Maths-

General

Easy

Question

# The difference between the sides of the right angles triangle containing the right angle is 7 cm, and its area is 60 cm2 . Find the perimeter of the triangle.

Hint:

### Sides containing right angle are height and base.

## The correct answer is: 40 cm

### It is given that difference between perpendicular and base = 7

i.e. b – h = 7 ⇒ b = 7 + h

Now, Area of triangle = 60 cm^{2}

= 60

= 120

h^{2} + 7h – 120 = 0

h^{2} +15h – 8h – 120 = 0

h(h + 15)-8(h + 15) = 0

(h + 15)(h – 8) = 0

h = 8 , - 15

Since, perpendicular is always positive so h = 8

Base, B = 7 + 8 = 15

Using Pythagoras theorem ,

H^{2} = B^{2} + P^{2}

H^{2} = 15^{2} + 8^{2} = 225 + 64 = 289

H = = 17 cm

Perimeter of triangle = Sum of all sides

= 17 + 8 + 15 = 40 cm

### Related Questions to study

Maths-

### An artist charges $20 constant fee plus $10 per extra hour he does. Write the linear model and find the charges taken by him if he works for 4 and a half hours.

Hint:-

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

### An artist charges $20 constant fee plus $10 per extra hour he does. Write the linear model and find the charges taken by him if he works for 4 and a half hours.

Maths-General

Hint:-

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

Maths-

### A square and equilateral triangle have equal perimeters. If the diagonal of the square is 12 cm, find the area of triangle.

It is given that diagonal of the square = 12 cm

Using Pythagoras theorem in triangle ACD

AD

(12)

288 = 2x

144 = x

12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

= 16

( = 1.73)

= 110.85 cm

Using Pythagoras theorem in triangle ACD

AD

^{2}= AC^{2}+ CD^{2}(12)

^{2}= x^{2}+ x^{2}= 2x^{2}288 = 2x

^{2}144 = x

^{2}12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

^{2}= 16

^{2}= 64( = 1.73)

= 110.85 cm

^{2}### A square and equilateral triangle have equal perimeters. If the diagonal of the square is 12 cm, find the area of triangle.

Maths-General

It is given that diagonal of the square = 12 cm

Using Pythagoras theorem in triangle ACD

AD

(12)

288 = 2x

144 = x

12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

= 16

( = 1.73)

= 110.85 cm

Using Pythagoras theorem in triangle ACD

AD

^{2}= AC^{2}+ CD^{2}(12)

^{2}= x^{2}+ x^{2}= 2x^{2}288 = 2x

^{2}144 = x

^{2}12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

^{2}= 16

^{2}= 64( = 1.73)

= 110.85 cm

^{2}Maths-

### Identify a line with an undefined slope .Application

a. y = x

b. y+ x =0

c. y = 10

d. x = 1

Hint:-

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

### Identify a line with an undefined slope .Application

a. y = x

b. y+ x =0

c. y = 10

d. x = 1

Maths-General

Hint:-

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

Maths-

### The lengths of the two sides of right triangle containing the right angle differ by 2 cm. If the area of triangle is 24 cm^{2}. Find the perimeter of triangle

It is given that difference between perpendicular and base = 2 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

= 24

= 48

h

h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

H

H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

^{2}= 24

= 48

h

^{2}+ 2h – 48 = 0h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

^{2}= B^{2}+ P^{2}H

^{2}= 8^{2}+ 6^{2}= 64 + 36 = 100H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

### The lengths of the two sides of right triangle containing the right angle differ by 2 cm. If the area of triangle is 24 cm^{2}. Find the perimeter of triangle

Maths-General

It is given that difference between perpendicular and base = 2 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

= 24

= 48

h

h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

H

H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

^{2}= 24

= 48

h

^{2}+ 2h – 48 = 0h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

^{2}= B^{2}+ P^{2}H

^{2}= 8^{2}+ 6^{2}= 64 + 36 = 100H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

Maths-

### Identify a line having zero slope. Application

a. x = y

b. x = 100

c. y = 100

d. x = 0

Hint:-

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

### Identify a line having zero slope. Application

a. x = y

b. x = 100

c. y = 100

d. x = 0

Maths-General

Hint:-

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

Maths-

### The cost of an ice cream scoop is $11 and the cost of additional toppings is $1.50. Find the cost of an ice cream with 4 toppings.

