In this article, we shall study to solve problems based on the area of the sector.

Example – 01:

Find the area of a sector of the circle which subtends an angle of 120° at the centre, if the radius of the circle is 6 cm.

Given: Angle subtended at centre = θ = 120° = 120 x (π/180) = (2π/3)^c , Radius of circle = r = 6 cm.

To find: Area of sector = A =?

Solution:

Area of sector = ½ r²θ= ½ x 6² x (2π/3) =12π sq. cm

Ans: The area of the sector is 12π sq. cm

Example – 02:

The area of the circle is 81π sq. cm. Find the length of its arc subtending an angle of 150° at the centre. Also, find the area of the corresponding sector.

Given: Area of circle = 81π sq. cm, Angle subtended at centre = θ = 150° = 150 x (π/180) = (5π/6)^c ,

To find: Length of arc = S = ?, Area of sector = A = ?

Solution:

Area of circle = πr²  = 81π

∴  r²  = 81

∴  r  = 9 cm

Length of arc = S = r θ = 9 x (5π/6) = 7.5 π cm

Area of sector = ½ r²θ= ½ x 9² x (5π/6) = 33.75 π sq. cm

Ans: length of the arc is 7.5 π cm and the area of the sector is 33.75 π sq. cm

Example – 03:

The area of a circle is 25π sq. cm. Find the length of its arc subtending an angle of 144° at the centre. Also, find the area of the corresponding sector.

Given: Area of circle = 25π sq. cm, Angle subtended at centre = θ = 144° = 144 x (π/180) = (4π/5)^c ,

To find: Length of arc = S = ?, Area of sector = A = ?

Area of circle = πr²  = 25π

∴  r²  = 25

∴  r  = 5 cm

Length of arc = S = r θ = 5 x (4π/5) = 4π cm

Area of sector = ½ r²θ= ½ x 5² x (4π/5) = 10π sq. cm

Ans: length of the arc is 4π cm and the area of the sector is 10π sq. cm

Example – 04:

The area of a circle is 81π sq. cm. Find the length of its arc subtending an angle of 300° at the centre. Also, find the area of the corresponding sector.

Given: Area of circle = 81π sq. cm, Angle subtended at centre = θ = 300° = 300 x (π/180) = (5π/3)^c ,

To find: Length of arc = S = ?, Area of sector = A = ?

Area of circle = πr²  = 81π

∴  r²  = 81

∴  r  = 9 cm

Length of arc = S = r θ = 9 x (5π/3) = 15π cm

Area of sector = ½ r²θ= ½ x 9² x (5π/3) = 67.5 π sq. cm

Ans: length of the arc is 15π cm and the area of the sector is 67.5π sq. cm

Example – 05:

The perimeter of a sector of a circle of area 25π sq. cm is 20 cm. Find the area of the sector.

Given: Area of circle = 25π sq. cm, Perimeter = 20 cm

To find: Area of sector = A = ?

Area of circle = πr²  = 25π

∴  r²  = 25

∴  r  = 5 cm

Perimeter of sector = r + r + s = 20

∴ 2r + r θ = 20

∴ r (2 + θ) = 20

∴ 5 (2 + θ) = 20

∴  2 + θ = 4

∴   θ = 2^c

Area of sector = ½ r²θ= ½ x 5² x 2 = 25 sq. cm

Ans: The area of the sector is 25 sq. cm

Example – 06:

The perimeter of a sector of a circle of area 64π sq. cm is 56 cm. Find the area of the sector.

Given: Area of circle = 64π sq. cm, Perimeter = 56 cm

To find: Area of sector = A = ?

Area of circle = πr²  = 64π

∴  r²  = 64

∴  r  = 8 cm

Perimeter of sector = r + r + s = 56

∴ 2r + r θ = 56

∴ r (2 + θ) = 56

∴ 8 (2 + θ) = 56

∴  2 + θ = 7

∴   θ = 5^c

Area of sector = ½ r²θ= ½ x 8² x 5 = 160 sq. cm

Ans: The area of the sector is 160 sq. cm

Example – 07:

Find the area of sector whose arc length is 30π cm and the angle of the sector is 40°.

Given: Length of arc = 30π cm, angle of sector = θ = 40° = 40 x π/180 = (2π/9)^c ,

To find: Area of sector = A = ?

Length of arc = S =  r θ

∴ 30π =  r x (2π/9)

∴ r   = 135 cm

Area of sector = ½ r²θ= ½ x 135² x (2π/9) = 2025π sq. cm

Ans: The area of the sector is 2025 sq. cm

Example – 08:

In a circle of radius 12 cm, an arc PQ subtends the angle of 30° at the centre. Find the area between arc PQ and chord PQ.

Given: radius of circle = r = 12 cm, angle subtended at the centre = θ = 30° = 30 x (π/180) = (π/6)^c ,

To find: the area between arc PQ and chord PQ.

Solution:

Area of sector  = ½ r²θ= ½ x 12² x (π/6) = 12π sq. cm

In Δ OOR, sin 30° = QR/OQ

∴  OR = OQ sin 30° = 12 x 1/2 = 6 cm

Area of Δ POQ = ½ x base x height = ½ x OP x QR = ½ x 12 x 6 = 36 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴  Area of shaded region = 12π – 36 = 12(π – 3) sq. cm

Ans: The area between arc PQ and chord PQ is 12(π – 3) sq. cm

Example – 09:

OPQ is the sector of a circle with centre O and radius 12 cm. if m ∠ POQ= 60°, find the difference between the areas of sector POQ and Δ POQ.

Given: radius of circle = r = 12 cm, angle subtended at the centre = θ = 60° = 60 x (π/180) = (π/3)^c ,

To find: the difference between the areas of sector POQ and Δ POQ.

Solution:

Area of sector  = ½ r²θ= ½ x 12² x (π/3) = 24π sq. cm

In Δ OQR, sin 60° = QR/OQ

∴  OR = OQ sin 60° = 12 x √3 /2 = 6√3 cm

Area of Δ POQ = ½ x base x height = ½ x OP x QR = ½ x 12 x 6√3 = 36√3 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴  Area of shaded region = 24π – 36√3 = 12(2π – 3√3) sq. cm

Ans: The difference between the areas of sector POQ and Δ POQ. is 12(2π – 3√3) sq. cm

Example – 10:

OPQ is a sector of a circle with centre O and radius 12 cm. if m∠OPQ = 30°, Find the area between arc PQ and chord PQ.

Given: radius of circle = r = 12 cm,

To find: the area between arc PQ and chord PQ.

Solution:

Δ OPQ is isosceles triangle

m∠ OPQ = m∠ OQP =  30°

m∠ POQ = θ = 120° = 120 x π/180 = (2π/3)^c

Area of sector  = ½ r²θ= ½ x 12² x (2π/3) = 48π sq. cm

Δ OQR is 30°-60°-90° triangle

OR = ½OQ = ½ x 12 = 6 cm

QR = √3 /2 OQ = √3 /2 x 12 = 6√3 

PQ = 2 QR = 2 x 6√3  = 12√3 

Area of Δ POQ = ½ x base x height = ½ x PQ x OR = ½ x 12√3  x 6 = 36√3 sq.cm.

Area of shaded region = Area of sector – Area of Δ POQ

∴  Area of shaded region = 48π – 36√3 = 12(4π – 3√3) sq. cm

Ans: The area between arc PQ and chord PQ. is 12(4π – 3√3) sq. cm

Example – 11:

Two circles each of radius 7 cm intersect each other such that the distance between their centres is 7√2 cm. Find area common to both the circles.

Given: radius of circle = r = 7 cm, Distance between centres = 7√2 cm

To find: the area of common portion = ?

Solution:

In quadrilateral ADBC

AD = DB = BC = CA = 7cm

diagonal AB = 7√2 cm

Hence quadrilateral ADBC is a square with each angle 90°

This is central angle subtended for sectors of both the circles = θ = 90° = 90 x π/180 = (π/2)^c

Area of common region = area of sector (A-CED) + Area of sector (B-CFD) – area of square ADBC

∴ Area of common region = ½ r²θ  + ½ r²θ  – r²

∴ Area of common region = r² θ  – r²

∴ Area of common region = r² ( θ – 1)

∴ Area of common region = 7² ( π/2 – 1)

∴ Area of common region = 49(π/2 – 1) sq. cm

Ans: The area common to both the circle is 49(π/2 – 1) sq. cm

Science > Mathematics > Trigonometry > Angle Measurement > Area of Sector

The post Area of Sector appeared first on The Fact Factor.

In this article, we shall solve problems based on the length of an arc (arc length).

Example – 01:

Find the length of the arc of a circle of diameter 10 cm, if the arc is subtending an angle of 36° at the centre.

Given: Diameter = 10 cm, radius = r = 10/2 = 5 cm, angle subtended = 36° = 36 x (π/180) = (π//5)^c

To Find: Length of arc = S = ?

Solution:

length of arc is given by

S = r θ  = 5 x (π//5) = π cm

Ans: The length of the arc is π cm

Example – 02:

Find the length of the arc of a circle which subtends an angle of 108° at the centre, if the radius of the circle is 15 cm.

Given: radius = r = 15 cm, angle subtended = 108° = 108 x (π/180) = (3π//5)^c

To Find: Length of arc = S = ?

Solution:

length of arc is given by

S = r θ  = 15 x (3π//5) = 9π cm

Ans: The length of the arc is 9π cm

Example – 03:

The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length equal to the length of the radius.

