Maths-
General
Easy

Question

If the length of the median of an equilateral triangle is x cm, then what is the area of triangle.

Hint:

In an equilateral triangle, median is the perpendicular of the base too.

The correct answer is: Equilateral triangle


    It is given that length of the median = x cm
    Let side of the triangle = a

    Since median divides the base ⇒ BD =
    Using Pythagoras in triangle ABD
    AB2 = AD2 + BD2
    a2 = x2 + ( a over 2 )2
    = x2  ⇒  fraction numerator square root of 3 a over denominator 2 end fraction = x
    a = fraction numerator 2 x over denominator square root of 3 end fraction
    Area of equilateral triangle = fraction numerator square root of 3 over denominator 4 end fraction a squared equals fraction numerator square root of 3 over denominator 4 end fraction open parentheses fraction numerator 2 x over denominator square root of 3 end fraction close parentheses squared
    fraction numerator square root of 3 over denominator 4 end fraction cross times fraction numerator 4 x squared over denominator 3 end fraction
    fraction numerator x squared over denominator square root of 3 end fraction

    Related Questions to study

    General
    Maths-

    Write an equation of the line with zero slope and passing through (3,7).Application

    Hint:-
    Equation of a line in slope point form-
    (y-y1) = m × (x-x1)
    tep-by-step solution:-
    As per given information-
    The given line passes through the point (3,7) and has a slope of 0.
    ∴ x1 = 3, y1 = 7 & m = 0
    Using Slope point form of a line, we can find the equation of the given line as-
    (y-y1) = m × (x-x1)
    ∴ y - 7 = 0 × (x - 3)
    ∴ y - 7 = 0
    ∴ y = 7

    Write an equation of the line with zero slope and passing through (3,7).Application

    Maths-General
    Hint:-
    Equation of a line in slope point form-
    (y-y1) = m × (x-x1)
    tep-by-step solution:-
    As per given information-
    The given line passes through the point (3,7) and has a slope of 0.
    ∴ x1 = 3, y1 = 7 & m = 0
    Using Slope point form of a line, we can find the equation of the given line as-
    (y-y1) = m × (x-x1)
    ∴ y - 7 = 0 × (x - 3)
    ∴ y - 7 = 0
    ∴ y = 7
    General
    Maths-

    The edges of a triangular board are 6 cm, 8 cm and 10 cm. Find the area of triangle.

    It is given that a = 6 cm , b = 8 cm and c = 10 cm
    Using Pythagoras theorem
    102 = 82 + 62
    100 = 64 + 36
    100 = 100  i.e. Pythagoras holds true
    Hence given triangle is a right angles triangle with Base, b = 8 cm . height, h = 6 cm and hypotenuse = 10 cm
    Area of a right angle triangle = 1 half cross times b cross times h
    1 half cross times 8 cross times 6 = 24 cm2

    The edges of a triangular board are 6 cm, 8 cm and 10 cm. Find the area of triangle.

    Maths-General
    It is given that a = 6 cm , b = 8 cm and c = 10 cm
    Using Pythagoras theorem
    102 = 82 + 62
    100 = 64 + 36
    100 = 100  i.e. Pythagoras holds true
    Hence given triangle is a right angles triangle with Base, b = 8 cm . height, h = 6 cm and hypotenuse = 10 cm
    Area of a right angle triangle = 1 half cross times b cross times h
    1 half cross times 8 cross times 6 = 24 cm2
    General
    Maths-

    An electrical geyser in cylindrical shape, having a diameter of 3.5 cm and height of 1.2 m neglecting the thickness of its walls. Calculate its capacity in litres?

