Straight lines

● System of Coordinates

● Angle Between Lines

● Equations of Straight Lines

● Some Formulae

● Family of Lines

● Concurrency Condition

In this section we will be exploring the following.

● Section formula

● Centers of triangle

● Area of triangle

● Locus

● Shifting of Origin

System of Coordinates

2 3

P

A (x1, y1)

B (x2, y2)P (h, k)

mn

A (1, 2) B (5, 9)

Internal Section Formula

Observation

A (x1, y1)B (x2, y2)

P (h, k)

A (x1, y1)

B (x2, y2)P (h, k)

m-n

A (x1, y1)

B (x2, y2)

P (h, k)

External Section Formula

Eg. Point which divides A (1, 2) and B (5, 7) in the ratio 2 : 3 externally

Q B (5, 7)A (1, 2)

−2 3

Q B (5, 7)A (1, 2)

Eg. Point which divides A (1, 2) and B (5, 7) in the ratio 2 : 3 externally

Find the ratio in which the segment joining the pointsA (2, –4) and B (4, 5) is divided by 2x + y + 1 = 0

A (2, –4)

k 1

P B (4, 5)

Solution :

A (2, –4)

k 1

P B (4, 5)

Solution :Alternate Solution

A(x1, y1) B(x2, y2)

Remarks

ax + by + c

k 1

1. Centroid

2. Incentre

3. Orthocentre

4. Circumcentre

Centres of a Triangle

It is the point of concurrence of the medians of a triangle.

F E

D

G

A (x1, y1)

C (x3, y3)B (x2, y2)

Result

Centroid divides any median in ratio 2 : 1, that is GA : GD = 2 : 1

Centroid

It is the point of concurrence of the internal angle bisectors of a triangle.

c b

a

I

A (x1, y1)

C (x3, y3)B (x2, y2)

Angle bisector divides opposite side in the ratio of angle containing sides

Recall

Incentre

If ⍺, β, 𝛾 are the roots of the equation x3 - 3px2 + 3qx - 1 = 0 then find the centroid of the triangle whose vertices are (⍺, β + 𝛾), (β, ⍺ + 𝛾), (𝛾, ⍺ + β)

A B C D(p, p) (p, 2p) (p, 0)(2p, p)

If ⍺, β, 𝛾 are the roots of the equation x3 - 3px2 + 3qx - 1 = 0 then find the centroid of the triangle whose vertices are (⍺, β + 𝛾), (β, ⍺ + 𝛾), (𝛾, ⍺ + β)

A B C D(p, p) (p, 2p) (p, 0)(2p, p)

Solution :

It is the point of concurrence of the altitudes of a triangle.

H

A (x1, y1)

B (x2, y2) C (x3, y3)

EF

D

Orthocentre

It is the point of concurrence of the perpendicular bisector of the sides of a triangle.

A

B C

O

E

D

F

(x3, y3)

(x, y)

(x1, y1)

(x2, y2)

Circumcentre

In any scalene triangle,

Result

2 : 1

GH

O

Note: In an equilateral triangle, G, I, O and H, all coincide.

Area of Triangle

Remark

We can also write the area of a triangle using determinants, like:

In general, the area of any n - sided polygon is given by

It is the path or curve traced by a moving point satisfying a given condition.

Locus

Steps to find Equation of Locus

(1) Assume the coordinates of the point whose locus is to be

found, to be (h, k).

(2) Write the given conditions involving h and k.

(3) Eliminate the variables, if any.

(4) Replace h by x and k by y in the eliminant to obtain the

equation of the locus.

We are generally interested in the equation of the locus of a point.

Locus

Find the locus of a point P such that ∠APB = 90°, where the coordinates of A and B are respectively (−a, 0) and (a, 0).

Solution :

A (1, 2) is a fixed point. A variable point B lies on x2 + y2 = 4. Find the locus of the midpoint of AB.

Solution :

Find the locus of a point whose coordinates are

a) x = cosθ, y = sinθ, where θ is a parameter

b)) x = t2, y = 2t, where t is a parameter

Solution :

Shifting of Origin

P (x, y) clearly, x‘ = x - a and y‘ = y - b

(0,0)

Observation: If the origin is shifted to (1, 3) then the new coordinates of the point (3, 6) will be (2, 3).