Given: radius = r = 9 cm, length of chord = r

To Find: Length of arc = S = ?

Solution:

Thus ΔOAB is equilateral triangle

angle subtended = 60° = 60 x (π/180) = (π//3)^c

We have

S = r θ  = 9 x π//3 = 3π cm

Ans: The length of the arc is 3π cm

Example – 04:

In a circle of diameter 40 cm, the length of the chord is 20 cm. Find the length of the minor arch of the chord.

Given: diameter = 4o cm, radius = r = 20 cm, length of chord = 20 cm

To Find: Length of arc = S = ?

Solution:

Thus ΔOAB is equilateral triangle

angle subtended = 60° = 60 x π/180 = (π//3)^c

We have

S = r θ  = 20 x π//3 = 20π/3 cm

Ans: The length of the minor arc is 20π/3 cm

Example – 05:

A pendulum of 14 cm long oscillates through an angle of 18°. Find the length of the path described by its extremity.

Given: radius = r = 14 cm, angle subtended = 18° = 18 x (π/180) = (π//10)^c

To Find: Length of arc = S = ?

Solution:

We have

S = r θ  = 14 x π/10 = 1.4π cm

Ans: The length of arc is 1.4π cm

Example – 06:

Find in radians and degrees the angle subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of circle is 25 cm.

Given: radius = r = 25 cm, Length of arc = 15 cm

To Find: angle subtended at the centre = θ = ?

Solution:

We have

S = r θ

∴  θ = S/r = 15/25 = (3/5)^c

(3/5)^c = (3/5) x (180/π) = (108/π)°

Ans: The angle subtended at the centre is (3/5)^c or (108/π)°

Example – 07:

Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm. Take ( π = 22/7)

Given: Length of arc = S = 37.4 cm, angle subtended = 60° = 60 x π/180 = (π//3)^c

To Find: Length of arc = S = ?

Solution:

We have

S = r θ

∴   r =S/θ = 37.4/(π//3) = (37.4 x 3) / (22/7)

∴   r = (37.4 x 3 x 7) / 22 = 35.7

Ans: The length of the arc is 35.7 cm

Example – 08:

A wire of length 10 cm is bent so as to form an arc of a circle of radius 4 cm. What is the angle subtended at the centre in degrees?

Given: radius = r = 4 cm, length of wire = length of arc = 10 cm

To Find: angle subtended at the centre = θ = ?

Solution:

We have

S = r θ

∴  θ = S/r = 10/4 = (2.5)^c

(2.5)^c = (2.5) x (180/π) = (450/π)

Ans: The angle subtended at the centre is (2.5)^c or (450/π)°

Example – 09:

Δ PQR is an equilateral triangle with a side 18 cm. A circle is drawn on segment QR as a diameter. Find the length of the arc of this circle intercepted within the triangle.

Given: Side of equilateral triangle = 18 cm, radius = 18/2 = 9 cm

To Find: length of the arc of circle intercepted = S =?

Solution:

s Δ PQR is an equilateral triangle, its each angle is 60°

Hence the triangles Δ QOE and Δ ROD are also equilateral triangles

∠ EOQ = ∠ ROD = 60°

EOD = 60° =  θ

Now, we have

S = r θ  = 9 x π//3 = 3π cm

Ans: The length of arc is 3π cm

Example – 10:

Two arcs of the same length subtend angle 60° and 75° at the centres of the circles. What is the ratio of the radii of the two circles?

Given: Angles subtended, θ₁ = 60° and θ₂ = 75°

To Find: Ratio of radii = r₁/r₂ = ?

Solution:

length of arc is given by S = r θ

For the first arc S₁ = r₁θ₁  ………. (1)

For the second arc S₂ = r₂θ₂  ………. (2)

Now length of two arcs is the same

S₁ = S₂

∴  r₁θ₁ =  r₂θ₂

∴  r₁/r₂ =  θ₂/ θ₁

∴  r₁/r₂ =  75°/ 60° = 5/4

Ans: The ratio of radii is 5:4

Example – 11:

Two arcs of the same length subtend angle 65° and 110° at the centres of the circles. What is the ratio of the radii of the two circles?

Given: Angles subtended, θ₁ = 65° and θ₂ = 110°

To Find: Ratio of radii = r₁/r₂ = ?

Solution:

length of arc is given by S = r θ

For the first arc S₁ = r₁θ₁  ………. (1)

For the second arc S₂ = r₂θ₂  ………. (2)

Now length of two arcs is the same

S₁ = S₂

∴  r₁θ₁ =  r₂θ₂

∴  r₁/r₂ =  θ₂/ θ₁

∴  r₁/r₂ =  110°/ 65° = 22/13

Ans: The ratio of radii is 22:13

Example – 12:

A train is running on a circular track of a radius of 1 km at a rate of 36 km per hour. Find the angle to the nearest minute, through which it will turn in 30 seconds.

Given: Radius of the arc = r = 1 km = 1000 m, Speed of train = v = 36 km per hour = 36 x 1000/3600 = 10 m/s, time taken = t = 30 s.

To Find: Angle through which the train turns = θ = ?

Solution:

Distance covered by train  i.e. length of arc =  speed x time = 10 x 30 = 300 m

S = r θ

∴  θ = S/r = 300/1000 = (0.3)^c

∴  θ = (0.3)^c = (0.3) x (180/π) = (54/3.142)° = 17.19°

∴  θ = 17° + 0.19° = 17° + 0.19 x 60′ = 17° + 11′ = 17°,11′

Ans: The train will turn through 17°,11′

Example – 13:

A train is running on a circular track of a radius of 1500 m at the rate of 66 km per hour. Find the angle to the in radian, through which it will turn in 10 seconds.

Given: Radius of the arc = r = 1500 m, Speed of train = v = 66 km per hour = 66 x 1000/3600 = 55/3 m/s, time taken = t = 10 s.

To Find: Angle through which the train turns = θ = ?

Solution:

Distance covered by train  i.e. length of arc =  speed x time = (55/3) x 10 = 550/3 m

S = r θ

∴  θ = S/r = (550/3)/1500 = 550/4500 = (11/90)^c

Ans: The train will turn through (11/90)^c

Example – 14:

A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 m when it traces an angle of 72° at the centre, find the length of the rope. Take π = 22/7.

Given: central angle = θ = 72° = 72 x π/180 = (2π/5)^c, arc length = S = 88 m

To Find: Length of rope = r = ?

S = r θ

∴  r = S/θ = 88/(2π/5) = 220/π = 220 x 7/22 = 70 m

Ans: The length of the rope is 70 m.

Example – 15:

If the perimeter of a sector of a circle is four times the radius of the circle, find the central angle of the corresponding sector in radians.

Given:  Perimeter = 4 x radius = 4r

To find: Central angle = θ = ?

Perimeter of sector = r + r + s = 4r

∴ 2r + r θ = 4r

∴ r θ = 2r

∴   θ = 2^c

Ans: The central angle  is 2^c

Science > Mathematics > Trigonometry > Angle Measurement > Length of an Arc

The post Length of an Arc appeared first on The Fact Factor.

In this article, we shall study the problems based on the interior angles of a polygon, and the angle between the hour hand and a minute hand of a clock.

Example – 01:

If x^c = 405° and y° = – (π/12)^c. Find x and y

Solution:

Given x^c = 405°

x containing term on R.H.S. is in radians. Hence we should convert L.H.S. into radians

∴  x^c = 405° = 405 x π/180 = (9π/4)^c

∴  x = 9π/4

Given y° = – (π/12)^cy containing term on R.H.S. is in degrees. Hence we should convert L.H.S. into degrees

∴  y° = – (π/12)^c = – (π/12) x (180/π) = 15°

∴  y = 15

Ans: x = 9π/4 and y = 15

Example – 02:

If θ° = – (5π/9)^c and Φ^c = 900°. Find θ and Φ

Solution:

Given θ° = – (5π/9)^c

θ containing term on R.H.S. is in degrees. Hence we should convert L.H.S. into degrees

∴  θ° = – (5π/9)^c = – (5π/9) x (180/π) = – 100°

∴  θ = -100

Given Φ^c = 900°

Φ containing term on R.H.S. is in radians. Hence we should convert L.H.S. into radians

∴  Φ^c = 900° = 900 x π/180 = (5π)^c

∴  Φ = 5π

Ans: θ = -100 and Φ = 5π

Example – 03:

Express following angles in radians

– 35°45’30”

Solution:

– 35°45’30” = – [35° + (45/60)° + (30/3600)°]

– 35°45’30” = – [35° + 0.75° + 0.0083°] = – 35.7583°

– 35°45’30” = – 35.7583 x π/180 = 0.1987 π

– 35°45’30” = 0.1987 x 3.142 = 0.6242 radian

50°37’30”

Solution:

50°37’30” = 50° + (37/60)° + (30/3600)°

50°37’30” = 50° + 0.6167° + 0.0083° = 50.625°

50°37’30” = 50.625 x π/180 = 0.2812 π

50°37’30” = 0.2812 x 3.142 = 0.8837 radian

– 10°40’30”

Solution:

10°40’30” = 10° + (40/60)° + (30/3600)°

10°40’30” = 10° + 0.6667° + 0.0083° = 10.675°

10°40’30” = 10.675 x π/180 = 0.0593 π

10°40’30” = 0.0593 x 3.142 = 0.1863 radian

Interior Angle of Regular Polygon:

Steps to Find Interior Angle of Polygon:

1. Find the measure of each exterior angle of regular polygon = 360°/No.of sides of polygon
2. Find the measure of each interior angle of polygon = 180° – measure of exterior angle