Hint:-

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

### The cost of an ice cream scoop is $11 and the cost of additional toppings is $1.50. Find the cost of an ice cream with 4 toppings.

Maths-General

Hint:-

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

Maths-

### The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one fifth of litre?

Hint:

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

So, 310 words use up 1.375 cm

Step 2 of 2:

1/5

Now, 1.375cm

200cm

Final Answer:

The required number of words is 45091.

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

^{3}So, 310 words use up 1.375 cm

^{3}of ink.Step 2 of 2:

1/5

^{th}of litre = 1/51000cm^{3}= 200cm^{3}Now, 1.375cm

^{3}of ink is used up by 310 words200cm

^{3}of ink will be used up by = 45090.9 45091 wordsFinal Answer:

The required number of words is 45091.

### The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one fifth of litre?

Maths-General

Hint:

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

So, 310 words use up 1.375 cm

Step 2 of 2:

1/5

Now, 1.375cm

200cm

Final Answer:

The required number of words is 45091.

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

^{3}So, 310 words use up 1.375 cm

^{3}of ink.Step 2 of 2:

1/5

^{th}of litre = 1/51000cm^{3}= 200cm^{3}Now, 1.375cm

^{3}of ink is used up by 310 words200cm

^{3}of ink will be used up by = 45090.9 45091 wordsFinal Answer:

The required number of words is 45091.

Maths-

### Find the equation of line that passes through the point (2, -9) and which is perpendicular to the line x = 5.Application

Hint:-

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

### Find the equation of line that passes through the point (2, -9) and which is perpendicular to the line x = 5.Application

Maths-General

Hint:-

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

Maths-

### Identify the parallel lines and perpendicular lines from the given set. Application

2x + y = 1

9x + 3y = 6

y = 3x

y = -3x

2y = 4x +6

Y = - x/2

Hint:-

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

### Identify the parallel lines and perpendicular lines from the given set. Application

2x + y = 1

9x + 3y = 6

y = 3x

y = -3x

2y = 4x +6

Y = - x/2

Maths-General

Hint:-

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

Maths-

### A line l touches the Y-axis at point 3 and has slope –3. Find the slope of the line perpendicular to the line l.

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.

### A line l touches the Y-axis at point 3 and has slope –3. Find the slope of the line perpendicular to the line l.

Maths-General

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.

Maths-

### Identify the equation of a line parallel to the line y = x – 1

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of paralle lines are equal.

Step-by-step solution:-

y = x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1 …...................................................................... (Equation i)

Now, we know that slopes of parallel lines are equal.

∴ Slope of line parallel to the given line = slope of given line

∴ Slope of line perpendicular to the given line = 1

∴ We need to find the line from the given options, whose slope = 1.

a. y = 2x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2 ≠ 1

b. x = y - 1

∴ x + 1 = y

i.e. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

c. x + y = 1

∴ y = -x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 1

d. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

Final Answer:-

∴ Option b i.e. x = y - 1 & d i.e. y = x + 1 are both the correct options as both the equations are actually the same.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of paralle lines are equal.

Step-by-step solution:-

y = x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1 …...................................................................... (Equation i)

Now, we know that slopes of parallel lines are equal.

∴ Slope of line parallel to the given line = slope of given line

∴ Slope of line perpendicular to the given line = 1

∴ We need to find the line from the given options, whose slope = 1.

a. y = 2x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2 ≠ 1

b. x = y - 1

∴ x + 1 = y

i.e. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

c. x + y = 1

∴ y = -x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 1

d. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

Final Answer:-

∴ Option b i.e. x = y - 1 & d i.e. y = x + 1 are both the correct options as both the equations are actually the same.