    Hint:
    We simply calculate the volume of the geyser to find its capacity.
    Explanations:
    Step 1 of 1:
    Given, base radius r = 3.5/2cm
    Height  h = 1.2m = 120cm
    Capacity of the geyser = volume of the geyser
    equals pi r squared h
    equals 22 over 7 cross times 35 over 20 cross times 35 over 20 cross times 120
    = 1155 cm3
    = 1.155 L (since 1 L = 1000 cm3)
    Final Answer:
    The capacity of the geyser is 1.155L

    An electrical geyser in cylindrical shape, having a diameter of 3.5 cm and height of 1.2 m neglecting the thickness of its walls. Calculate its capacity in litres?

    Maths-General
    Hint:
    We simply calculate the volume of the geyser to find its capacity.
    Explanations:
    Step 1 of 1:
    Given, base radius r = 3.5/2cm
    Height  h = 1.2m = 120cm
    Capacity of the geyser = volume of the geyser
    equals pi r squared h
    equals 22 over 7 cross times 35 over 20 cross times 35 over 20 cross times 120
    = 1155 cm3
    = 1.155 L (since 1 L = 1000 cm3)
    Final Answer:
    The capacity of the geyser is 1.155L
    parallel
    General
    Maths-

    Write an equation of the line with undefined slope and passing through (5,11).Application

    Hint:-
    1. X-intercept is the point on a line at which the given line intersects the x-axis.
    2. At this point y-coordinate = 0.
    3. A  verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope.
    Step-by-step solution:-
    The given line has an undefined slope.
    ∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.
    We know that any line that is parallel to the Y-axis can be represented as-
    x = a ....................................................................................... (Equation i)
    where a = x-intercept.
    Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.
    Since the given line passes through point (5,11) where the x-coordinate = 5
    The x-coordinate at any point on the line will be 5
    i.e. for y= 0 also, x = 5
    and x-intercept is the point at which y = 0
    ∴ x-intercept for the given line = a = 5 ................................ (Equation ii)
    ∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)

    Write an equation of the line with undefined slope and passing through (5,11).Application

    Maths-General
    Hint:-
    1. X-intercept is the point on a line at which the given line intersects the x-axis.
    2. At this point y-coordinate = 0.
    3. A  verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope.
    Step-by-step solution:-
    The given line has an undefined slope.
    ∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.
    We know that any line that is parallel to the Y-axis can be represented as-
    x = a ....................................................................................... (Equation i)
    where a = x-intercept.
    Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.
    Since the given line passes through point (5,11) where the x-coordinate = 5
    The x-coordinate at any point on the line will be 5
    i.e. for y= 0 also, x = 5
    and x-intercept is the point at which y = 0
    ∴ x-intercept for the given line = a = 5 ................................ (Equation ii)
    ∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)
    General
    Maths-

    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.

    It is given that difference between perpendicular and base = 7
    i.e. b – h = 7  ⇒  b = 7 + h
    Now, Area of triangle = 60 cm2
    1 half cross times b cross times straight h= 60
    left parenthesis 7 plus h right parenthesis cross times h= 120
    h2 + 7h – 120 = 0
    h2 +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 ,
    H2 = B2 + P2
    H2 = 152 + 82 = 225 + 64 = 289
    H = square root of 289 = 17 cm
    Perimeter of triangle = Sum of all sides
    = 17 + 8 + 15 = 40 cm

    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.

    Maths-General
    It is given that difference between perpendicular and base = 7
    i.e. b – h = 7  ⇒  b = 7 + h
    Now, Area of triangle = 60 cm2
    1 half cross times b cross times straight h= 60
    left parenthesis 7 plus h right parenthesis cross times h= 120
    h2 + 7h – 120 = 0
    h2 +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 ,
    H2 = B2 + P2
    H2 = 152 + 82 = 225 + 64 = 289
    H = square root of 289 = 17 cm
    Perimeter of triangle = Sum of all sides
    = 17 + 8 + 15 = 40 cm
    General
    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.

