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.

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:

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))
(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
(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


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
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
Part – C
Soln:
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 m2 x2 = 0
m1 m2 x2 – (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.

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
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
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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