Example 1 :
Find the equation of the straight line which passes through the midpoint of the line segment joining (4, 2) and (3, 1) whose angle of inclination is 30° .
Solution :
Let us draw a rough diagram based on the given information.
From the above diagram, we come to know that we have to find the line passing through the point C.
Midpoint of AB = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2
A(4, 1) B(3, 2)
= (4 + 3)/2 , (1 + 2)/2
= C (7/2, 3/2)
From the angle of inclination of the required line, we have to find the slope.
m = tan θ ==> tan 30° ==> 1/√3
Equation of the required line :
(y - y_{2}) = m(x - x_{1})
(y - (3/2)) = (1/√3) (x - (7/2))
√3 (y - (3/2)) = 1 (x - (7/2))
√3 (2y - 3) = 1 (2x - 7)
2√3 y - 3√3 = 2x - 7
2x - 2√3y - 7 + 3√3 = 0
Hence equation of the required line is 2x - 2√3y + (3√3 - 7) = 0
Example 2 :
Find the equation of the straight line passing through the points
(i) (-2, 5) and (3, 6)
Solution :
Equation of the line passing through two given points.
(y - y_{1})/(y_{2 }- y_{1}) = (x - x_{1})/(x_{2 }- x_{1})
Substitute (x_{1}, y_{1}) = (-2, 5) and (x_{2}, y_{2}) = (3, 6).
(y-5)/(6 - 5) = (x - (-2))/(3 - (-2))
(y-5)/1 = (x+2)/(3 +2)
(y-5)/1 = (x +2)/5
5(y - 5) = 1(x + 2)
5y - 25 = x + 2
x - 5y + 2 + 25 = 0
x - 5y + 27 = 0
(ii) (0, -6) and (-8, 2)
Equation of the line passing through two given points.
(y - y_{1})/(y_{2 }- y_{1}) = (x - x_{1})/(x_{2 }- x_{1})
Substitute (x_{1}, y_{1}) = (0, -6) and (x_{2}, y_{2}) = (-8, 2).
(y-(-6))/(2 - (-6)) = (x - 0)/(-8 - 0)
(y+6)/(2 + 6) = x/(-8)
(y+6)/8 = x/(-8)
(y + 6) = -x
x + y + 6 = 0
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