YY’

(0, y)(0, y)’

(a, b) (x’, 0)X’

X(x, 0)

Shifting of Origin

Note 1: Here, α is the angle made by the line with the positive direction of the X-axis in the anti-clockwise sense.

If a straight line makes an angle α with the X-axis, then it’s slope m is

defined as tan α, where 0 ≤ ⍺ < 𝜋.

Slope of a Straight Line

If a straight line makes an angle α with the X-axis, then it’s slope m is

defined as tan α, where 0 ≤ ⍺ < 𝜋.

Note 2: (x2, y2)

(x1, y1)

Note 1: Here, α is the angle made by the line with the positive direction of the X-axis in the anti-clockwise sense.

Slope of a Straight Line

Before we study various equations of a straight line, let’s first see two

particular cases.

(1) Horizontal Line

(2) Vertical Line

Equations of Horizontal and Vertical Lines

3.

2. 1.

4.

Equation of a Horizontal Line

As on a horizontal line, the y-coordinate does not change, therefore it’s equation is of the form y = constant.

(0, b) (a, b)

(1, 2) (-2, 1)

(y = b) (y = b)

(y = 2) (y = 1)

Equation of a Vertical Line

As on a vertical line, the x-coordinate does not change, therefore it’s equation is of the form x = constant.

3.

2. 1.

4.

(x = a)

(a, 0)

(x = a)

(a, b)

(x = -2)

(-2, 1)(x = 1)(1, 2)

Note: Equations 1 and 2 are the most frequently used forms.

1. Slope Intercept form 2. Point Slope form

3. Two Point form 4. Intercept form

5. Normal form 6. Parametric form

7. General form

Various forms of equations of a line:

1.

2.

slope = m

slope = m

3.

4.

(0, c)

y = mx + c

(x1, y1)

(x2, y2)

(x1, y1)

y - y1 = m(x - x1)

(0, b)

(a, 0)

Eg. Equation of line passing through (1, 2) and having slope 4 is

Eg. Equation of line bisecting segment joining (1, 2) and (3, 4) and

making an angle of 45° with X-axis is

Eg. Equation of line through (1, 2) and (3, 5) is

Eg. Equation of line passing through (1, 2) and having slope 4 is

y - 2 = 4(x - 1) , that is 4x - y - 2 = 0 .

Eg. Equation of line bisecting segment joining (1, 2) and (3, 4) and

making an angle of 45° with X-axis is y - 3 = (1) (x - 2), ie x - y + 1 = 0.

Eg. Equation of line through (1, 2) and (3, 5) is that is 3x - 2y + 1 = 0.

X

Y

O

p

Normal form Parametric form

(x1, y1)

(x, y)

m = tan θ

r

x cos 𝛼 + y sin 𝛼 = p

Where, 0 ≤ 𝛼 < 360°

𝛼

Convert x+y = 2 into normal form.

Solution :

Try to comment on the point P in the following case.

(1, 2)5

P

X30°

Various forms of Equation of a Straight Line

On the line y = x+1, find the points 2 units away from (2, 3).

45° X

Q

(2, 3)2

2

P

Solution :

Any linear equation in x and y represents a straight line, that is, the

equation ax + by + c = 0 is the general form of the equation of a line.

Note: Slope of this general form is -a/b

General Form

where θ is the acute angle between the two lines

θ

m1

m2

Angle between two Lines

Note1. Lines are parallel if m1 = m2 2. Lines are perpendicular if m1 m2 = -1, provided both m1 and m2

are finite.

Note(a) A line parallel to the line ax + by + c = 0 has an equation of the

form ax + by + d = 0.

(b) A line perpendicular to the line ax + by + c = 0 has an equation of

the form bx − ay + d = 0.

2x + 3y + 1 = 0 and 3x + 2y + 2 = 0

x − y − 2 = 0 and y = x

7x − 5y + 13 = 0 and y = 3x + 8

Which of the following pairs of lines are mutually perpendicular?

4x + 3y + 5 = 0 and 3x − 4y − 5 = 0

A

B

C

D

Which of the following pairs of lines are mutually perpendicular?