Example – 04:

Find Interior angles of following regular polygons in degrees and radians

Pentagon:

Solution:

Pentagon has 5 sides

Each exterior angle  = 360°/5 = 72°

Each interior angle = 180° – 72° = 108° = 108 x π/180 = (3π/5)^c

Ans: The interior angle of a regular pentagon is 72° or (3π/5)^c

Hexagon:

Solution:

Hexagon has 6 sides

Each exterior angle  = 360°/6 = 60°

Each interior angle = 180° – 60° = 120° = 120 x π/180 = (2π/3)^c

Ans: The interior angle of a regular hexagon is 120° or (2π/3)^c

Octagon:

Solution:

Octagon has 8 sides

Each exterior angle  = 360°/8 = 45°

Each interior angle = 180° – 45° = 135° = 135 x π/180 = (3π/4)^c

Ans: The interior angle of a regular octagon is 135° or (3π/4)^c

A Polygon with 20 sides:

Solution:

Polygon has 20 sides

Each exterior angle  = 360°/20 = 18°

Each interior angle = 180° – 18° = 162° = 162 x π/180 = (9π/10)^c

Ans: The interior angle of a regular polygon with 20 sides is 162° or (9π/10)^c

A Polygon with 15 sides:

Solution:

Polygon has 15 sides

Each exterior angle  = 360°/15 = 24°

Each interior angle = 180° – 24° = 156° = 156 x π/180 = (13π/15)^c

Ans: The interior angle of a regular polygon with 15 sides is 156° or (13π/15)^c

A Polygon with 12 sides:

Polygon has 12 sides

Each exterior angle  = 360°/12 = 30°

Each interior angle = 180° – 30° = 150° = 150 x π/180 = (5π/6)^c

Ans: The interior angle of a regular polygon with 12 sides is 150° or (5π/6)^c

Example – 05:

Find the number of sides of polygon if each of its interior angle is (3π/4)^c.

Solution:

Each interior angle = (3π/4)^c = (3π/4) x (180/π) = 135°

Hence each exterior angle = 180° – 135° = 45°

Number of sides of polygon = 360°/each exterior angle = 360°/45 = 8°

Ans: Thus the polygon has 8 sides

Angle Between Hour Hand and Minute Hand:

Example – 06:

Find the degree and radian measure of the angle between the hour hand and minute hand of a clock at the following timings.

Twenty minutes past seven:

Solution:

At twenty minutes past seven, the minute hand is at 4 and hour hand crossed 7

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 20 minutes = 0.5° x 20 = 10°

Thus the hour hand is 10° ahead of 7 th Mark

The angle between 4 and 7 is 90°

Thus angle between hour hand and minute hand = 90° + 10° = 100°

100° = 100 x π/180 = (5π/9)^c

Ans: The angle between the hour hand and the minute hand is 100° or (5π/9)^c

Twenty minutes past two:

Solution:

At twenty minutes past two, the minute hand is at 4 and hour hand crossed 2

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 20 minutes = 0.5° x 20 = 10°

Thus the hour hand is 10° ahead of 2 nd Mark

The angle between 2 and 4 is 60°

Thus angle between hour hand and minute hand = 60° – 10° = 50°

50° = 50 x π/180 = (5π/18)^c

Ans: The angle between the hour hand and the minute hand is 50° or (5π/18)^c

Quarter past six:

Solution:

At quarter past six, the minute hand is at 3 and hour hand crossed 6

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 15 minutes = 0.5° x 15 = 7.5°

Thus the hour hand is 7.5° ahead of 6th Mark

The angle between 3 and 6 is 90°

Thus angle between hour hand and minute hand = 90° + 7.5° = 97.5°

97.5° = 97.5 x π/180 = (13π/24)^c

Ans: The angle between the hour hand and the minute hand is 97.5° or (13π/24)^c

Ten past eleven:

Solution:

At ten past eleven, the minute hand is at 2 and hour hand crossed 11

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by hour hand in 10 minutes = 0.5° x 10 = 5°

Thus the hour hand is 5° ahead of 11th Mark

The angle between 11 and 2 is 90°

Thus angle between hour hand and minute hand = 90° – 5° = 85°

85° = 85 x π/180 = (17π/36)^c

Ans: The angle between the hour hand and the minute hand is 85° or (17π/36)^c

Example – 07:

Show that the minute hand of a clock gains 5°30′ on hour hand in one minute.

Solution:

Angle traced by hour hand in 1 minute = 0.5°

Angle traced by minute hand in 1 minute = 6°

Thus angle between hour hand and minute hand = 6° – 0.5° = 5.5° = 5°30′

Ans: Thus the minute hand of a clock gains 5°30′ on hour hand in one minute.

Example – 08:

Determine which of the following pairs of angles are coterminal.

210° and – 150°

– 150° = – 150° + 360° = 210°

Thus the two angles have the same initial arm and terminal arm.

Hence the angles 210° and – 150° are coterminal angles.

330° and – 60°

– 60° = – 60° + 360° = 300°

Thus the two angles do not have the same initial arm and terminal arm.

 Hence the angles 330° and – 60° are not coterminal angles.

405° and – 675°

405° = 405° – 360° = 45°

– 675° + 360° x 2 = 45°

Thus the two angles have the same initial arm and terminal arm.

 Hence the angles 405° and – 675° are coterminal angles.

1230° and – 930°

1230° = 1230° – 360° x 3 = 150°

– 930° + 360° x 3 = 150°

Thus the two angles have the same initial arm and terminal arm.

 Hence the angles 1230° and – 930° are coterminal angles.

Example – 09:

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in 1 second?

Given: No. of revolutions = 360 per minute

To Find: Radians per second =?

Solution:

No. of revolutions per second = 360/60 = 6

In one revolution the wheel turns through 2π radians

Radians per second = 2π x 6 = 12π^c

Ans: The wheel will turn through 12π^c in 1 second

Science > Mathematics > Trigonometry > Angle Measurement > Angle Measurement

The post Angle Measurement appeared first on The Fact Factor.

In this article, we shall study to find the equation of a line using a two-point form.

Example 01:

Find the equation of a line passing through the points

• (5, 4) and (3, -2)

Solution:

Let the points be A(5, 4) ≡ (x₁, y₁) and (3, -2) ≡ (x₂, y₂)

By two-point form

∴ (y – 4)/(4 + 2) = (x – 5)/(5 – 3)

∴ (y – 4)/(6) = (x – 5)/(2)

∴ 2(y – 4) = 6 (x – 5)

∴ 2y – 8 = 6 x – 30

 ∴ 6x – 2y – 30 + 8 =0

∴ 6x – 2y – 22 =0

Dividing throughout by 2

∴ 3x –y – 11 =0

Ans: The equation of the line is 3x –y – 11 =0

• (2, -3) and (-4, 6)

Solution:

Let the points be A(2, -3) ≡ (x₁, y₁) and (-4, 6) ≡ (x₂, y₂)

By two-point form

∴ (y + 3)/(-3 – 6) = (x – 2)/(2 + 4)

∴ (y + 3)/(-9) = (x – 2)/(6)

∴ 6(y + 3) = -9(x – 2)

∴ 6y + 18 =  – 9x + 18

 ∴ 9x + 6y =0

Dividing equation throughout by 3

∴ 3x + 2y =0

Ans: The equation of the line is 3x + 2y =0

• (0, 0) and (6, 4)

Solution:

Let the points be A(0, 0) ≡ (x₁, y₁) and (6, 4) ≡ (x₂, y₂)

By two-point form

∴ (y – 0)/(0 – 4) = (x – 0)/(0 – 6)

∴ (y)/(-4) = (x)/(-6)

∴ – 6y = -4x

∴ 4x – 6y = 0

Dividing equation throughout by 2

∴ 2x – 3y =0

Ans: The equation of the line is 2x – 3y =0

• (3, 4) and (4, 3)

Solution:

Let the points be A(3, 4) ≡ (x₁, y₁) and (4, 3) ≡ (x₂, y₂)

By two-point form

∴ (y – 4)/(4 – 3) = (x – 3)/(3 – 4)

∴ (y – 4)/(1) = (x – 3)/(-1)

∴ y – 4 =-x + 3

∴ x + y -4 -3 = 0

∴ x + y – 7 = 0

Ans: The equation of line is x + y – 7 = 0

• (1, -2) and (2, 3)

Solution:

Let the points be A(1, -2) ≡ (x₁, y₁) and (2, 3) ≡ (x₂, y₂)

By two-point form

∴ (y + 2)/(-2 – 3) = (x – 1)/(1 – 2)

∴ (y + 2)/(-5) = (x – 1)/(-1)

∴ -1(y + 2) = -5(x – 1)

∴ – y – 2 = -5x + 5

∴ 5x – y – 2 – 5 = 0

∴ 5x – y – 7 = 0

Ans: The equation of the line is 5x – y – 7 = 0

Example 02:

Find the equation of a line passing through the points (-4, 6) and (8, -3) and also find the slope and its intercepts on coordinate axes.

