If two lines are parallel, then the slopes are equal.
m_{1} = m_{2}
If two lines are perpendicular, then the product of the slopes is equal to -1.
m_{1}⋅ m_{2 }= -1
Example 1 :
Find the equation of a straight line passing through the point P(-5, 2) and parallel to the line joining the points Q(3, -2) and R(-5, 4).
Solution :
The required line is passing through the point P(-5, 2) and parallel to the line joining the points Q and R.
Since the line joining the points QR is parallel to the required line, the slopes will be equal.
Slope of QR = (y_{2} - y_{1})/(x_{2} - x_{1})
= (4 - (-2))/(-5 - 3)
= (4 + 2)/(-8)
= 6/(-8)
Slope of QR = -3/4
Slope of the required line = -3/4
Equation of the line passing through the point P.
(y - y_{1}) = m(x - x_{1})
(y - 2) = (-3/4) (x - (-5))
4(y - 2) = -3 (x + 5)
4y - 8 = -3x - 15
3x + 4y - 8 + 15 = 0
3x + 4y + 7 = 0
So, the equation of the required line is 3x + 4y + 7 = 0
Example 2 :
Find the equation of a line passing through (6, –2) and perpendicular to the line joining the points (6, 7) and (2, -3).
Solution :
The required line is perpendicular to the line joining the points (6, 7) and (2, -3).
Slope of the line joining the above points
m = (y_{2} - y_{1})/(x_{2} - x_{1})
m = (-3-7)/(2-6)
m = -10/(-4)
m = 5/2
Slope of the required line = -1/(5/2)
= -2/5
The required line is passing through the point (6, -2)
(y - y_{1}) = m(x - x_{1})
(y - (-2)) = (-2/5) (x - 6)
5(y + 2) = -2 (x - 6)
5y + 10 = -2x + 12
2x + 5y + 10 - 12 = 0
2x + 5y - 2 = 0
Example 3 :
Find the equation of the line in the xy-plane that contains the point (3, 2) and that is parallel to the line y = 4x − 1.
Solution :
Since the required line is parallel to the line y = 4x - 1,
m = 4
(y - y_{1}) = m(x - x_{1})
(y - 2) = 4(x - 3)
y - 2 = 4x - 12
4x - y - 12 + 2 = 0
4x - y - 10 = 0
Example 4 :
Find the equation of the line that contains the point (2, 3) and that is parallel to the line containing the points (7, 1) and (5, 6).
Solution :
Slope of the line containing the points (7, 1) and )(5, 6)
m = (y_{2} - y_{1})/(x_{2} - x_{1})
m = (6 - 1)/(5 - 7)
m = 4/(-2)
m = -2
Equation of the line :
(y - y_{1}) = m(x - x_{1})
(y - 3) = -2(x - 2)
y - 3 = -2x + 4
2x + y - 3 - 4 = 0
2x + y - 7 = 0
Example 5 :
Find the equation of the line in the xy-plane that contains the point (4, 1) and that is perpendicular to the line whose equation is y = 3x + 5.
Solution :
The required line is perpendicular to the line y = 3x + 5
Slope (m) = 3
Slope of required line = -1/3
Equation of the line :
(y - y_{1}) = m(x - x_{1})
Point => (4, 1) and slope m = -1/3
(y - 1) = (-1/3)(x - 4)
3(y - 1) = -1(x - 4)
3y - 3 = -x + 4
x + 3y - 3 - 4 = 0
x + 3y - 7 = 0
Example 6 :
Find a number t such that the line in the xy plane containing the points (t, 4) and (2,−1) is perpendicular to the line y = 6x − 7.
Solution :
Slope of the line containing the points (t, 4) and (2, -1)
m = (y_{2} - y_{1})/(x_{2} - x_{1})
m = (-1 - 4) / (2 - t)
m = -5/(2 - t)
Slope of the given line :
y = 6x − 7
m = 6
Slope of line perpendicular to the line y = 6x - 7 is -1/6
-1/6 = -5/(2 - t)
-1(2 - t) = -5(6)
-2 + t = -30
t = -30 + 2
t = -30
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