Kendriya Vidyalaya Karwar

Math’s Project

2012-2013

Welcome to my PowerPoint presentation

Topic : Straight Line

By: Atit S Gaonkar

IndexA

EQUATIONS

A GLANCE AT X MATHS

A LINESLOPE OF

INDEXA

A Glance At ‘X’ Maths

• Distance Formula :

Let P(x1, y1) & Q (x2, y2)

PQ = ( [x2-x1]2 + [y2-y1]

2 )

• Section formula :

If The Line Joining the Points P(x1, y1) & Q (x2, y2)

and in ratio m : n , the coordinates are

( [mx2+ nx1] / [m + n] , [my2+ ny1] / [m + n] )

• If m = n ; the coordinates of point are

([x2+ x1] / 2 , [y2+ y1] / 2 )

• The Area Of the Triangle whose vertices are

P(x1, y1) , Q (x2, y2) & R (x3, y3) equals

1/2 | x1(y2-y3) + x2(y3-y1) + x3(y1-y2) | .

• If the area equals zero then, the three points are

collinear.

• The Angle θ made by the line l with the positive

direction of x-axis and measured anti-clockwise, this

is called the slope or inclination.

• m = tan θ , where θ ≠ 90◦ .

Slope Of A Line

• Slope Of A Line When Any Two Points Of A Line

Are Given.

• ( [y2-y1] / [x2-x1] ).

• So, the slope of the line through the points P(x1, y1)

& Q (x2, y2) is given by

( [y2-y1] / [x2-x1] )

• Conditions For Parallelisms .

• Let we consider line l1 with it’s slope m1, and

another line l2 with it’s slope m2 .

• So, for the lines l1 & l2 to be parallel m1 should be

equal to m2 .

• i.e. m1 = m2

• Conditions For Perpendicularity .

• Let we consider line l1 with it’s slope m1, and

another line l2 with it’s slope m2 .

• So, for the lines l1 & l2 to be perpendicular the

products of the slopes should be equal to -1.

• i.e. m1 * m2 = -1

• Angle Between Two Lines.

• If we consider any line l1 passing through another

line l2 then there can be two angles :

• θ & (180 - θ ) = Φ

• tan θ = | [ m1 - m2 ] / [1 + m1m2 ] |

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Equations of straight line

• If A Line Passes Through P(x1, y1) the equation of

the line is

m = ( [y-y1] / [x-x1] )

[y-y1] = m [x-x1]

1Point

Slope Form

• Thus the point P(x1, y1) lies on the line with slope m

through the fixed points (x1, y1) if and only if, its

coordinates satisfy the equation

• [y-y1] = m [x-x1].

TwoPoint Form

• If a line passes from two points

P(x1, y1) & Q (x2, y2), then the equation of the line

passing through these points is

( [y-y1] / [x-x1] ) = [y2 -y1] / [x2-x1]

• ( [y-y1]) = ( [y2 -y1] [x-x1] / [x2-x1] )

2

Slope Intercept Form

Suppose a line l with slope m, cuts the y-axis at a

distance ‘c’ from the origin, then ‘c’ is called the

y-intercept .

• So the equation of the line will be

[y-c] = m [x-0]

y = mx + c

3

Suppose a line l with slope m, cuts the x-axis at a

distance ‘d’ from the origin, then ‘d’ is called the

x-intercept .

• So the equation of the line will be :

y = m (x – d)

Intercept form• Suppose a line makes x-intercept

‘a’ and y-intercept ‘b’ on the axes, so qbviously

the line meets at (x, 0) & (0, y). So by two point

form the equation of the line is:

x / a + y / b = 1

4

Normal Form• Suppose a non vertical line is known to us with the

following data :

Length of the perpendicular (normal) ‘p’ from the

origin to the line.

Angle θ which normal makes with the x-axis in

positive direction.

The slope of the line will be m = - ( cos θ ) / ( sin θ )

5

• So the equation of the line will be :

x cos ω + y sin ω = p.

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Equations of line

• The equation of first degree in two variables in the

form :

Ax + By + c = 0 , Where A, B & C are real

constants

1The General form

Slope Intercept Form

• y = mx + c ; So in general equation form the

equation will be :

y = (-A / B ) x – ( C / B ) , where

m = (-A / B ) & c = – ( C / B )

2

Intercept Form

x / a + y / b = 1, The equation in the general

form will be :

x / (-A / B) + y / (-C / B) = 1, where

x-intercept is (-A / B) & y-intercept is (-C / B)

3

Normal form• x cos ω + y sin ω = p , the equation in the

general form will be :

A/( A2 + B2 )1/2 x + B/( A2 + B2 ) 1/2 y = C/( A2 + B2 )

where cos ω = ± A/( A2 + B2 )1/2

sin ω = ± B/( A2 + B2 ) 1/2

c = ± C/( A2 + B2 ) 1/2

4

• The distance of a line Ax + By + C = 0

perpendicular to the point P(x1, y1) is given by ‘d’.

d = | (Ax1 + By1 + C ) / (A2 + B2 ) 1/2 |

Distance Of A Point

From A Line

• Let two lines be :

y = mx + c1 .

y = mx + c2 .

Then the perpendicular distance between the two

line is given by ‘d’.

d = | c1 – c2 | / ( 1 + m2 ) 1/2

Distance Between

two parallel Line

• Let two lines be :

Ax + By + C1 = 0

Ax + By + C2 = 0

Then the perpendicular distance between the two

line is given by ‘d’.

d = | C1 – C2 | / ( A2 + B2 ) 1/2