Solution:

Let the points be A(-4, 6) ≡ (x₁, y₁) and (8, -3) ≡ (x₂, y₂)

By two-point form

∴ (y – 6)/(6 + 3) = (x + 4)/(-4 – 8)

∴ (y – 6)/(9) = (x + 4)/(-12)

∴ – 12(y – 6) = 9(x + 4)

∴ – 12y + 72 = 9x + 36

∴ 9x + 12y + 36 – 72 = 0

∴ 9x + 12y – 36 = 0

Dividing equation throughout by 3

∴ 3x + 4y – 12 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (3/4) = – 3/4

To find x-intercept of line substitute y = 0 in the equation of line

∴ 3x + 4(0) – 12 = 0

∴ 3x = 12

∴ x = 4

To find y-intercept of line substitute x = 0 in the equation of line

∴ 3(0) + 4 y -12 = 0

∴  4y = 12

∴  y = 3

Ans: The equation of line is 3x + 4y – 12 = 0, Its lope is – 3/4, x-intercept = 4, and y-intercept = 3

Example 03:

Find equation of a line passing through the points (2, 6) and (5, 1) and also find the slope and its intercepts on coordinate axes.

Solution:

Let the points be A(2, 6) ≡ (x₁, y₁) and (5, 1) ≡ (x₂, y₂)

By two-point form

∴ (y – 6)/(6 – 1) = (x – 2)/(2 – 5)

∴ (y – 6)/(5) = (x – 2)/( – 3)

∴ – 3(y – 6) = 5(x – 2)

∴ – 3y + 18 = 5x – 10

∴ 5x + 3y – 10 – 18 = 0

∴ 5x + 3y – 28 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (5/3) = – 5/3

To find x-intercept of line substitute y = 0 in the equation of line

∴ 5x + 3(0) – 28 = 0

∴ 5x = 28

∴ x = 28/5

To find y-intercept of line substitute x = 0 in the equation of line

∴ 5(0) + 3y – 28 = 0

∴  3y = 28

∴  y = 28/3

Ans: The equation of line is 5x + 3y – 28 = 0, Its lope is – 5/3, x-intercept = 28/5, and y-intercept = 28/3.

Example 04:

Find ‘k’ if the point (7, k) lies on a line passing through the points (3, 6) and (-5, -2).

Solution:

Let the points be A(3, 6) ≡ (x₁, y₁) and (-5, -2) ≡ (x₂, y₂)

By two-point form

∴ (y – 6)/(6 + 2) = (x – 3)/(3 + 5)

∴ (y – 6)/(8) = (x – 3)/(8)

∴ y – 6 = x  – 3

∴ x – y – 3 + 6 = 0

∴ x – y + 3 = 0  ………. (1)

Thus the equation of the line is x – y + 3 = 0

Now point (7, k) lies on this line

Substituting x = 7 and y = k in equation (1)

7 – k + 3 = 0

∴ k = 10

Example 05:

Find ‘k’ if the point (2, b) lies on a line passing through the points (5, 4) and (3, -2).

Solution:

Let the points be A(5, 4) ≡ (x₁, y₁) and (3, -2) ≡ (x₂, y₂)

By two-point form

∴ (y – 4)/(4 + 2) = (x – 5)/(5 – 3)

∴ (y – 4)/(6) = (x – 5)/(2)

∴ 2(y – 4) = 6 (x – 5)

∴ 2y – 8 = 6 x – 30

 ∴ 6x – 2y – 30 + 8 =0

∴ 6x – 2y – 22 =0

Dividing throughout by 2

∴ 3x –y – 11 =0 ………. (1)

Thus the equation of the line is 3x –y – 11 = 0

Now point (2, b) lies on this line

Substituting x = 2 and y = b in equation (1)

3(2) – b – 11 =0

6 – b – 11 =0

∴ b = – 5

Science > Mathematics > Coordinate Geometry > Straight Lines > Equation of Line: Two Point Form

The post Equation of Line: Two Point Form appeared first on The Fact Factor.

In this article, we shall study the concept of intercepts of line on coordinate axes.

If a straight line intercepts X-axis at the point A and Y-axis at point B, then the x-coordinate of A and y-coordinate of B are called x-intercept and y-intercept of the line on the axes respectively.

Generally, x-intercept is represented by letter ‘a’ and y-intercept is represented by letter ‘b’.

Example 01:

Find slopes and intercepts of the following lines.

• 3x + 4y = 12

Solution:

Equation of line is 3x + 4y – 12 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (3/4) = – 3/4

To find x-intercept of line substitute y = o in the equation of line

∴  3x + 4 (0) -12 = 0

∴  3x = 12

∴  x = 4

To find y-intercept of line substitute x = 0 in the equation of line

∴  3(0) + 4 y -12 = 0

∴  4y = 12

∴  y = 3

Ans: Slope of line = – 3/4, x-intercept = 4, and y-intercept = 3

• 2x + 3y = 6

Solution:

Equation of line is 2x + 3y – 6 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (2/3) = – 2/3

To find x-intercept of line substitute y = o in the equation of line

∴  2x + 3(0) – 6 = 0

∴  2x = 6

∴  x = 3

To find y-intercept of line substitute x = 0 in the equation of line

∴  2(0) + 3y – 6 = 0

∴  3y = 6

∴  y = 2

Ans: Slope of line = – 2/3, x-intercept = 3, and y-intercept = 2

• Y = 3x – 4

Solution:

Equation of line is 3x –y – 4 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (3/-1) = 3

To find x-intercept of line substitute y = o in the equation of line

∴  3x – (0) – 4 = 0

∴  3x = 4

∴  x = 4/3

To find y-intercept of line substitute x = o in the equation of line

∴  3(0) – y -4 = 0

∴  – y = 12

∴  y = – 4

Ans: Slope of line = 3, x-intercept = 4/3, and y-intercept = – 4

• x/4 – y/3  = 2

Solution:

Equation of line is x/4 – y/3  = 2

Multiplying throughout by 12

∴  3x – 4y = 24

∴  3x – 4y – 24 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (3/-4) = 3/4

To find x-intercept of line substitute y = o in the equation of line

∴  3x – 4(0) – 24 = 0

∴  3x = 24

∴  x = 8

To find the y-intercept of line substitute x = o in the equation of line

∴  3(0) – 4y – 24 = 0

∴  – 4y = 24

∴  y = – 6

Ans: Slope of line = 3/4, x-intercept = 8, and y-intercept = – 6

• x/2 – y/3  = 1/4

Solution:

Equation of line is x/2 – y/3  = 1/4

Multiplying throughout by 12

∴  6x – 4y = 3

∴  6x – 4y – 3 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (6/-4) = 3/2

To find x-intercept of line substitute y = o in the equation of line

∴  6x – 4(0) – 3 = 0

∴  6x = 3

∴  x = 1/2

To find y-intercept of line substitute x = o in the equation of line

∴  6(0) – 4y – 3 = 0

∴  – 4y = 3

∴  y = – 3/4

Ans: Slope of line = 3/2, x-intercept = 1/2, and y-intercept = – 3/4

• 2x/3 + y/4  = 5

Solution:

Equation of line is 2x/3 + y/4 = 5

Multiplying throughout by 12

∴  8x + 3y = 60

∴  8x + 3y – 60 = 0

Slope of line = – (coefficient of x/coefficient of y)

∴  Slope of line = – (8/3) = – 8/3

To find x-intercept of line substitute y = o in the equation of line

∴  8x + 3(0) – 60 = 0

∴  8x = 60

∴  x = 60/8 = 15/2

To find y-intercept of line substitute x = o in the equation of line

∴  8(0) + 3y – 60 = 0

∴  3y = 60

∴  y = 20

Ans: Slope of line = – 8/3, x-intercept = 15/2, and y-intercept = 20

Example – 02:

Find the Slope of a line which makes equal intercepts on the axes.

Solution:

Let line PQ makes an equal intercepts on both the axes.

Thus in right-angled DPOQ, side OP = side OQ

Thus ΔPOQ is right angled isosceles triangle.

Thus m ∠ PQO = 45^o

Thus line PQ makes an angle of 135^o with the positive direction of x-axis

θ = 135^o

Slope of line PQ = m = tan θ = tan 135^o = -1

Thus the slope of a line making equal intercepts on coordinate axes is -1

Example – 03:

Find the Slope of a line which makes equal intercepts but of opposite signs on the axes.

Solution:

Let line PQ makes an equal but opposite in signs intercepts on both the axes.

Thus in right-angled DPOQ, side OP = side OQ

Thus DPOQ is right angled isosceles triangle.

Thus m ∠ PQO = 45^o

Thus line PQ makes an angle of 45^o with the positive direction of x-axis

θ = 45^o

Slope of line PQ = m = tan θ = tan 45^o = 1

Thus the slope of a line making equal but opposite in signs intercepts on coordinate axes is 1

Science > Mathematics > Coordinate Geometry > Straight Lines > Intercepts of Line on the Axes

The post Intercepts of a Line on the Axes appeared first on The Fact Factor.

In this article, we shall study to find slopes of altitude, median, and perpendicular bisectors of sides of triangle and also the mrthod to prove that given triagle is a right angled triangle using slopes of its sides.

Example – 01:

A(2, 3), B(-2, 1) and C(4, -3) are the vertices of ΔABC. Find the slope of (i) side AB (ii) altitude through A (iii) median through A and (iv) perpendicular bisector of AB.

