Tamil Nadu Diploma Engineering Mathematics – II Unit – I ( 1.1 – Analytical Geometry – I) material 2020-21 ( N-Scheme)

UNIT – I ANALYTICAL GEOMETRY

1.1 ANALYTICAL GEOMETRY I

Straight Line:

When a variable point moves in accordance with a geometrical law, the point will trace some curve. This curve is known as the locus of the variable point.

If a relation in x and y represent a curve then

(i) The co-ordinates of every point on the curve will satisfy the relation.

(ii) Any point whose co-ordinates satisfy the relation will lie on the curve.

Straight line is a locus of a point.

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Slope or gradient of a straight line:

The tangent of the angle of inclination of the straight line is called slope or gradient of the line. If θ is the angle of inclination then slope = tan θ and is denoted by m.

i.e m = tan θ

Equation of a straight line:

When ‘c’ is the y intercept and slope is ‘m’, the equation of the straight line is

y = mx+c

Formulae

1) The length of the perpendicular distance from (x₁,y₁) to the line ax + by + c = 0 is

± (ax₁ + by₁ + c)/ √(a² + b² )

2) The distance between the parallel lines ax + by + c₁ = 0 and ax + by + c₂ = 0 is

± (c₁– c₂ )/ √(a² + b² )

ANGLE BETWEEN TWO STRAIGHT LINES:

Book Work:

Find the angle between the lines y = m₁ x+c₁ and y = m₂x+c₂. Deduce the conditions for the lines to be

(i) parallel (ii) perpendicular.

Proof:

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Let θ₁ be the angle of inclination of the line y = m₁x+c₁.

Slope of this line is m₁ = tanθ₁.

Let ‘θ’₂ be the angle of inclination of the line y=m₂x+c₂ .

Slope of this line is m₂ = tanθ₂ .

Let ‘θ’ be the angle between the two lines, then θ₁ = θ₂ + θ

Θ = θ₁– θ₂

∴tan θ = tan (θ₁-θ₂)

= (tan θ1- tan θ2)/(1 + tan θ1 tan θ2 )

tan θ = (m1 – m2)/(1 + m1 m2)

θ = tan^(-1)⁡((m1 – m2)/(1 + m1 m2))

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(i) Condition for two lines to be parallel:

If the two lines are parallel then the angle between the two lines is zero

∴ tan θ = tan 0 = 0

(i.e) = 0

m₁ – m₂ = 0

∴ m₁= m₂

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(ii)Condition for two lines to be perpendicular:

If the two lines are perpendicular then the angle between them

Θ = 90⁰

∴ tan θ = tan 90⁰ = ∞ = 1/0

(m1 – m2)/(1 + m1 m2) = 1/0

1 + m₁ m₂ = 0

m₁ m₂ = – 1

∴ For perpendicular lines, product of the slopes will be -1

Note: 1) Any line parallel to the line ax+by+c = 0

will be of the form ax+by+ k = 0 (differ only constant term)

2) Any line perpendicular to the line ax+by+c =0

will be of the form bx – ay + k = 0

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WORKED EXAMPLES

Part – A

. 1. Find the perpendicular distance from the point (2,3) to the straight line 2x+y+3=0.

Soln: W.K.T The length of the perpendicular distance from (x₁,y₁) to the line ax + by + c = 0 is

± (ax₁ + by₁ + c)/ √(a² + b² )

Given straight line is 2x+y+3=0

Given point (x₁,y₁) = (2,3)

i.e 2(2 ) + 1(3) + 3/ √(2² + 1² ) = 10 / √5

2. Find the distance between the line 2x+3y+4 = 0 and 2x+3y -1 = 0

Soln: W.K.T the distance between the parallel lines

ax + by + c₁ = 0 and ax + by + c₂ = 0 is

± (c₁– c₂ )/ √(a² + b² )

Here a = 2, b = 3 , c₁ = 4 and c₂ = – 1

i.e 4 + 1/√(2² + 3² ) = 5 /√13

3. Show that the lines 6x+y-11 =0 and 12x+2y+14 =0 are parallel

Soln: 6x+y-11 = 0 (1)

12x+2y+14 = 0 (2)

Slope of the line (1) = m₁ = – a/b

= – 6/1

m₁ = – 6

Slope of the line (2) = m₂ = – a/b

= – 12/2

= – 6

m₁= m₂

∴ The lines are parallel.

