Distance of a Point from a Line and Distance between two Parallel Lines | Mathematics (Maths) Class 11 - Commerce PDF Download

F. Distance Between Point & Line And Two Parallel Lines 

(1) Length of the Perpendicular from a Point on a Line : 

The length of the perpendicular from P(x₁, y₁) on ax + by + c = 0 is

The length of the perpendicular from origin on ax + by + c = 0 is

(2) The distance between two parallel lines : 

The distance between two parallel lines : ax + by + c₁ = 0 and ax + by + c₂ = 0 is 

(3) Area of parallelogram with given sides :

(4) Condition of parallelogram as shown becomes a rhombus :

(5) Reflection (Image) of a point P(a, b) about a line (ax + by + c = 0)

(6) Foot of perpendicular from a point 

(7) Shifting of the origin :

x, y ⇒ old co-ordinates axes

X, Y ⇒ New co-ordinate axes

X = 0 ⇒ x - a = 0 ⇒ x = a

Y = 0 ⇒ y - b = 0 ⇒ y = b

Slope and area of closed figure remains unchanged under the translation of co-ordinate axes.

Ex.13 Three lines x + 2y + 3 = 0, x + 2y - 7 = 0 and 2x - y - 4 = 0 form 3 sides of two squares. Find the equation of remaining sides of these squares.

Sol. 

Distance between the two parallel lines is     The equations of sides A and C are of the form 2x – y + k = 0. Since distance between sides A and B = distance between sides 

Hence the fourth sides of the two squares are (i) 2x – y + 6 = 0, (ii) 2x – y – 14 = 0

Ex.14 Find the foot of the perpendicular drawn from the point (2, 3) to the line 3x - 4y + 5 = 0. Also, find the image of (2, 3) in the given line.

Sol.

Let AB ≡ 3x – 4y + 5 = 0, P ≡ (2, 3) and PM ⊥ AB .

Slope of AB = 3/4 

slope of  PM  = -4/3 = tanθ 

Which is the foot of the perpendicular.  

Let Q be the image of P 

G. Bisectors Of The Angles Between Two Given Lines

Angular bisector is the locus of a point which moves in such a way so that its distance from two intersecting lines remains same.

The equation of the two bisectors of the angles between the lines a₁x + b₁y + c₁ = 0 and

a₂ x + b₂y + c₂ = 0 are .

If the two given lines are not perpendicular i.e. a₁ a₂ + b₁ b₂ ¹ 0, then one of these equation is the equation of the bisector of the acute angle and the other that of the obtuse angle.

Note : Whether both lines are perpendicular or not but the angular bisectors of these lines will always be mutually perpendicular.

 (1) The bisectors of the acute and the obtuse angles :

Take one of the lines angle let its slope be m₁ and take one of the bisectors and let its slope be m₂. If q be the acute angle between them, then find tanθ =  .

If  then the bisector taken is the bisector of the obtuse angle and the other one will be the bisector of the acute angle. If  then the bisector taken is the bisector of the acute angle and the other one will be the bisector of the obtuse angle.

If two lines are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 , then    will  represent the equation of the bisector of the acute or obtuse angle between the lines according as c₁c₂ (a₁a₂+ b₁b₂ ) is negative or positive.

(2) The equation of the bisector of the angle containing the origin 

Write the equations of the two lines so that the constants c₁ and c₂ become positive. Then the equation  is the equation of the bisector containing the origin.

Notes : 

(i) If , then the origin will lie in the acute angle and if , then origin will lie in the obtuse angle.

(ii) The note (i) is helpful in finding the equation of bisector of the obtuse angle or acute angle directly.

 (3) The equation of the bisector of the angle which contains a given point

The equation of the bisector of the angle between the two lines containing the point  is

 or  according as  and  are of the same sings or of opposite signs.

Ex.15 For the straight line  and , find the equation of the

(i) bisector of the obtuse angle between them.

(ii) bisector of the actus angle between them.

(iii) bisector of the angle which contains (1, 2).