### Identify the equation of a line parallel to the line y = x – 1

Maths-General

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of paralle lines are equal.

Step-by-step solution:-

y = x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1 …...................................................................... (Equation i)

Now, we know that slopes of parallel lines are equal.

∴ Slope of line parallel to the given line = slope of given line

∴ Slope of line perpendicular to the given line = 1

∴ We need to find the line from the given options, whose slope = 1.

a. y = 2x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2 ≠ 1

b. x = y - 1

∴ x + 1 = y

i.e. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

c. x + y = 1

∴ y = -x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 1

d. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

Final Answer:-

∴ Option b i.e. x = y - 1 & d i.e. y = x + 1 are both the correct options as both the equations are actually the same.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of paralle lines are equal.

Step-by-step solution:-

y = x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1 …...................................................................... (Equation i)

Now, we know that slopes of parallel lines are equal.

∴ Slope of line parallel to the given line = slope of given line

∴ Slope of line perpendicular to the given line = 1

∴ We need to find the line from the given options, whose slope = 1.

a. y = 2x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2 ≠ 1

b. x = y - 1

∴ x + 1 = y

i.e. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

c. x + y = 1

∴ y = -x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 1

d. y = x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 1

Final Answer:-

∴ Option b i.e. x = y - 1 & d i.e. y = x + 1 are both the correct options as both the equations are actually the same.

Maths-

### The sum of the radius and height of a cylinder is 21 cm and the T.S.A of the cylinder is 660 sq. cm. Find the height and volume of the cylinder?

Hint:

We take suitable variables for height and radius. Put it in the formula for TSA and find the required parameters.

Explanations:

Step 1 of 2:

Let the height of the cylinder be h .

Sum of the radius and height is 21 cm, so the radius will be cm.

Given, TSA of the cylinder = 660

So, r = 21 – 16= 5 cm

Step 2 of 2:

Volume of the cylinder =

cm

Final Answer:

The height of the cylinder is 16 cm and the volume is 400 cm

We take suitable variables for height and radius. Put it in the formula for TSA and find the required parameters.

Explanations:

Step 1 of 2:

Let the height of the cylinder be h .

Sum of the radius and height is 21 cm, so the radius will be cm.

Given, TSA of the cylinder = 660

So, r = 21 – 16= 5 cm

Step 2 of 2:

Volume of the cylinder =

cm

^{3}Final Answer:

The height of the cylinder is 16 cm and the volume is 400 cm

^{3}.### The sum of the radius and height of a cylinder is 21 cm and the T.S.A of the cylinder is 660 sq. cm. Find the height and volume of the cylinder?

Maths-General

Hint:

We take suitable variables for height and radius. Put it in the formula for TSA and find the required parameters.

Explanations:

Step 1 of 2:

Let the height of the cylinder be h .

Sum of the radius and height is 21 cm, so the radius will be cm.

Given, TSA of the cylinder = 660

So, r = 21 – 16= 5 cm

Step 2 of 2:

Volume of the cylinder =

cm

Final Answer:

The height of the cylinder is 16 cm and the volume is 400 cm

We take suitable variables for height and radius. Put it in the formula for TSA and find the required parameters.

Explanations:

Step 1 of 2:

Let the height of the cylinder be h .

Sum of the radius and height is 21 cm, so the radius will be cm.

Given, TSA of the cylinder = 660

So, r = 21 – 16= 5 cm

Step 2 of 2:

Volume of the cylinder =

cm

^{3}Final Answer:

The height of the cylinder is 16 cm and the volume is 400 cm

^{3}.Maths-

### Identify the equation of a line perpendicular to the line x/2 + y = -1

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

x/2 + y = -1

∴ y = -1 - x/2

i.e. Y = -1/2 x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 …...................................................................... (Equation i)

Now, we know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line perpendicular to the given line = -1/ slope of given line

∴ Slope of line perpendicular to the given line = -1/ -1/2

∴ Slope of line perpendicular to the given line = 2

∴ We need to find the line from the given options, whose slope = 2.