2x + 3y + 1 = 0 and 3x + 2y + 2 = 0

x − y − 2 = 0 and y = x

7x − 5y + 13 = 0 and y = 3x + 8

4x + 3y + 5 = 0 and 3x − 4y − 5 = 0

A

B

C

D

Solution :

(a) Find the equation of the perpendicular bisector of the line segment joining the points A (1, 3) and B (5, 7).(b) If the vertices of a triangle are (2, 5), (3, 9) and (-4, 0), then find the equation of the altitude through (2, 5).

(a) Find the equation of the perpendicular bisector of the line segment joining the points A (1, 3) and B (5, 7).

Solution :

(b) If the vertices of a triangle are (2, 5), (3, 9) and (-4, 0), then find the equation of the altitude through (2, 5).

Solution :

If A (0, 0), B (4, –3) and C (4, 2) are the vertices of Δ ABC,

then find its circumcentre.

OF D

A (0, 0)

B (4, −3) C (4, 2)

Solution :

NoteTry to see as a relation between m1, m2 and θ

Now that we have understood angle between two lines, let’s see one

very important category of questions, where we find the equations of

a line making a given angle with a given line.

m2

m1

θ

Angle between two Lines

(a) Find the equation of a line through the point (1, 2)

making an angle of 30° angle with the line .

Solution :

Note: A point on y = mx + c is assumed as (t, mt + c).

Sometimes we are required to assume a point on a line or you may

say, sometimes, assuming a point on line facilitates solving the

question easily.

Find points on x + y = 1 which are at 2 units distance from

(0, 0) .

Solution :

Distance of a Point from a Line

Special case : Distance of origin from ax + by + c = 0 is

(x1, y1)

Eg :Distance of (1, 2) from 3x - 4y + 2 = 0 is

ax + by + c = 0

d

Some Formulae

Eg : Distance between x + y + 2 = 0 and x + y + 4 = 0 is

ax + by + c = 0

ax + by + d = 0

D

Distance between two Parallel Lines

(x1, y1)

Eg : Foot of perpendicular of (2, 3) on x + 2y - 1 = 0 is given by

ax + by + c = 0(⍺, β)

Foot of Perpendicular from a Point to a Line

(x1, y1)

(⍺’, β’)

ax + by + c = 0

Image of a Point in a Line

Note: Any line through intersection point of L1 = 0 and L2 = 0 (that is a member of their family) has equation of the form L1 + λ L2 = 0

Given any two lines L1 = 0 and L2 = 0, all the lines passing through their point of intersection constitutes family of lines of L1 = 0 and L2 = 0.

L2 = 0

L1 + λ L2 = 0

L1 = 0

Family of Lines

Eg. Any line through intersection point of 2x + y + 3 = 0 and x - y + 1 = 0 will have equation of the form 2x + y + 3 + λ (x - y + 1) = 0

Family of Lines

Eg. Any line through intersection point of 2x + y + 3 = 0 and x - y + 1 = 0 will have equation of the form 2x + y + 3 + λ (x - y + 1) = 0

Eg. If equation of line is of the form x + y - 2 + λ (x + 2y) = 0, then for

sure it passes through point of intersection of x + y - 2 = 0 and

x + 2 y = 0

Family of Lines

If a, b, c are in AP then prove that the variable straight line whose equation is ax + by + c = 0 always passes through a fixed point.

Note: This is standard variety of question. So, you better remember the approach.

Solution :

Alternate Solution:

Solution :

Find the equation of a line which passes through the intersection point of the lines 3x − 4y + 6 = 0 and x + y + 2 = 0, that is farthest from the point P (2, 3).

Note: These is standard variety of question. So, you better remember the approach.

Solution :

4.1 Condition for Concurrency of three Lines

Consider three lines as follows.

L1 = a1x + b1y + c1 = 0

L2 = a2x + b2y + c2 = 0

L3 = a3x + b3y + c3 = 0

The condition for their concurrency is

Basically, it’s the condition for the existence of a common solution of the three linear equations.

Condition for Concurrency of three Lines

Check whether the lines 2x + 3y = 0 , x − 5y + 9 = 0 and 3x − 4y − 11 = 0 are concurrent or not.

Solution :

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