Sol. Equations of bisectors of the angles between the given lines are

 and .

 is the acute angle between the line  and the bisector , then

Hence

(i) The bisector of the obtuse angle is

(ii) The bisector of the acute angle is

(iii) The bisector of the angle containing the origin

(i) For the point (1, 2), 4x + 3y – 6 = 4 × 1 + 3 × 2 – 6 > 0  ⇒ 5x + 12y + 9 = 5 × 1 + 12 × 2 + 9 > 0

Hence equation of the bisector of the angle containing the point (1, 2) is

Alternative : 5 lines. Similarly bisector of obtuse angle is 9x - 7y - 41 = 0.

(4) The equation of reflected ray :

Let L₃ be the reflected ray from the line L₂. Clearly L₂ will be one of the bisectors of the angles between L₁ and L₃. Since L₃ passes through A,

Let (h, k) be a point on L₂. Then

Note : Some times the reflected ray L₃ is also called the mirror image of L₁ in L₂.

H. Family of Lines

The general equation of the family of lines through the point of intersection of two given lines is

 where L = 0 and L' = 0 are the two given lines, and  Conversely, any line of the form  passes through a fixed point which is the point of intersection of the lines
L₁ = 0 and L₂ = 0.
The family of lines perpendicular to a given line ax + by + c = 0 is given by bx - ay + k = 0, where k is a parameter.

The family of lines parallel to a given line ax + by + c = 0 is given by ax + by + k = 0, where k is a parameter.

 Ex.16 Show that all the chords of the curve 3x² - y² - 2x + 4y = 0, which subtend a right angle at the origin pass through a fixed point. Find that the point. 

Sol. Let the equation of chord be lx + my = 1. So equation of pair of straight line joining origin to the points of intersection of chord and curve.

3x² – y² – 2x (lx + my) + 4y(l + my) = 0, which subtends right angle at origin.

⇒ (3 – 2l + 4m – 1) = 0 ⇒ 1 = 2m + 1. Hence chord becomes (2m + 1)x + my = 1

x – 1 + m(2x + y) = 0

Which will pass through point of intersection of L₁ = 0 and L₂ = 0. 

⇒ x = 1, y = – 2. Hence fixed point is (1, – 2).

(1) One Parameter Family of Straight Lines

If a linear expression L₁ contains an unknown coefficient, then the line L₁ = 0 can not be a fixed line. Rather it represents a family of straight lines known as one parameter family of straight lines. e.g. family of lines parallel to the x-axis i.e. y = c and family of straight lines passing through the origin i.e. y = mx.

Each member of the family passes a fixed point. We have two methods to find the fixed point.

Method (i) :

Let the family of straight lines of the form ax + by + c = 0 where a, b, c are variable parameters satisfying the condition al + bm + cn = 0, where I, m, n, are given and n ¹ 0. Rewriting the condition as a  and comparing with the given family of straight lines, we find that each member of it passes through the fixed point

Ex.17 If the algebraic sum of perpendiculars from n given points on a variable straight line is zero then prove that the variable straight line passes through a fixed point.

Sol. Let n given points be (x₁, y₁) where i = 1, 2 .......... n and the variable line is ax + by + c = 0, Given that

Hence the variable straight line always passes through the fixed point .

 Method (ii) :

If a family of straight lines can be written as  where L₁, L₂ are two fixed lines and  parameter, then each member of it will pass through a fixed point given by point of intersection of L₁ = 0 and L₂ = 0.

Note : If L₁= 0 an L₂ = 0 are parallel lines, they will meet at infinity.

 Ex.18 Prove that each member of the family of straight lines

 passes through a fixed point.