a. y = -1/2 x

∴ y = -1/2 x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 ≠ 2

b. y = -x

∴ y = - x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 2

c. y - 2x = 0

∴ y = 2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2

d. y + 2x = 0

∴ y = -2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -2 ≠ 2

Final Answer:-

∴ Option c i.e. y - 2x = 0 is the correct option.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

x/2 + y = -1

∴ y = -1 - x/2

i.e. Y = -1/2 x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 …...................................................................... (Equation i)

Now, we know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line perpendicular to the given line = -1/ slope of given line

∴ Slope of line perpendicular to the given line = -1/ -1/2

∴ Slope of line perpendicular to the given line = 2

∴ We need to find the line from the given options, whose slope = 2.

a. y = -1/2 x

∴ y = -1/2 x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 ≠ 2

b. y = -x

∴ y = - x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 2

c. y - 2x = 0

∴ y = 2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2

d. y + 2x = 0

∴ y = -2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -2 ≠ 2

Final Answer:-

∴ Option c i.e. y - 2x = 0 is the correct option.

### Identify the equation of a line perpendicular to the line x/2 + y = -1

Maths-General

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

x/2 + y = -1

∴ y = -1 - x/2

i.e. Y = -1/2 x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 …...................................................................... (Equation i)

Now, we know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line perpendicular to the given line = -1/ slope of given line

∴ Slope of line perpendicular to the given line = -1/ -1/2

∴ Slope of line perpendicular to the given line = 2

∴ We need to find the line from the given options, whose slope = 2.

a. y = -1/2 x

∴ y = -1/2 x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 ≠ 2

b. y = -x

∴ y = - x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 2

c. y - 2x = 0

∴ y = 2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2

d. y + 2x = 0

∴ y = -2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -2 ≠ 2

Final Answer:-

∴ Option c i.e. y - 2x = 0 is the correct option.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

x/2 + y = -1

∴ y = -1 - x/2

i.e. Y = -1/2 x - 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 …...................................................................... (Equation i)

Now, we know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line perpendicular to the given line = -1/ slope of given line

∴ Slope of line perpendicular to the given line = -1/ -1/2

∴ Slope of line perpendicular to the given line = 2

∴ We need to find the line from the given options, whose slope = 2.

a. y = -1/2 x

∴ y = -1/2 x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1/2 ≠ 2

b. y = -x

∴ y = - x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -1 ≠ 2

c. y - 2x = 0

∴ y = 2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = 2

d. y + 2x = 0

∴ y = -2x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get-

m = -2 ≠ 2

Final Answer:-

∴ Option c i.e. y - 2x = 0 is the correct option.

Maths-

### A line passes through the points (-6, 4) and (-2,8). Where does the line intersect the X-axis and the Y-axis? Application

Hint:-

1. X-intercept is the point on a line at which that line intersects the X-axis and at this point, y-coordinate = 0.

2. Y-intercept is the point on a line at which that line intersects the Y-axis and at this point, x-coordinate = 0.

3. A line is said to be passing through a point when the coordinates of such point satisfy equation of the given line.

4. When we have 2 points that lie on a given line then we can find the equation of the said line by using the 2-point formula-

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

Step-by-step solution:-

The given line passes through the point (-6,4) & (-2,8).

Hence, x1 = -6, y1 = 4, x2 = -2 & y2 = 8

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

∴ (y-4) = (8-4) × [x- (-6)]

[-2 - (-6)]

∴ y - 4 = 4 × (x + 6)

-2 + 6

∴ y - 4 = 4 × (x + 6)

4

∴ y - 4 = 1 × (x + 6)

∴ y - 4 = x + 6

∴ y = x + 6 + 4

∴ y = x + 10 ............................................ (Equation i)

We need to find the point at which the given line intersects x-axis & y-axis i.e. we need to find the x-intercept and y-intercept of the given line.

X-intercept is the point at which y-coordinate is 0.

∴ we substitute y = 0 in Equation i-

y = x + 10

∴ 0 = x + 10

-10 = x

Y-intercept is the point at which x-coordinate is 0.

