Equation of the line :
y = mx + b
Here m = slope and b = y-intercept
Equation of the line :
y − y_{1} = m(x − x_{1})
Here m = slope and (x_{1}, y_{1}) is the given point.
Equation of the line :
(y−y_{1})/(y_{2}−y_{1}) = (x−x_{1})/(x_{2}-x_{1})
Here (x_{1}, y_{1}) and (x_{2}, y_{2}) are the given points.
Equation of the line :
(x/a) + (y/b) = 1
Here a = x-intercept and b = y-intercept.
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.
Equation of the line :
(x − x_{1})/cos θ = (y − y_{1})/sin θ = r
Where the parameter r is the distance between (x1, y_{1}) 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 - y_{1}) = m (x - x_{1})
(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.
x_{1} = 1, y_{1} = 1
x_{2} = -2, y_{2} = 3
(y−y_{1})/(y_{2}−y_{1}) = (x−x_{1})/(x_{2}-x_{1})
(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 - y_{1}) = m (x - x_{1})
(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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