4. Find ‘p’ such that the lines 7x-4y+13 = 0 and px = 4y+6 are parallel.

Soln: 7x-4y+13 = 0 (1)

px – 4y+ 6 = 0 (2)

Slope of the line (1) = m₁ = – 7/-4

m₁ = 7/4

Slope of the line (2) = m₂ = – p/-4

m₂ = p/4

Since (1) and (2) are parallel lines

m₁= m₂

7/4 = p/4

4p = 28

∴ p = 7

5. Show that the lines 2x+3y-7 =0 and 3x- 2y+4 =0 are perpendicular.

Soln: 2x+3y-7 =0 (1)

3x- 2y+4 =0 (2)

Slope of the line (1) = m₁ = – a/b

m₁ = – 2/3

Slope of the line (2) = m₂ = – 3/-2

m₂ = 3/2

m₁ m₂ = (- 2/3 ) × (3/2)

= – 1

∴ The lines (1) and (2) are perpendicular.

6. Find the value of m if the lines 2x + my = 4 and x + 5y – 6 = 0 are perpendicular.

Soln: 2x + my – 4 = 0 (1)

x + 5y – 6 = 0 (2)

Slope of the line (1) = m₁ = – 2/m

Slope of the line (2) = m₂ = – 1/5

Since (1) and (2) are perpendicular

m₁ m₂ = – 1

(- 2/m ) (- 1/5) = -1

( 2/5m ) = – 1

– 5m = 2

∴ m = – 2/5

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Part – B

1. Find the angle between the lines y = √3 x and x- y = 0

Sol: y = √3 x

√3 x – y = 0 (1)

x- y = 0 (2)

Slope of the line (1) = m₁ = – coefficient of x / coefficient of y

= – √3 / – 1

m₁ = √3

tan θ₁ = √3 ⇒ θ₁ = 60⁰

Slope of the line (2) = m₂ = – coefficient of x / coefficient of y

= – 1 / – 1

m₂ = 1

tan θ₂ = 1 ⇒ θ₂ = 45⁰

∴ Θ = θ₁– θ₂

Θ = 60⁰– 45⁰

Θ = 15⁰

2. Find the angle between the lines 3x + 6y = 8 and 2x = – y + 5

Sol: 3x + 6y = 8

3x + 6y – 8 = 0 (1)

2x + y – 5 = 0 (2)

Slope of the line (1) = m₁ = – coefficient of x / coefficient of y

= – 3 / 6

m₁ = – 1 / 2

Slope of the line (2) = m₂ = – coefficient of x / coefficient of y

= – 2 / 1

m₂ = – 2

W.K.T tan θ = (m1 – m2)/(1 + m1 m2)

tan θ = (1 / 2 + 2)/(1 + ( – 1/2 ) ( – 2 ))

tan θ = 3/2/2

tan θ = 3/4

3. Find the equation of the straight line parallel to the line 3x + 2y -7=0 and passing through the point (1, -2).

Soln: Let the equation of line parallel to 3x + 2y -7=0 (1)

is 3x + 2y +k = 0 (2)

Equation (2) passes through ( 1, -2 )

put x= 1, y =-2 in equation (2)

3(1) + 2(-2) +k = 0

3 – 4 + k = 0

k = 1

∴ Required line is 3x + 2y +1 = 0

4. Find the equation of the line passing through (3,- 3) and

perpendicular to 4x – 3y – 10 = 0.

Soln: Let the equation of line perpendicular to 4x – 3y – 10 = 0 (1)

is – 3x – 4y + k = 0 (2)

Equation (2) passes through (3,-3)

put x = 3, y =- 3 in equation (2)

-3 (3) – 4(-3) +k = 0

-9 + 12+ k = 0

k = – 3

– 3x – 4y – 3 = 0

∴ Required line is 3x +4y + 3 = 0

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Part – C

\[1.\ If\ the\ two\ straight\ lines\ 2x – 3y + 9 = 0\ and\ 6x + ky + 9 = 0\ are\ parallel\ then\ find\ the\ value\ of\ ‘k’\]\[ and\ also\ find\ the\ distance\ between\ them\]

