# EQUATION OF STRAIGHT LINE IN VARIOUS FORMS

## Slope (m) and y - intercept

Equation of the line :

y = mx + b

Here m = slope and b = y-intercept

## Slope (m) and point

Equation of the line :

y − y1 = m(x − x1)

Here m = slope and (x1, y1) is the given point.

## Two points (x1, y1) and (x2, y2)

Equation of the line :

(y−y1)/(y2−y1)  =  (x−x1)/(x2-x1)

Here (x1, y1) and (x2, y2) are the given points.

## x-intercept (a) and y-intercept (b)

Equation of the line :

(x/a) + (y/b)  =  1

Here a = x-intercept and b = y-intercept.

## Normal length (p) , angle (α)

Equation of the line :

x cos α + y sin α = p

If p is positive in all positions of the line and if α is always measured from x-axis in the positive direction.

## Parametric form

Equation of the line :

(x − x1)/cos θ = (y − y1)/sin θ = r

Where the parameter r is the distance between (x1, y1) and any point (x, y) on the line. This is called the symmetric form or parametric form of the line.

Question  :

Find the equation of the lines passing through the point (1,1)

(i) with y-intercept (−4)

Solution :

Equation of the line :

(x/a) + (y/b)  =  1

Since the point (1, 1) passing through the line, we may apply 1 instead of x and y respectively.

(1/a) + (1/-4) = 1

(1/a)  =  1 + (1/4)

1/a  =  5/4  ==>  a  =  4/5

x/(4/5) + y/(-4)  =  1

(5x/4) - (y/4) = 1

5x - y - 4  =  0

(ii) with slope 3

Solution :

Since we have a slope and a point passing through the line, we may use the point slope form to find the equation of the line.

(y - y1)  =  m (x - x1)

(y - 1)  =  3 (x - 1)

y - 1  =  3x - 3

y  =  3x - 2

(iii) and (-2, 3)

Solution :

Since we have two points passing through the line, we may use the two point form to find the equation of the line.

x1  =  1,  y1  =  1

x2  =  -2,  y2  =  3

(y−y1)/(y2−y1)  =  (x−x1)/(x2-x1)

(y−1)/(3−1)  =  (x+2)/(-2-1)

(y−1)/2  =  (x+2)/(-3)

-3(y - 1)  =  2(x + 2)

-3y + 3  =  2x + 4

2x + 3y + 4 - 3  =  0

2x + 3y + 1  =  0

(iv) and the perpendicular from the origin makes an angle 60 with x- axis

Solution :

m = tan θ  ==> tan 60◦  =  √3

(y - y1)  =  m (x - x1)

(y - 1)  =  √3 (x - 1)

√3x - y - (√3 - 1) = 0

Perpendicular line :

x + √3y + k  =  0

1 + √3(1) + k  =  0

k  =  - 1 - √3  ==> - (1 + √3)

x + √3y  - (1 + √3)  =  0

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