Solution:

Slope of side AB:

Slope of side AB = (1 – 3)/(-2 – 2) = (-2)/(-4) = 1/2

Slope of Altitude Through A:

Slope of BC = (-3 -1)/(4 + 2) = (-4)/(6) = – 2/3

Now AD ⊥ side BC

∴  Slope of AD = 3/2

Median Through A:

M is midpoint of side BC, By midpoint formula

M ≡ ((-2 + 4)/2, (1 -3)/2) = M(2/2, -2/2) = M(1, -1)

End points of median are A(2, 3) and M (1, -1)

Slope of median through A = (-1 – 3)/(1 – 2) = (-4)/(-1) = 4

Perpenicular bisector of AB:

Slope of side AB = (1 – 3)/(-2 – 2) = (-2)/(-4) = 1/2

line l is perpendicular bisector of side AB

line l ⊥ side AB

Slope of line l = – 2

Example – 02:

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Solution:

Let A(4, 4), B(3, 5), and C(-1, -1) be the given points

Slope of AB = (5 – 4)/(3 – 4) = (1)/(-1) = – 1  ………….. (1)

Slope of BC = (-1 – 5)/(-1 – 3) = (-6)/(-4) = 3/2  ………….. (2)

Slope of AC = (-1 – 4)/(-1 – 4) = (-5)/(-5) = 1  ………….. (3)

From equations (1) and (3) we have

Slope of AB x Slope of AC = (-1) x (1) = -1

Thus AB ⊥  AC

Hence ΔABC is right angled at A

∴   The points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Example – 03:

Without using the Pythagoras theorem, show that the points (7, 10), (-2, 5) and (3, –4) are the vertices of a right-angled triangle.

Solution:

Let A(7, 10), B(-2, 5), and C(3, -4) be the given points

Slope of AB = (5 – 10)/(-2 – 7) = (-5)/(-9) = 5/9  ………….. (1)

Slope of BC = (-4 – 5)/(3 + 2) = (-9)/(5) = – 9/5  ………….. (2)

Slope of AC = (-4 – 10)/(3 – 7) = (-14)/(-4) = 7/2  ………….. (3)

From equations (1) and (2) we have

Slope of AB x Slope of BC = (5/9) x (- 9/5) = -1

Thus AB ⊥  BC

Hence ΔABC is right angled at B

∴   The points (7, 10), (-2, 5), and (3, -4) are the vertices of a right-angled triangle.

Example – 04:

Show that the points (6, 1), (-1, 8), and (3, -2) form a right angled triangle using slope method.

Solution:

Let A(6, 1), B(-1, 8), and C(3, -2) be the given points

Slope of AB = (8 – 1)/(-1 – 6) = 7/(-7) = – 1  ………….. (1)

Slope of BC = (-2 – 8)/(3 + 1) = – 10/4 = – 5/2  ………….. (2)

Slope of AC = (- 2 – 1)/(3 – 6) = (-3)/(-3) = 1   ………….. (3)

From equations (1) and (3) we have

Slope of AB x Slope of AC = (-1) x (1) = -1

Thus AB ⊥ AC

Hence ΔABC is right angled at A

∴   The points (6, 1), (-1, 8), and (3, -2) form a right angled triangle.

Example – 05:

Show that the points (-1, -4), (4, 6), and (-4, 10) form a right angled triangle using slope method.

Solution:

Let A(-1, -4), B(4, 6), and C(-4, 10) be the given points

Slope of AB = (6 + 4)/(4 + 1) = 10/5 = 2  ………….. (1)

Slope of BC = (10 – 6)/(-4 – 4) = 4/(-8) = – 1/2  ………….. (2)

Slope of AC = (10 + 4)/(- 4 + 1) = (14)/(-3) = – 14/3   ………….. (3)

From equations (1) and (2) we have

Slope of AB x Slope of BC = (2) x (- 1/2) = -1

Thus AB ⊥ BC

Hence ΔABC is right angled at B

∴   The points (-1, -4), (4, 6), and (-4, 10) form a right angled triangle.

Example – 06:

Using the concept of slopes prove that the points (1, 3), (3, -1) and (-5, -5) are the vertices of a right angled triangle.

Solution:

Let A(1, 3), B(3, -1) and C(-5, -5) be the given points

Slope of AB = (-1 – 3)/(3 – 1) = -4/2 = – 2  ………….. (1)

Slope of BC = (-5 + 1)/(-5 – 3) = – 4/(-8) = 1/2  ………….. (2)

Slope of AC = (-5 – 3)/(- 5 – 1) = (-8)/(-6) = 4/3   ………….. (3)

From equations (1) and (2) we have

Slope of AB x Slope of BC = (2) x (- 1/2) = -1

Thus AB ⊥ BC

Hence ΔABC is right angled triangle right angled at B

∴  The points (1, 3), (3, -1) and (-5, -5) are the vertices of a right-angled triangle.

Science > Mathematics > Coordinate Geometry > Straight Lines > Slope Problems Related With Triangles

The post Slope Problems Related With Triangles appeared first on The Fact Factor.

In this article, we shall study the Condition for parallelism and the condition of perpendicularity of lines.

Condition for Parallelism:

If two lines are parallel, then their slopes are equal. Let slopes of the two parallel lines be m₁ and m₂, then by the condition of parallelism,

m₁ = m₂.

Condition for Perpendicularity:

If two lines are perpendicular, then the product of their slopes is – 1. Let slopes of the two perpendicular lines be m₁ and m₂, then by the condition of parallelism,

m₁ x m₂. = – 1

Example – 01:

Find the value of k so that the line joining (3, k) and (2, 7) is parallel to the line through (-1, 4) and (0, 6).

Solution:

Let A(3, k), B(2, 7), and C(-1, 4), D(0, 6) be the given points

Given line AB ∥ line CD

∴  Slope of line AB = Slope of line CD

∴   (7 – k)/(2 – 3) = (6 – 4)/(0 + 1)

∴   (7 – k)/(-1) = 2/1

∴   (7 – k) = 2

∴   k = 7 + 2

∴   k = 9

Example – 02:

Find the value of k so that the line joining (2, k) and (-3, 5) is parallel to the line through (-1, 5) and (0, 7).

Solution:

Let A(2, k), B(-3, 5), and C(-1, 5), D(0, 7) be the given points

Given line AB ∥ line CD

∴ Slope of line AB = Slope of line CD

∴   (5 – k)/(-3 – 2) = (7 – 5)/(0 + 1)

∴   (5 – k)/(-5) = 2/1

∴   (5 – k) = 2(-5)

∴   5 – k = – 10

∴    k = 5 + 10

∴   k = 15

Example – 03:

Find the value of k so that seg AB is parallel to Seg CD where A(-3, 1), B(4, 3), C(k, 3), and D(2, 5).

Solution:

Given points are A(-3, 1), B(4, 3), C(k, 3), and D(2, 5)

Given seg AB ∥ seg CD

∴ Slope of seg AB = Slope of seg CD

∴   (3 – 1)/(4 + 3)) = (5 – 3)/(2 – k)

∴   2/7 = 2/(2 – k)

∴   1/7 = 1/(2 – k)

∴   2 – k = 7

∴    k = 2 – 7

∴    k = – 5

Example – 04:

A(0, 9), B(1, 11), C(3, 13) and D(7, k) are the four points if AB ⊥ CD then find k

Solution:

Given points are A(0, 9), B(1, 11), and C(3, 13), D(7, k)

Slope of AB = (11 – 9)/(1 – 0) = 2/1 = 2

Slope of CD = (k – 13)/(7 – 3) = (k – 13)/4

Given line AB ⊥ line CD

∴  Slope of line AB x Slope of line CD = -1

∴   2  x  ((k – 13)/4) = – 1

∴   (k – 13)/2 = – 1

∴   k – 13 = – 2

∴   k  = – 2 + 13

∴   k = 11

Example – 05:

A(0, 9), B(1, 5), C(-2, 7) and D(4, k) are the four points if AB ⊥ CD then find k.

Solution:

Given points are A(0, 9), B(1, 5), C(-2, 7) and D(4, k)

Slope of AB = (5 – 9)/(1 – 0) = – 4/1 = – 4

Slope of CD = (k – 7)/(4 + 2) = (k – 7)/6

Given line AB ⊥ line CD

∴  Slope of line AB x Slope of line CD = -1

∴   – 4  x  ((k – 7)/6) = – 1

∴   – 4(k – 7) = – 6

∴   – 4k + 28 = – 6

∴   – 4k  = – 34

∴   k  = 34/4

∴   k = 17/2

Example – 06:

A(3, 2), B(-2, 7), C(-5, -4) and D(-2, k) are the four points if AB ⊥ CD then find k

Solution:

Given points are A(3, 2), B(-2, 7), C(-5, -4) and D(-2, k)

Slope of AB = (7 – 2)/(-2 – 3) = 5/(-5) = – 1

Slope of CD = (k + 4)/(-2 + 5) = (k + 4)/3

Given line AB ⊥ line CD

∴  Slope of line AB x Slope of line CD = -1

∴   – 1  x  ((k + 4)/3) = – 1

∴   k + 4 = 3

∴   k = – 1

Example – 07:

The line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.

Solution:

Let A(-2, 6), B(4, 8), and C(8, 12), D(x , 24) be the given points

Slope of AB = (8 – 6)/(4 + 2) = 2/6 = 1/3

Slope of CD = (24 – 12)/(x – 8) = 12/(x – 8)

Given line AB ⊥ line CD

∴  Slope of line AB x Slope of line CD = -1

∴   (1/3  x  (12/(x – 8)) = – 1

∴   12/(x – 8)= – 3

∴   12 = – 3x + 24∴  3x  = 12

∴   x = 4

Example – 08:

If A(6, 4) and B(2, 12) are two given points then find the slope of the line (i) parallel to AB and (ii) perpendicular to AB

Solution:

Given A(6, 4) and B(2, 12)

Slope of AB = (12 – 4)/(2 – 6) = (8)/(- 4) = – 2

Let l be the line parallel to line AB

As lines are parallel their slopes are equal

∴  Slope of line l = slope of line AB = – 2

Let m be the line perpendicular to line AB

As lines are perpendicular to each other, product of their slopes is -1

∴  Slope of line m =  – 1/slope of line AB = – (1/- 2) = 1/2

The slope of the line parallel to AB is – 2 and that of the line perpendicular to AB is 1/2

Example – 09:

2x + 3y + 7 = 0 and 4x + 6y + 2 = 0are two straight lines. Are they parallel to each other?

