EQUATION OF STRAIGHT LINE IN VARIOUS FORMS

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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