∴ we substitute x = 0 in Equation i-

y = x + 10

∴ y = 0 + 10

∴ y = 10

1. X-intercept is the point on a line at which that line intersects the X-axis and at this point, y-coordinate = 0.

2. Y-intercept is the point on a line at which that line intersects the Y-axis and at this point, x-coordinate = 0.

3. A line is said to be passing through a point when the coordinates of such point satisfy equation of the given line.

4. When we have 2 points that lie on a given line then we can find the equation of the said line by using the 2-point formula-

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

Step-by-step solution:-

The given line passes through the point (-6,4) & (-2,8).

Hence, x1 = -6, y1 = 4, x2 = -2 & y2 = 8

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

∴ (y-4) = (8-4) × [x- (-6)]

[-2 - (-6)]

∴ y - 4 = 4 × (x + 6)

-2 + 6

∴ y - 4 = 4 × (x + 6)

4

∴ y - 4 = 1 × (x + 6)

∴ y - 4 = x + 6

∴ y = x + 6 + 4

∴ y = x + 10 ............................................ (Equation i)

We need to find the point at which the given line intersects x-axis & y-axis i.e. we need to find the x-intercept and y-intercept of the given line.

X-intercept is the point at which y-coordinate is 0.

∴ we substitute y = 0 in Equation i-

y = x + 10

∴ 0 = x + 10

-10 = x

Y-intercept is the point at which x-coordinate is 0.

∴ we substitute x = 0 in Equation i-

y = x + 10

∴ y = 0 + 10

∴ y = 10

### A line passes through the points (-6, 4) and (-2,8). Where does the line intersect the X-axis and the Y-axis? Application

Maths-General

Hint:-

1. X-intercept is the point on a line at which that line intersects the X-axis and at this point, y-coordinate = 0.

2. Y-intercept is the point on a line at which that line intersects the Y-axis and at this point, x-coordinate = 0.

3. A line is said to be passing through a point when the coordinates of such point satisfy equation of the given line.

4. When we have 2 points that lie on a given line then we can find the equation of the said line by using the 2-point formula-

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

Step-by-step solution:-

The given line passes through the point (-6,4) & (-2,8).

Hence, x1 = -6, y1 = 4, x2 = -2 & y2 = 8

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

∴ (y-4) = (8-4) × [x- (-6)]

[-2 - (-6)]

∴ y - 4 = 4 × (x + 6)

-2 + 6

∴ y - 4 = 4 × (x + 6)

4

∴ y - 4 = 1 × (x + 6)

∴ y - 4 = x + 6

∴ y = x + 6 + 4

∴ y = x + 10 ............................................ (Equation i)

We need to find the point at which the given line intersects x-axis & y-axis i.e. we need to find the x-intercept and y-intercept of the given line.

X-intercept is the point at which y-coordinate is 0.

∴ we substitute y = 0 in Equation i-

y = x + 10

∴ 0 = x + 10

-10 = x

Y-intercept is the point at which x-coordinate is 0.

∴ we substitute x = 0 in Equation i-

y = x + 10

∴ y = 0 + 10

∴ y = 10

1. X-intercept is the point on a line at which that line intersects the X-axis and at this point, y-coordinate = 0.

2. Y-intercept is the point on a line at which that line intersects the Y-axis and at this point, x-coordinate = 0.

3. A line is said to be passing through a point when the coordinates of such point satisfy equation of the given line.

4. When we have 2 points that lie on a given line then we can find the equation of the said line by using the 2-point formula-

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

Step-by-step solution:-

The given line passes through the point (-6,4) & (-2,8).

Hence, x1 = -6, y1 = 4, x2 = -2 & y2 = 8

(y-y1) = (y2-y1) × (x-x1)

(x2-x1)

∴ (y-4) = (8-4) × [x- (-6)]

[-2 - (-6)]

∴ y - 4 = 4 × (x + 6)

-2 + 6

∴ y - 4 = 4 × (x + 6)

4

∴ y - 4 = 1 × (x + 6)

∴ y - 4 = x + 6

∴ y = x + 6 + 4

∴ y = x + 10 ............................................ (Equation i)

We need to find the point at which the given line intersects x-axis & y-axis i.e. we need to find the x-intercept and y-intercept of the given line.