Soln:

\[Given\ 2x – 3y + 9 = 0\ ———- —————– (1)\]

\[Given\ 6x + ky + 9 = 0\ ———- —————– (2)\]

\[From\ (1)\ m_1 = \frac{3}{2}\]

\[From\ (2)\ m_2 = \frac{-k}{6}\]

\[Given\ (1)\ and\ (2)\ are\ parallel\ i.e\ m_1=m_2\]

\[\frac{3}{2} = \frac{-k}{6}\]

\[18 = -2k\]

\[k = -9\]

\[Equation\ (2)\ becomes\ 6x – 9y + 9 = 0\]

\[3(2x – 3y + 3) = 0\]

\[To\ find\ distance\ between\ parallel\ lines\ \hspace10cm\]

\[d = \frac{c_1 – c_2}{\sqrt{a^2 + b^2}}\]

\[Here\ a = 2,\ b = -3,\ c_1 = 9,\ c_2 = 3\]

\[d = \frac{9 – 3}{\sqrt{2^2 + (-3)^2}}\]

\[d = \frac{6}{\sqrt{4 + 9}}\]

\[d = \frac{6}{\sqrt{13}}\]

PAIR OF STRAIGHT LINES

Pair of Straight lines passing through origin

The general equation of pair of straight lines passing through origin is

ax² + 2hxy + by² = 0 ———- ( A )

Let y = m₁x

y – m₁x = 0 ————– ( 1 )

and y = m₂x

y – m₂x = 0 ————– ( 2 )

(y – m₁x) (y – m₂x ) = 0

y² – (m₁ + m₂ ) xy + m₁ m₂x² = 0

m₁ m₂x² – (m₁ + m₂ ) xy + y² = 0 ————– ( B )

Equations (A) and ( B ) represent the same pair of straight lines. Hence the ratios

of the corresponding coefficients of like terms are proportional.

m₁ m₂ / a = – (m₁ + m₂ ) / 2h = 1 / b

m₁ + m₂ = – 2h / b and m₁ m₂ = a / b

i.e sum of slopes = – 2h / b and product of slopes = a / b

Formulae

1 ) The angle between the pair of straight lines ax² + 2hxy + by² = 0 is

tanθ = ± (2√h² – ab ) / a + b

2) The condition for the pair of lines to be parallel is h² – ab = 0.

3) The condition for the pair of lines to be perpendicular is a + b = 0.

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Part – A

1. Write down the combined equation of the pair of lines

x − 2y = 0 and 3x + 2y = 0

Soln: The two separate lines are x − 2y = 0 and 3x + 2y = 0

The combined equation is

(x − 2y) (3x + 2y) = 0

3x² + 2xy – 6xy – 4 y² = 0

3x²– 4xy – 4 y² = 0

2. Write down the separate equations of the pair of lines 12x²+ 7xy – 10y² = 0

Soln: 12x²+ 7xy – 10y² = 0

12x²+ 15xy – 8xy – 10y² = 0

3x( 4x + 5 ) – 2y( 4x – 5y) = 0

(3x – 2y) ( 4x – 5y) = 0

∴ The separate equations are3x – 2y = 0 and 4x – 5y = 0

3. Show that the two lines represented by 4x²+ 4xy + y² = 0 are parallel to each

other.

Soln: 4x²+ 4xy + y² = 0

This is of the form ax² + 2hxy + by² = 0

a = 4, 2h = 4 h = 2, b = 1

If the lines are parallel then h² – ab = 0

h² – ab = (2)² – (4) ( 1 )

= 4 – 4

= 0

∴ pair of lines are parallel

4. Find the value of ‘p’ if the pair of lines 4x²+ pxy + 9y² = 0 are parallel to each

Other.

Soln: 4x²+ pxy + 9y² = 0

This is of the form ax² + 2hxy + by² = 0

a = 4, 2h = p h = p / 2, b = 9

Given lines are parallel

h² – ab = 0

( p /2)² – ( 4 ) ( 9 ) = 0

P² / 4 – 36 = 0

P² = 144

∴ p = ± 12

5. Prove that the two lines represented by 7x²– 48xy – 7y² = 0 are perpendicular to

each Other.

Soln: 7x²– 48xy – 7y² = 0

This is of the form ax² + 2hxy + by² = 0

a = 7, 2h = – 48 h = – 24, b = – 7

If the lines are perpendicular then a + b = 0

a + b = 7 – 7

= 0

∴ pair of lines are perpendicular.

6. . Find the value of ‘p’ if the pair of lines px²– 5xy – 7y² = 0 are perpendicular

to each Other.

Soln: px²– 5xy – 7y² = 0

This is of the form ax² + 2hxy + by² = 0

a = p, 2h = – 5 h = – 5 / 2, b = – 7

Given lines are perpendicular

a + b = 0

p – 7 = 0

∴ p = 7

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Part – B

1. Find the separate equations of the line 2x²– 7xy + 3y² = 0. Also find the angle

between them.