Solution:

Equation of first line is 2x + 3y + 7 = 0

Slope of first line = – (coefficient of x / Coefficient of y)

Slope of first line = – (2/3) = -2/3  ………. (1)

Equation of second line is 4x + 6y + 2 = 0

Slope of second line = – (coefficient of x / Coefficient of y)

Slope of second line = – (4/6) = -2/3  ………. (2)

From equation (1) and (2) we have

Slope of first line = Slope of second line

Hence the two lines are parallel to each other.

Example – 10:

Show that the lines 3x – 2y + 6 = 0 and 2x + 3y – 1 = 0are perpendicular to each other?

Solution:

Equation of first line is 3x – 2y + 6 = 0

Slope of first line = – (coefficient of x / Coefficient of y)

Slope of first line = – (3/-2) = 3/2  ………. (1)

Equation of second line is 2x + 3y – 1 =  0

Slope of second line = – (coefficient of x / Coefficient of y)

Slope of second line = – (2/3) = -2/3  ………. (2)

Slope of first line x Slope of second line = (3/2) x (-2/3) = -1

Hence the two lines are perpendicular to each other.

Example – 11:

Show that the lines 5x + 6y -1 = 0 and 6x – 5y + 3 = 0are perpendicular to each other?

Solution:

Equation of first line is 5x + 6y -1 = 0

Slope of first line = – (coefficient of x / Coefficient of y)

Slope of first line = – (5/6) = – 5/6  ………. (1)

Equation of second line is 6x – 5y + 3 = 0

Slope of second line = – (coefficient of x / Coefficient of y)

Slope of second line = – (6/-5) = 6/5  ………. (2)

Slope of first line x Slope of second line = (- 5/6) x (6/5) = -1

Hence the two lines are perpendicular to each other.

Example – 12:

Find value of ‘k’, if the lines kx – 6y = 9 and 6x + 5y = 13 are perpendicular to each other.

Solution:

Equation of first line is kx – 6y – 9 = 0

Slope of first line = – (coefficient of x / Coefficient of y)

Slope of first line = – (k/-6) = k/6   ………. (1)

Equation of second line is 6x + 5y – 13 = 0

Slope of second line = – (coefficient of x / Coefficient of y)

Slope of second line = – (6/5) = – 6/5  ………. (2)

Given the two lines are perpendicular to each other

Slope of first line x Slope of second line = -1

(k/6) x (-6/5) = -1

k/5 = 1

k = 5

Example – 13:

Find value of ‘k’, if the lines 3x + 4py + 8 = 0 and 3py – 9x + 10 = 0 are perpendicular to each other.

Solution:

Equation of first line is 3x + 4py + 8 = 0

Slope of first line = – (coefficient of x / Coefficient of y)

Slope of first line = – (3/4p) ………. (1)

Equation of second line is – 9x + 3py + 10 = 0

Slope of second line = – (coefficient of x / Coefficient of y)

Slope of second line = – (-9/3p) = 9/3p  ………. (2)

Given the two lines are perpendicular to each other

Slope of first line x Slope of second line = -1

(- 3/4p) x (9/3p) = -1

9/4p² = 1

p² = 9/4

p = ± 3/2

Science > Mathematics > Coordinate Geometry > Straight Lines > Condition for Parallelism and Perpendicularity

The post Conditions for Parallelism and Perpendicularity of Lines appeared first on The Fact Factor.

In this article, we shall study to find the equation of a straight line using slope point form.

Example – 01:

Find the equation of following lines

• passing through (3, -2) with slope -2

Line passes through point (3, -2) = (x₁, y₁)

Slope of line = m = – 2

By slope point form

y – y₁ = m(x – x₁)

y + 2 = – 2(x – 3)

y + 2 = – 2x + 6

2x + y – 4 = 0

Ans: The equation of the line is 2x + y – 4 = 0

Passing through (1, 2) with slope – 3/2

Line passes through point (1, 2) = (x₁, y₁)

Slope of line = m = -3/2

By slope point form

y – y₁ = m(x – x₁)

y – 2 = -3/2(x – 1)

2y – 4 = -3x + 3

3x + 2y – 7 = 0

Ans: The equation of the line is 3x + 2y – 8 = 0

• Passing through (6, 2) with slope – 3

Line passes through point (6, 2) = (x₁, y₁)

Slope of line = m = -3

By slope point form

y – y₁ = m(x – x₁)

y – 2 = -3 (x – 6)

y – 2 = -3x + 18

3x + y – 20 = 0

Ans: The equation of the line is 3x + y – 20 = 0

• Passing through the point (– 4, 3) with slope 1/2.

Line passes through point (-4, 3) = (x₁, y₁)

Slope of line = m = 1/2

By slope point form

y – y₁ = m(x – x₁)

y – 3 = 1/2(x + 4)

2y – 6 = x + 4

x – 2y + 10 = 0

Ans: The equation of the line is x – 2y + 10 = 0

• Passing through (0, 0) with slope m.

Line passes through point (0, 0) = (x₁, y₁)

Slope of line = m

By slope point form

y – y₁ = m(x – x₁)

y – 0 = m(x – 0)

y = mx

Ans: The equation of the line is y = mx

• Passing through (-4, – 3) and parallel to the x-axis

Line passes through point (-4, -3) = (x₁, y₁)

Line is parallel to the x-axis

Slope of line = m = 0

By slope point form

y – y₁ = m(x – x₁)

y + 3 = 0(x + 4)

y + 3 = 0

Ans: The equation of the line is y + 3 = 0

• Having x-intercept 4 and tan θ = 1/2, where θ is inclination of line.

x- intercept of line is 4

Line passes through point (4, 0) = (x₁, y₁)

Slope of line = m = tan θ = 1/2

By slope point form

y – y₁ = m(x – x₁)

y  – 0 = 1/2(x – 4)

2y  = x – 4

x – 2y – 4 = 0

Ans: The equation of the line is x – 2y – 4 = 0

• Having y – intercept 2 and slope – 2

y- intercept of line is 2

Line passes through point (0, 2) = (x₁, y₁)

Slope of line = m = – 2

By slope point form

y – y₁ = m(x – x₁)

y  – 2 = -2(x – 0)

y  – 2 = -2x

2x + y – 2 = 0

Ans: The equation of the line is 2x + y – 2 = 0

• Having y – intercept 5 and slope – 2

y- intercept of line is 5

Line passes through point (0, 5) = (x₁, y₁)

Slope of line = m = – 2

By slope point form

y – y₁ = m(x – x₁)

y  – 5 = -2(x – 0)

y  – 5 = -2x

2x + y – 5 = 0

Ans: The equation of the line is 2x + y – 5 = 0

• Having y – intercept -4 and slope – 1/3

y- intercept of line is -4

Line passes through point (0, -4) = (x₁, y₁)

Slope of line = m = – 2

By slope point form

y – y₁ = m(x – x₁)

y  + 4 = -1/3(x – 0)

3y  + 12  = – x

x + 3y + 12 = 0

Ans: The equation of the line is x + 3y + 12 = 0

• Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.

x- intercept of line is – 3

Line passes through point (-3, 0) = (x₁, y₁)

Slope of line = m = – 2

By slope point form

y – y₁ = m(x – x₁)

y  – 0 = -2(x + 3)

y  = -2x – 6

2x + y + 6 = 0

Ans: The equation of the line is 2x + y + 6 = 0

• Intersecting the x-axis at a distance of 4 units to the left of origin with slope –3.

x- intercept of line is – 4

Line passes through point (-4, 0) = (x₁, y₁)

Slope of line = m = – 3

By slope point form

y – y₁ = m(x – x₁)

y  – 0 = -3(x + 4)

y  = -3x – 12

3x + y + 12 = 0

Ans: The equation of the line is 3x + y + 12 = 0

• Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with the positive direction of the x-axis

y- intercept of line is + 2

Line passes through point (0, 2) = (x₁, y₁)

Slope of line = m = tan θ = tan 30^o = 1/√3

By slope point form

y – y₁ = m(x – x₁)

y  – 2 = 1/√3(x – 0)

√3y  -2√3 = x

x – √3y + 2√3 = 0

Ans: The equation of the line is x – √3y + 2√3 = 0

• It cuts intercept of length 3 on y axis and having slope 5.

y- intercept of line is 3

Line passes through point (0, 3) = (x₁, y₁)

Slope of line = m = 5

By slope point form

y – y₁ = m(x – x₁)

y  – 3 = 5(x – 0)

5x – y + 3 = 0

Ans: The equation of the line is 5x – y + 3 = 0

• Passing through (2, -3) and making the inclination of 135^o.

Line passes through point (2, -3) = (x₁, y₁)

Slope of line = m = tan θ = tan 135^o = – 1

By slope point form

y – y₁ = m(x – x₁)

y + 3 = -1(x – 2)

y + 3 = – x + 2

x + y + 1 = 0

Ans: The equation of the line is x + y + 1 = 0

• Passing through (3, -1) and making inthe clination of 60^o.