    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.
    parallel
    General
    Maths-

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

    It is given that diagonal of the square = 12square root of 2 cm
    Using Pythagoras theorem in triangle ACD
    AD2 = AC2 + CD2
    (12square root of 2)2 = x2 + x2 = 2x2
    288 = 2x2
    144 = x2
    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 = fraction numerator square root of 3 over denominator 4 end fraction a2
    = fraction numerator square root of 3 over denominator 4 end fraction 162 = 64
    (square root of 3 = 1.73)
    = 110.85 cm2

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

    Maths-General
    It is given that diagonal of the square = 12square root of 2 cm
    Using Pythagoras theorem in triangle ACD
    AD2 = AC2 + CD2
    (12square root of 2)2 = x2 + x2 = 2x2
    288 = 2x2
    144 = x2
    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 = fraction numerator square root of 3 over denominator 4 end fraction a2
    = fraction numerator square root of 3 over denominator 4 end fraction 162 = 64
    (square root of 3 = 1.73)
    = 110.85 cm2
    General
    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.

    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.
    General
    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 cm2. 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 cm2
    1 half cross times b cross times straight h= 24
    left parenthesis 2 plus h right parenthesis cross times h= 48
    h2 + 2h – 48 = 0
    h = fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction
    h = fraction numerator negative 2 plus-or-minus square root of 2 squared minus 4 left parenthesis negative 48 right parenthesis end root over denominator 2 end fraction equals fraction numerator negative 2 plus-or-minus square root of 196 over denominator 2 end fraction equals fraction numerator negative 2 plus-or-minus 14 over denominator 2 end fraction
    h = 6 , - 8
    Since, perpendicular is always positive so h = 6
    Base, B = 2 + 6 = 8
    Using Pythagoras theorem ,
    H2 = B2 + P2
    H2 = 82 + 62 = 64 + 36 = 100
    H = square root of 100 = 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 cm2. 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 cm2
    1 half cross times b cross times straight h= 24
    left parenthesis 2 plus h right parenthesis cross times h= 48
    h2 + 2h – 48 = 0
    h = fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction
    h = fraction numerator negative 2 plus-or-minus square root of 2 squared minus 4 left parenthesis negative 48 right parenthesis end root over denominator 2 end fraction equals fraction numerator negative 2 plus-or-minus square root of 196 over denominator 2 end fraction equals fraction numerator negative 2 plus-or-minus 14 over denominator 2 end fraction
    h = 6 , - 8
    Since, perpendicular is always positive so h = 6
    Base, B = 2 + 6 = 8
    Using Pythagoras theorem ,
    H2 = B2 + P2
    H2 = 82 + 62 = 64 + 36 = 100
    H = square root of 100 = 10 cm
    Perimeter of triangle = Sum of all sides
    = 8 + 10 + 6 = 24 cm
    parallel
    General
    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.

    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.
    General
    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.

    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.
    General
    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
    equals pi r squared h
    equals pi left parenthesis 0.25 right parenthesis squared cross times 7
    equals 1.375 cm3
    So, 310 words use up 1.375 cm3 of ink.
    Step 2 of 2:
    1/5th of  litre  = 1/51000cm3 = 200cm3
    Now, 1.375cm3 of ink is used up by 310 words
    therefore 200cm3 of ink will be used up by fraction numerator 310 over denominator 1.375 end fraction cross times 200 = 45090.9 almost equal to 45091 words
    Final 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
    equals pi r squared h
    equals pi left parenthesis 0.25 right parenthesis squared cross times 7
    equals 1.375 cm3
    So, 310 words use up 1.375 cm3 of ink.
    Step 2 of 2:
    1/5th of  litre  = 1/51000cm3 = 200cm3
    Now, 1.375cm3 of ink is used up by 310 words
    therefore 200cm3 of ink will be used up by fraction numerator 310 over denominator 1.375 end fraction cross times 200 = 45090.9 almost equal to 45091 words
    Final Answer:
    The required number of words is 45091.
    parallel
    General
    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

    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
    General
    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)

    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)
    General
    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.

    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.
    parallel

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