We will find the equation of a straight line passing through a given point A(x_{1}, y_{1}) and having the slope m.
Let P(x, y) be any point other than A on the given line.
Slope of the line joining A(x_{1}, y_{1}) and P(x, y) is given by
m = (y - y_{1})/(x - x_{1})
m(x - x_{1}) = (y - y_{1})
Hence, equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Example 1 :
Find the point-slope form equation of a line passing through the point (1, 2) with slope -4.
Solution :
Given : Point = (1, 2) and slope m = -4.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (1 , 2) and m = -4.
y - 2 = -4(x - 1)
Example 2 :
Find the point-slope form equation of a line passing through the point (-2, 3) with slope 1/3.
Solution :
Given : Point = (-2, 3) and slope m = 1/3.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (-2 , 3) and m = 1/3.
y - 3 = (1/3)[x - (-2)]
Example 3 :
Find the point-slope form equation of the straight line passing through the point (-5, -4) with slope 2/3.
Solution :
Given : Point = (-5, -4) and slope m = 2/3.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (-5 , -4) and m = 2/3.
y - (-4) = (2/3)[x - (-5)]
Example 4 :
Find the point-slope form equation of the straight line passing through the point (1, 2) and parallel to the line whose equation is x + 2y + 3 = 0.
Solution :
Write the equation of the line 'x + 2y + 3 = 0' in slope intercept form.
x + 2y + 3 = 0
2y = -x - 3
y = (-1/2)x - 3/2
So, slope of the given line is -1/2.
Because the required line is parallel to the given line, the slopes are equal.
Then, slope of the required line is -1/2.
The required line is passing through (1, 2) with slope -1/2.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (1, 2) and m = -1/2.
y - 2 = (-1/2)(x - 1)
Example 5 :
Find the point-slope form equation of the straight line passing through the point (-2, 3) and perpendicular to the line whose equation is x - 2y - 6 = 0.
Solution :
Write the equation of the line 'x - 2y - 6 = 0' in slope intercept form.
x - 2y - 6 = 0
-2y = -x + 6
2y = x - 6
y = (1/2)x - 3
So, slope of the given line is 1/2.
Because the required line is perpendicular to the given line, product of the slopes is equal to -1.
Let 'm' be the slope of the required line.
m x (1/2) = -1
m/2 = -1
m = -2
So, the required line is passing through (-2, 3) with slope -2.
Equation of line in point-slope form :
y - y_{1} = m(x - x_{1})
Substitute (x_{1} , y_{1}) = (-2, 3) and m = -2.
y - 3 = -2[x - (-2)]
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