Line passes through point (3, -1) = (x₁, y₁)

Slope of line = m = tan θ = tan 60^o = √3

By slope point form

y – y₁ = m(x – x₁)

y + 1 = √3(x – 3)

y + 1 =  √3 x + 3 √3

√3 x – y – 1 + 3√3 = 0

√3 x – y – (3√3 – 1) = 0

Ans: The equation of the line is √3 x – y – (3√3 – 1) = 0

• Passing through (-5, 6) and making the inclination of 150^o.

Line passes through point (-5, 6) = (x₁, y₁)

Slope of line = m = tan θ = tan 150^o = – 1/√3

By slope point form

y – y₁ = m(x – x₁)

y – 6 = -1/√3(x + 5)

√3y – 6√3 =  – x – 5

x + √3y – 6 √3  + 5 = 0

x + √3y + (5 – 6√3) = 0

Ans: The equation of the line is  x + √3y + (5 – 6 √3) = 0

• Passing through (-2, 3) and making the inclination of 45^o.

Line passes through point (-2, 3) = (x₁, y₁)

Slope of line = m = tan θ = tan 45^o = 1

By slope point form

y – y₁ = m(x – x₁)

y – 3 = 1(x + 2)

y – 3 = x + 2

x  – y + 5 = 0

Ans: The equation of the line is root x  – y + 5 = 0

• Passing through (2, 2√3)and inclined with the x-axis at an angle of 75°.

Line passes through point (2, 2√3) = (x₁, y₁)

Slope of line = m = tan θ = tan 75^o = 2 + √3

By slope point form

y – y₁ = m(x – x₁)

y – 2√3 = (2 + √3)(x – 2)

y – 2√3 = (2 + √3)x – 4 – 2√3

(2 + √3)x – y – 4 = 0

Ans: The equation of the line is root (2 + √3)x – y – 4 = 0

Science > Mathematics > Coordinate Geometry > Straight Lines > Slope Point Form

The post Equation of Line: Slope Point Form appeared first on The Fact Factor.

In this article, we shall study to find the slope of line AB, when points A(x₁, y₁) and B(x₂, y₂) are given.

Type – I: To Find Slopeof a Line When Two Points on the Line are Given:

Example – 01:

Find the slopes of the lines which pass through the points

• (2, 5) and (-4, -4)

Let the points be A(2, 5) ≡ (x₁, y₁) and (-4, -4) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (- 4 – 5)/(- 4 – 2) = (-9)/(-6)

∴  Slope of line AB = 3/2

• (-2, 3) and (4, -6)

Let the points be A(-2, 3) ≡ (x₁, y₁) and (4, -6) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (- 6 – 3)/(4  + 2) = (-9)/(6)

∴  Slope of line AB = – 3/2

• (1, -3) and (-1, -1)

Let the points be A(1, -3) ≡ (x₁, y₁) and (-1, -1) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (- 1 + 3)/(- 1  – 1) = (2)/(-2)

∴  Slope of line AB = – 1

• (2, √3) and (3, 2√3)

Let the points be A(2, √3) ≡ (x₁, y₁) and (3, 2√3) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (2√3 – √3)/(3  – 2) = (√3)/(1)

∴  Slope of line AB = √3

• (3, 4) and (2, 5)

Let the points be A(3, 4) ≡ (x₁, y₁) and (2, 5) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (5 – 4)/(2 – 3) = (1)/(-1)

∴  Slope of line AB = – 1

• (2, 5) and (4, -6)

Let the points be A(2, 5) ≡ (x₁, y₁) and (4, -6) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (- 6 – 5)/(4  – 2) = (- 11)/(2)

∴  Slope of line AB = – 11/2

• (-1, 4) and (2, 2)

Let the points be A(-1, 4) ≡ (x₁, y₁) and (2, 2) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (2 – 4)/(2 + 1) = (-2)/(3)

∴  Slope of line AB = – 2/3

• (a, 0) and (0, b)

Let the points be A(a, 0) ≡ (x₁, y₁) and (0, b) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (b – 0)/(0 – a) = (b)/(-a)

∴  Slope of line AB = – b/a

• (at₁², 2at₁) and (at₂², 2at₂)

Let the points be A(at₁², 2at₁) ≡ (x₁, y₁) and (at₂², 2at₂) ≡ (x₂, y₂)

∴  Slope of line AB = (y₂ – y₁)/ (x₂ – x₁) = (2at₂ – 2at₁)/(at₂² – at₁²)

∴  Slope of line AB = 2a(t₂ – t₁)/a(t₂² – t₁²)

∴  Slope of line AB = 2(t₂ – t₁)/(t₂ + t₁)(t₂ – t₁)

∴  Slope of line AB = 2)/(t₂ + t₁)

• The origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Midpoint of segment AB is M( (0 + 8)/2, (-4, 0)/2) = M(8/2, -4/2) = M(4, – 2)

Now slope of OM = (-2 – 0)/ (4 – 0) = -2/4 = – 1/2

Type – II: Find Value of Arbitrary Constant, When Slope is Given:

Example – 02:

Find the value of ‘k’ if the slope of the line passing through the points

• (2, 4), (5, k) is 5/3

Let the points be A(2, 4) and B(5, k)

Slope of AB = (k – 4)/(5 – 2) = 5/3

∴  (k – 4)/3 = 5/3

∴  k – 4 = 5

∴  k = 9

• (2, 5), (k, 3) is 2

Let the points be A(2, 5) and B(k, 3)

Slope of AB = (3 – 5)/(k – 2) = 2

∴  – 2  = 2k – 4

∴  2k = 2

∴  k = 1

• (k, 2), (-6, 8) is – 5/4

Let the points be A(k, 2) and B(-6, 8)

Slope of AB = (8 – 2)/(-6 – k) = -5/4

∴  6/(-6 – k) = -5/4

∴ 24 = 30 + 5k

∴ 5k = – 6

∴  k = – 6/5

Type – III: To Show Given Points are Collinear:

Example – 03:

By using slopes show that the following points are collinear.

• A(0,4), B(2,10) and C(3,13)

Given points are A(0,4), B(2,10) and C(3,13)

Slope of line AB = (10 – 4)/(2 – 0) = (6)/(2) = 3 …………. (1)

Slope of line BC = (13 – 10)/(3 – 2) = (3)/(1) = 3 …………. (2)

From equations (1) and (2)

Slope of line AB = Slope of line BC

B is common point

Hence points A, B and C are collinear

• P(5, 0), Q(10, -3) and R(-5, 6)

Given points are P(5, 0), Q(10, -3) and R(-5, 6)

Slope of line PQ = (-3 – 0)/(10 – 5) = (-3)/(5) = – 3/5 …………. (1)

Slope of line QR = (6 + 3)/(-5 – 10) = (9)/(- 15) = – 3/5 …………. (2)

From equations (1) and (2)

Slope of line PQ = Slope of line QR

Q is common point

Hence points P, Q and R are collinear

• L(2, 5), M(5, 7) and N(8, 9)

Given points are L(2, 5), M(5, 7) and N(8, 9)

Slope of line LM = (7 – 5)/(5 – 2) = (2)/(3) = 2/3 …………. (1)

Slope of line MN = (9 – 7)/(8 – 5) = (2)/(3) = 2/3 …………. (2)

From equations (1) and (2)

Slope of line LM = Slope of line MN

M is common point

Hence points L, M, and N are collinear

• D(5, 1), E(1, -1) and F(11, 4)

Given points are D(5, 1), E(1, -1) and F(11, 4)

Slope of line DE = (-1 – 1)/(1 – 5) = (-2)/(-4) = 1/2 …………. (1)

Slope of line EF = (4 + 1)/(11 – 1) = (5)/(10) = 1/2 …………. (2)

From equations (1) and (2)

Slope of line DE = Slope of line EF

E is common point

Hence points D, E, and F are collinear

Type IV: To Find Value of K if Given Points are Collinear:

Example – 04:

If the following points are collinear find value of k using slopes.

• (-2, -3), (k, 4) and (5, 5)

Let given points be A(-2, -3), B(k, 4) and C(5, 5)

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (4 + 3)/(k + 2) = (5 – 4)/(5 – k)

∴  7/(k + 2) = 1/(5 – k)

∴  7(5 – k) = 1(k + 2)

∴  35 – 7k = k + 2

∴  33 = 8k

∴  k = 33/8

• (5, k), (-3, 1) and (-7, -2)

Let given points be A(5, k), B(-3, 1) and C(-7, -2)

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (1 – k)/(-3 – 5) = (-2 – 1)/(-7 + 3)

∴  (1 – k)/(-8) = (-3)/(-4)

∴  (1 – k)/(-8) = (3/4)  x (-8)

∴  1 – k = – 6

∴  k = 7

• (2, 1), (4, 3) and (0, k)

Let given points be A(2, 1), B(4, 3) and C(0, k)

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (3 – 1)/(4 – 2) = (k – 3)/(0 – 4)

∴  2/2 = (k – 3)/(- 4)

∴ 1 (-4) = (k – 3)

∴  k – 3  = – 4

∴  k = – 1

• (-4, 5), (-3, 5) and (-1, k)

Let given points be A(-4, 5), B(-3, 5) and C(-1, k)

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (5 – 5)/(-3 + 4) = (k – 5)/(-1 + 3)∴  0/1 = (k – 5)/2

∴ 0 = (k – 5)/2

∴  k – 5  = 0

∴  k = 5

• (k, 1), (2, -3) and (3, 4)

Let given points be A(k, 1), B(2, -3) and C(3, 4)

Points A, B, and C are collinear

∴  Slope of line AB = Slope of line BC

∴  (-3 – 1)/(2 – k) = (4 + 3)/(3 – 2)

∴  (-4)/(2 – k) = 7/1

∴ -4 = 14 – 7k

∴  7k = 18

∴  k = 18/7

Type V: To Prove Given Condition When Collinearity of Points is Given:

Example – 05:

If three points (h, 0), (a, b) and (0, k) lie on a line, show that a/h + b/k = 1

Solution:

Let A(h, 0), B(a, b) and C(0, k) be the points lying on the line.

