ = 45^{o}

Intercept of y – axis :

y-intercept (c) = 5 / 2

**Question 2: The hill is in the form of a triangle and has its foot at (19, 3). The inclination of the hill to the ground is 45 ^{o}. Find the equation of the hill joining the foot and top.**

**Solution:**

Equation of the hill joining the foot and top :

slope (m) = tan 45^{o} = 1

y – y_{1} = m(x – x_{1})

y – 3 = 1(x – 19)

y – 3 = x – 19

x + y – 19 + 3 = 0

x + y – 16 = 0

**Question 3: A cat is located at the point (-6, -4) in the xy plane. A bottle of milk is kept at (5, 11). The cat wishes to consume milk travelling through the shortest possible distance. Find the equation of the path it needs to take its milk.**

**Solution:**

Equation of the path :

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

(-6, -4) and (5, 11)

(y + 4) / (11 + 4) = (x + 6) / (5 + 6)

(y + 4) / 15 = (x + 6) / 11

11(y + 4) = 15(x + 6)

11y + 44 = 15x + 90

15x – 11y + 90 – 44 = 0

15x – 11y + 46 = 0

**Question 4: Find the equation of a straight line which has slope -5 / 4 and passes through the point (–1, 2).**

**Solution:**

Slope = -5 / 4

Equation of the line passing through the point (-1, 2)

y – y_{1} = m(x – x_{1})

(y – 2) = (-5 / 4) (x – (-1))

4(y – 2) = – 5(x + 1)

4y – 2 = -5x – 5

5x + 4y – 2 + 5 = 0

5x + 4y + 3 = 0

### Samacheer Kalvi 10th Maths Chapter 5: Coordinate Geometry Book Exercise 5.4 Questions and Solutions

**Question 1: If the straight lines 12y = −(p + 3)x +12 , 12x − 7y = 16 are perpendicular then find ‘p’.**

**Solution:**

If two lines are perpendicular, then the product of their slopes = -1

The slope of the 1^{st} line :

12y = −(p + 3)x + 12

By dividing the entire equation by 12,

y = [−(p + 3) / 12]x + (12 / 12)

y = [−(p + 3) / 12]x + 1

m_{1} = -(p + 3) / 12

The slope of the 2^{nd} line :

12x − 7y = 16

m = -12 / (-7)

m_{2} = 12 / 7

m_{1} ⋅ m_{2} = -1

(-(p + 3) / 12) (12 / 7) = -1

-(p + 3) / 7 = -1

p + 3 = 7

p = 7 – 3

p = 4

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

**Solution:**

Perpendicular bisector means the line will pass through the midpoint of the line segment AB and makes a 90^{o} angle.

Midpoint = (x_{1} + x_{2}) / 2, (y_{1} + y_{2}) / 2

= (-4 + 6) / 2, (2 – 4) / 2

= 2 / 2, -2 / 2

= (1, -1)

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

= (-4 -2) / (6 + 4)

= -6 / 10

= -3 / 5

Equation of perpendicular bisector :

(y – y_{1}) = (-1 / m) (x – x_{1})

(y + 1) = (5 / 3)(x – 1)

3(y + 1) = 5(x – 1)

3y + 3 = 5x – 5

5x – 3y – 5 – 3 = 0

3x + 5y – 8 = 0

**Question 3: 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:**

Find the point of intersection of the given lines 7x + 3y = 10, 5x − 4y = 1.

7x + 3y = 10 —- (1)

5x − 4y = 1 —- (2)

4* (1) + 3 * (2)

28x + 12y = 40

15x – 12y = 3

——————

43x = 43

x = 1

By applying the value of x in (1),

7(1) + 3y = 10

7 + 3y = 10

3y = 10 – 7

3y = 3

y = 1

So, the point of intersection of the given lines is (1, 1).

The required line is passing through the point (1, 1) and parallel to the line 13x + 5y + 12 = 0.

The slope of the required line = -13 / 5

(y – y_{1}) = m(x – x_{1})

(y – 1) = (-13 / 5) (x – 1)

5(y – 1) = -13 (x – 1)

5y – 5 = -13x + 13

13x + 5y – 5 – 13 = 0

13x + 5y – 18 = 0

**Question 4: 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:**

8x + 3y = 18 —-(1)

4x + 5y = 9 —-(2)

5 * (1) – 3 * (2)

40x + 15y = 90

12x + 15y = 27

(-) (-) (-)

——————–

28x = 63

x = 63 / 28

By applying the value of x in (1), we get

8(63 / 28) + 3y = 18

3y = 18 – (126 / 7)

3y = (126 – 126) / 7

y = 0

Point of intersection of the given lines is (63 / 28, 0).

(5, –4) and (–7, 6)

Midpoint = (5 – 7) / 2, (-4 + 6) / 2

= -2 / 2, 2 / 2

= (-1, 1)

Equation of the line passing through the points (-1, 1) and (63 / 28, 0)

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

(y – 1) / (0 – 1) = (x + 1) / ((63 / 28) + 1)

(y – 1) / (- 1) = (x + 1) / (91 / 28)

91(y – 1) = -28(x + 1)

91y – 91 = -28x – 28

28x + 91y – 91 + 28 = 0

28x + 91y – 63 = 0

Dividing the entire equation by 7,

4x + 13y – 9 = 0

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