Sol. The given family of straight lines can be rewritten as (3x + 2y + 1) sinθ +(4x - 7y + 2) cosθ = 0

or, (4x - 7y + 2) + tanθ (3x + 2y + 1) = 0 which is of the form

Hence each member of it will pass through a fixed point which is the intersection of 4x - 7y + 2 = 0 and

(2) Concurrency of Straight Lines : The condition for 3 lines a₁x + b₁y + c₁ = 0, a₂x + b₂ + c₂ = 0, a₃x + b₃x + b₃y + c₃ = 0 to be concurrent is

(ii) There exist 3 constants l, m, n (not all zero at the same time) such that L₁ + mL₂ + nL₃ = 0, where L₁ = 0, L₂ = 0 and L₃ = 0 are the three given straight lines.

(iii) the three lines are concurrent if any one of the lines passes through the pint of intersection of the other two lines.

I. Pair of Straight lines

The combined equation of pair of straight lines L₁ = a₁x + b₁y + c₁ = 0 and L₂ = a₂x + b₂y + c₂ = 0 is (a₁x + b₁y + c₁) (a₂x + b₂y + c₂) = 0 i.e. L₁L₂ = 0. Opening the brackets and comparing the terms with the terms of general equation of 2^nd degree ax² + 2hxy + by² + 2gx + 2fy + c = 0, we can get all the following results for a pair of straight lines.

The general equation of second degree ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of straight     and h² ≥ ab.

⇒ abc + 2fgh – al² – bg² – ch² = 0 and h² ≥ ab.

The homogeneous second degree equation ax² + 2hxy + by² = 0 represents a pair of straight lines through the origin if

If the lines through the origin whose joint equation is ax² + 2hxy + by² = 0, are y = m₁x and

The lines are perpendicular if a + b = 0 and coincident if h² = ab.

In the more general case, the lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0 will be perpendicular if a + b = 0, parallel if the terms of second degree make a perfect square i.e. ax² + 2hxy + by² gets converted into (l₁x + 2hxy + by² gets converted into (l₁x ± m₁y)², coincident if the whole equation makes a perfect square i.e. ax² + 2hxy + by² + 2fy + c can be written as (lx + my + n)².

Note : Point of intersection of the two lines represented by ax² + 2hxy + by² + 2gx + 2fy +c = 0 is        

the derivative of f with respect to y, keeping x constant. The fact can be used in splitting ax² + by² + 2hxy + 2gx + 2fy + c = 0 into equations of two straight lines. With the above method, the point of intersection can be found. Now only the slopes need to be determined.

If should be noted that the line ax + hy + g = 0 and hx + by + f = 0 are not the lines represented by ax² + 2hxy + by² + 2gx + 2fy + c = 0. These are the lines concurrent with the lines represented by given equation.

(Homogenization) Joint equation of pair of lines joining the origin and the points of intersection of a curve and a line :

If the line lx + my + n = 0, ((n # 0) i.e. the line does not pass through origin) cuts the curve ax² + 2hxy + by² + 2gx + 2fy + c = 0 at two points A and B, then the joint equation of straight lines passing through A and B and the origin is given by homogenizing the equation of the curve by the equation of the line. i.e.

  is the equation of the line OA and OB

Ex.19 If the lines ax + y + 1 = 0, x + by + 1 = 0 and x + y + c = 0(a, b and c being distinct and different from 1) are concurrent, then prove that  = 1.

Sol. 

Since the given lines are concurrent 

Operating C₂ → C₂ – C₁ and C₃ → C₃ – C₁ , we get 

a (b – 1) (c – 1) – (b – 1) (1 – a) –(c – 1) (1 – 0) = 0 

Ex.20 The chord √6 y = √8 ρx + √2 of the curve ρy² + 1 = 4x subtends a right angle at origin then find the value ofρ.

Sol. 

√3 y - 2ρx = 1 is the given chord. Homogenizing the equation of the curve, we get,

ρy² – 4x( 3 y – 2ρx) + (√ 3 y – 2ρx)² = 0

 (4ρ² + 8ρ)x² + (ρ + 3)y² - 4√ 3 xy - 4√ 3 ρxy = 0

Now, angle at origin is 90º

coefficient of x² + coefficient of y² =

4ρ² + 8ρ + ρ + 3 = 0

4ρ² + ρ+ 3 = 0

=