∴ Slope of AB = Slope of BC

∴ (b – 0)/(a – h) = (k – b)/(0 – a)

∴ b/(a – h) = (k – b)/(-a)

∴ – ab   = (a – h)(k – b)

∴  ak – kh + hb = 0

∴  ak + hb = kh

Dividing both sides of equation by kh

∴  ak/kh + hb/kh = kh/kh

∴  a/h + b/k = 1  (proved as required)

Example – 08:

A line passes through (x1, y1) and (h, k). If the slope of the line is m, show that k – y₁ = m (h – x₁).

Line passes through (x₁, y₁) and (h, k)

Slope of line = (k – y₁)/(h – x₁) = m

∴  k – y₁ = m (h – x₁)  (Proved as required)

Example – 09:

Three points P (h, k), Q (x₁, y₁) and R (x₂, y₂) lie on a line. Show that (h – x₁) (y₂ – y₁) = (k – y₁) (x₂ – x₁).

As the points lie on a line

Slope of line PQ = Slope of line QR

(k – y₁)/(h – x₁) = (y₂– y₁)/(x₂ – x₁)

∴ (k – y₁) (x₂ – x₁)= (h – x₁) (y₂ – y₁)

∴  (h – x₁) (y₂ – y₁) = (k – y₁) (x₂ – x₁)    (Proved)

Type – VI: Interpretation of Graph:

Example – 10:

Consider the following population and year graph, find the slope of the line AB, and using it, find what will be the population in the year 2010?

Solution:

Slope of line AB = (97 – 92)/(1995 -1985) = 5/10 = 1/2

Let the populatiomn in year 2010 be k

Thus point P(2010, k) lies on the line

Slope of AP = Slope of AB

Slope of AP = (k – 92)/(2010 – 1985) = 1/2

∴ (k – 92)/25 = 1/2

∴ 2k – 184 = 25

∴ 2k = 209

∴ k = 104.5

Hence population in 2010 will be 104.5 crore

Example – 11:

In the figure, the time and distance graph of linear motion is given. Two positions of time and distance are recorded as when T = 0, D = 2 and when T = 3, D = 8. Using the concept of slope, find the law of motion, i.e., how distance depends upon time.

Slope of AB = Slope of BC

∴ (8 – 2)/(3 – 0) = (D – 8)(T – 3)

∴ 6/3 = (D – 8)(T – 3)

∴ 2 = (D – 8)(T – 3)

∴ 2(T – 3) = D – 8

∴ 2T – 6 = D – 8

∴ D = 2T – 6 + 8

∴ D = 2T + 2

This relation is known as the law of motion.

Science > Mathematics > Coordinate Geometry > Straight Lines > The slope of a Line

The post The slope of a Line appeared first on The Fact Factor.

In this article, we shall study the concept of inclination of a line and the slope of the line

The inclination of a Line:

A line in a coordinate plane forms two angles with the x-axis, which are supplementary.  The angle (say) θ made by the line l with the positive direction of the x-axis and measured anti-clockwise is called the inclination of the line. Thus 0° ≤ θ ≤ 180°.

• The lines parallel to the x-axis, or the coinciding with the x-axis, have the inclination of 0°.
• The inclination of a vertical line (parallel to or coinciding with y-axis) is 90°.

The slope of a Line:

If θ is the inclination of a line l, then tan θ, (θ ≠ 90°) is called the slope or gradient of the line l. Note that the slope of a line whose inclination is 90° is not defined. The slope of a line is denoted by the letter, ‘m’. Thus by definition m = tan θ, θ ≠ 90°

• The slope of the x-axis is zero and the slope of the y-axis is not defined

Equation for Slope of a Line:

Let P(x₁, y₁) and Q(x₂, y₂) be two points on non-vertical line l whose inclination is θ. As the line is vertical, x₁ ≠ x₂, The inclination of the line l may be acute or obtuse. Let us consider both of these cases.

Case – 1: When the inclination is acute

Draw perpendicular QR to the x-axis and PM perpendicular to RQ as shown

MQ = y₂ – y₁ and MP = x₂ – x₁∠MPQ = θ ………… (1)

Therefore, the slope of line l = m = tan θ.

But in ∆MPQ, we have

∴ Slope of line l = (y₂ – y₁)/(x₂ – x₁)

Case – 2: When the inclination is obtuse

Draw perpendicular QR to the x-axis and PM perpendicular to RQ as shown

MQ = y₂ – y₁ and MP = x₁ – x₂∠MPQ =  π – θ

Therefore, the slope of line l = m = tan θ = – tan (π – θ).

But in ∆MPQ, we have

∴ Slope of line l = (y₂ – y₁)/(x₂ – x₁)

Thus, in either case, the slope of the line is m = (y₂ – y₁)/(x₂ – x₁)

Sign of Slope:

The slope can be positive, zero or negative

• Positive slope: This means the angle of inclination θ is such that 0° < θ < 90°. i.e. the angle of inclination is acute.
• Zero slope: This means the line is parallel to the x-axis.
• Negative slope: This means the angle of inclination θ is such that 90° < θ < 180°. i.e. the angle of inclination is obtuse.
• If the line is perpendicular to the x-axis its slope is not defined.

Conditions for parallelism of lines in terms of their slopes

In a coordinate plane, suppose that non-vertical lines l₁ and l₂ have slopes m₁ and m₂, respectively. Let their inclinations be α and β, respectively.

If the line l₁ is parallel to l₂ (Fig 10.4), then their inclinations are equal,

i.e., α = β, and hence, tan α = tan β

Therefore m₁ = m₂, i.e., their slopes are equal.

Conversely, if the slope of the two lines l₁ and l₂ are the same, i.e., m₁ = m₂.
By the property of tangent function (between 0° and 180°), α = β. Therefore, the lines are parallel. Hence, two non-vertical lines l₁ and l₂ are parallel if and only if their slopes are equal

Conditions for perpendicularity  of lines in terms of their slopes:

If the lines l₁ and l₂ are perpendicular,  then by exterior angle property, 

β = α + 90°.

∴  tan β = tan (α + 90°)

∴  tan β = – cotα

∴  tan β = –  1/ tan α

∴  tanα . tan β = –  1

∴  m₁ . m₂ = – 1

Conversely, if m₁ m₂ = – 1, i.e., tan α tan β = – 1.

Then tanα = – cot β = tan (β + 90°) or tan (β – 90°)

Therefore,α and β differ by 90°.

Thus, lines l₁ and l₂ are perpendicular to each other.

Hence, two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other, or the product of their slopes is – 1.

Slope of Line from Inclination of a Line:

Tangent Functions of Some Angles:

Example – 01:

Find the slope of lines whose inclinations are

• 45°

Given θ = 45°

∴  The slope of a line = m = tan θ = tan 45° = 1

• 60°

Given θ = 60°

∴  The slope of a line = m = tan θ = tan 60° = √3

• 30°

Given θ = 30°

∴  The slope of a line = m = tan θ = tan 30° = 1/√3

• 120°

Given θ = 120°

∴  The slope of a line = m = tan θ = tan 120°

∴  m = tan (90° + 30°) = – cot 30° = – √3

• (3π/4)^c

Given θ = (3π/4)^c

∴  The slope of a line = m = tan θ = tan (3π/4)^c

∴  m = tan (π/2 + π/4)^c = – tan (π/4)^c = – 1/√2

• (5π/6)^c

Given θ = (5π/6)^c

∴  The slope of a line = m = tan θ = tan (5π/6)^c

∴  m = tan (π – π/6)^c = – tan (π/6)^c = – 1/√3

• 105°

Given θ = 105°

∴  The slope of a line = m = tan θ

= tan 105° = tan (60° + 45°)

• The line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise.

The angle made by line with the positive direction of the y-axis = – 30° (anticlockwise)

The angle made by line with positive direction of the x-axis = 90° – 30° = 60°

Now θ = 60°

∴  The slope of a line = m = tan θ = tan 60° = √3

Example – 02:

Find the inclinations of lines whose slopes are

• 1

The slope of line = m = tan θ =1

∴  θ = tan^-1(1) = 45° or (π/4)^c

• – 1

The slope of line = m = tan θ = -1

∴  θ = tan^-1(-1) = 135° or (3π/4)^c

• √3

The slope of line = m = tan θ =√3

∴  θ = tan^-1(√3) = 60° or (π/3)^c

•  – √3

The slope of line = m = tan θ = -√3

∴  θ = tan^-1(- √3) = 120° or (2π/3)^c

• 1/√3

The slope of line = m = tan θ =1/√3

∴  θ = tan^-1(1/√3) = 30° or (π/6)^c

•  – 1/√3

The slope of line = m = tan θ = – 1/√3

∴  θ = tan^-1(- 1/√3) = 150° or (5π/6)^c

Science > Mathematics > Coordinate Geometry > Straight Lines > The Inclination of a Line

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