Soln: 2x²– 7xy + 3y² = 0

2x²– 6xy – xy + 3y² = 0

2x(x – 3y ) – y( x – 3y) = 0

(2x – y) ( x – 3y) = 0

∴ The separate equations are 2x – y = 0 and x – 3y = 0

2x²– 7xy + 3y² = 0

This is of the form ax² + 2hxy + by² = 0

a = 2, 2h = – 7 h = – 7 / 2, b = 3

Let θ be the angle between the two straight lines

W.K.T tanθ = ± (2√h² – ab ) / a + b

= ± (2√ (( – 7 / 2)² – (2) (3) )/ (2 + 3)

= ± (2 √ (49 / 4 – 6 ) / 5

= ± (2 √ (49 – 24) / 4 ) / 5

= ± (2 √ 25 / 4 ) / 5

= ± (2 ( 5 / 2 ) / 5

tanθ = ± 1

tan θ = tan 45, ∴ θ = 45°

2. The slope of one of the lines ax² + 2hxy + by² = 0 is thrice that of the other. Show that 3h²= 4ab

Soln: ax² + 2hxy + by² = 0 ——————– ( 1 )

Let y = m₁x and y = m₂x be the separate equations of equation ( 1 )

m₁+ m₂ = – 2h / b ——————— ( 2 )

m₁m₂ = a / b ———————- ( 3 )

Slope of one of the line = thrice slope of the other line

i.e m₁= 3m₂

Equation (2) becomes

3m₂ + m₂ = – 2h / b

4m₂ = – 2h / b

m₂ = – 2h / 4b

m₂ = – h / 2b

Substitute m₁= 3m₂ in equation (3 )

3m₂m₂ = a / b

3 (m₂ )² = a / b

3 (- h / 2b )² = a / b

3 (h² / 4b² ) = a / b

3h² / 4b² = a / b

3h²b = 4ab²

(i.e) 3h² = 4ab

PAIR OF STRAIGHT LINES NOT PASSING THROUGH THE ORIGIN

Formulae

1) The angle between the pair of straight lines ax² + 2hxy + by² + 2gx + 2fy + c = 0

is tanθ = (2√h² – ab ) / a + b

2) The condition for the pair of lines to be parallel is h²– ab = 0.

3) The condition for the pair of lines to be perpendicular is a + b = 0.

4) Condition for the second degree equation ax² + 2hxy + by² + 2gx + 2fy + c = 0 to

represent a pair of straight lines is

abc + 2fgh – af² – bg² – ch² = 0

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Part – C

1. Show that the equation 3x² + 7xy + 2y² + 5x + 5y + 2= 0 represents a pair of

straight lines.

Soln: Given 3x² + 7xy + 2y² + 5x + 5y + 2= 0

Comparing with ax² + 2hxy + by² + 2gx + 2fy + c = 0

a = 3, 2h = 7 , b = 2, 2g = 5 , 2f = 5 , c = 2

h = 7/ 2 g = 5/ 2 f = 5/ 2

To show abc + 2fgh – af² – bg² – ch² = 0

abc + 2fgh – af² – bg² – ch² =

(3) (2) (2) + 2 (5/2) (5/2) (7/2) – 3 (5/ 2)² – 2 (5/ 2)² – 2 (7/ 2)²

= 12 + 175/4 – 75/4 – 50/4 – 98/4

= 12 + ( 175 – 75 – 50 – 98)/4

= 12 + ( 175 – 223)/4

= 12 + ( -48 /4 )

= 12 – 12 = 0

Hence, the given equation represents pair of straight lines.

2. Find K if 2x² – 7xy + 3y² + 5x – 5y + k= 0 represents a pair of

straight lines.

Soln: Given 2x² – 7xy + 3y² + 5x – 5y + k= 0 ————– ( 1 )

Comparing with ax² + 2hxy + by² + 2gx + 2fy + c = 0

a = 2, 2h = – 7 , b = 3, 2g = 5 , 2f = – 5 , c = k

h = – 7/ 2 g = 5/ 2 f = – 5/ 2

Given equation ( 1 ) represents a pair of straight lines

i.e abc + 2fgh – af² – bg² – ch² = 0

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

6k + 175/4 – 50/4 – 75/4 – 49k/4 = 0

(24k – 49k )/4 + ( 175 – 50 – 75)/4 = 0

– 25k/4 + 50/4 = 0

( -25k + 50 ) / 4 = 0

-25 k = – 50

k = 2

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Tamil Nadu Diploma Engineering Mathematics – II Unit – I ( 1.1 – Analytical Geometry – I) material 2020-21 ( N-Scheme)

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