X-intercept is the point at which y-coordinate is 0.

∴ we substitute y = 0 in Equation i-

y = x + 10

∴ 0 = x + 10

-10 = x

Y-intercept is the point at which x-coordinate is 0.

∴ we substitute x = 0 in Equation i-

y = x + 10

∴ y = 0 + 10

∴ y = 10

Maths-

### Find the equation of a line which passes through (1, 2) and is perpendicular to the line with an equation y = x – 1.

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Parallel lines have equal slopes.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

Comparing the equation y = x – 1 with standard form of a straight line i.e. y = mx + c, we get-

m = 1 ….................................................................................................................... (Equation i)

Now, line l is perpendicular to this line (y = x – 1) .............................................. (Given)

and we know that- Slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line l = -1 / slope of line (y = x - 1)

∴ Slope of line l = -1 / 1 = -1 ....................... (From Equation i) ............................ (Equation ii)

We are given that line l passes through the point (1,2) and we know its slope.

i.e. x1 = 1 & y1 = 2 & m = -1

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ (y - 2) = -1 (x-1)

∴ y - 2 = -x + 1

∴ x + y = 1 + 2

∴ x + y = 3

Final Answer:-

∴ x + y = 3 is the equation of the given line.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Parallel lines have equal slopes.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

Comparing the equation y = x – 1 with standard form of a straight line i.e. y = mx + c, we get-

m = 1 ….................................................................................................................... (Equation i)

Now, line l is perpendicular to this line (y = x – 1) .............................................. (Given)

and we know that- Slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line l = -1 / slope of line (y = x - 1)

∴ Slope of line l = -1 / 1 = -1 ....................... (From Equation i) ............................ (Equation ii)

We are given that line l passes through the point (1,2) and we know its slope.

i.e. x1 = 1 & y1 = 2 & m = -1

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ (y - 2) = -1 (x-1)

∴ y - 2 = -x + 1

∴ x + y = 1 + 2

∴ x + y = 3

Final Answer:-

∴ x + y = 3 is the equation of the given line.

### Find the equation of a line which passes through (1, 2) and is perpendicular to the line with an equation y = x – 1.

Maths-General

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Parallel lines have equal slopes.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

Comparing the equation y = x – 1 with standard form of a straight line i.e. y = mx + c, we get-

m = 1 ….................................................................................................................... (Equation i)

Now, line l is perpendicular to this line (y = x – 1) .............................................. (Given)

and we know that- Slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line l = -1 / slope of line (y = x - 1)

∴ Slope of line l = -1 / 1 = -1 ....................... (From Equation i) ............................ (Equation ii)

We are given that line l passes through the point (1,2) and we know its slope.

i.e. x1 = 1 & y1 = 2 & m = -1

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ (y - 2) = -1 (x-1)

∴ y - 2 = -x + 1

∴ x + y = 1 + 2

∴ x + y = 3

Final Answer:-

∴ x + y = 3 is the equation of the given line.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Parallel lines have equal slopes.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

Comparing the equation y = x – 1 with standard form of a straight line i.e. y = mx + c, we get-

m = 1 ….................................................................................................................... (Equation i)

Now, line l is perpendicular to this line (y = x – 1) .............................................. (Given)

and we know that- Slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of line l = -1 / slope of line (y = x - 1)

∴ Slope of line l = -1 / 1 = -1 ....................... (From Equation i) ............................ (Equation ii)

We are given that line l passes through the point (1,2) and we know its slope.

i.e. x1 = 1 & y1 = 2 & m = -1

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ (y - 2) = -1 (x-1)

∴ y - 2 = -x + 1

∴ x + y = 1 + 2

∴ x + y = 3

Final Answer:-

∴ x + y = 3 is the equation of the given line.