Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be two given distinct points.
Slope of the straight line passing through these points is given by
m = (y_{2} - y_{1})/(x_{2} - x_{1})
Using point-slope form equation of a line,
(y - y_{1}) = [(y_{2} - y_{1})/(x_{2} - x_{1})](x - x_{1})
Hence, the equation of line in two-point form :
(y - y_{1})/(y_{2} - y_{1}) = (x - x_{1})/(x_{2} - x_{1})
Example 1 :
Find the general equation of a line passing through the points (-2, 1) and (4, -7).
Solution :
Given : Two points on the straight line : (-2, 1) and (4, -7).
Equation of line in two-point form :
(y - y_{1})/(y_{2} - y_{1}) = (x - x_{1})/(x_{2} - x_{1})
Substitute (x_{1} , y_{1}) = (-2, 1) and (x_{2}, y_{2}) = (4, -7).
(y - 1)/(-7 - 1) = (x + 2)/(4 + 2)
(y - 1)/(-8) = (x + 2)/6
6(y - 1) = -8(x + 2)
Distribute.
6y - 6 = -8x - 16
Simplify.
8x + 6y + 10 = 0
Divide each side by 2.
4x + 3y + 5 = 0
Example 2 :
Find the equation of a line in slope-intercept form which passing through the points (-2, 5) and (3, 6).
Solution :
Given : Two points on the straight line : (-2, 5) and (3, 6).
Equation of line in two-point form :
(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 - 1)/1 = (x + 2)/5
5(y - 1) = x + 2
Distribute.
5y - 5 = x + 2
Add 5 to each side.
5y = x + 7
Divide each side by 5.
y = x/5 + 7/5
y = (1/5)x + 7/5
Example 3 :
The vertices of a triangle ABC are A(2, 1), B(-2, 3) and C(4, 5). Find the equation of the median through the vertex A.
Solution :
Median is a straight line joining a vertex and the midpoint of the opposite side.
In ΔABC above, midpoint of BC :
= D((-2 + 4)/2, (3 + 5)/2)
= D(1, 4)
The median through A is the line joining two points A (2, 1) and D(1, 4).
Equation of the median through A :
(y - y_{1})/(y_{2} - y_{1}) = (x - x_{1})/(x_{2} - x_{1})
Substitute (x_{1} , y_{1}) = (2, 1) and (x_{2}, y_{2}) = (1, 4).
(y - 1)/(4 - 1) = (x - 2)/(1 - 2)
(y - 1)/3 = (x - 2)/(-1)
-1(y - 1) = 3(x - 1)
-y + 1 = 3x - 3
3x + y - 4 = 0
Example 4 :
Two buildings of different heights are located at opposite sides of each other. If a heavy rod is attached joining the terrace of the buildings from (6, 10) to (14, 12), find the equation of the rod joining the buildings ?
Solution :
The equation of the rod is the equation of the line passing through the two points (6, 10) and (14, 12).
Equation of the line in two-point form :
(y - y_{1})/(y_{2} - y_{1}) = (x - x_{1})/(x_{2} - x_{1})
Substitute (x_{1} , y_{1}) = (6, 10) and (x_{2}, y_{2}) = (14, 12).
(y - 10)/(12 - 10) = (x - 6)/(14 - 6)
(y - 10)/2 = (x - 6)/8
Least common multiple of the denominators 2 and 8 is 8.
Multiply each side by 8.
4(y - 10) = x - 6
4y - 40 = x - 6
x - 4y + 34 = 0
Hence, equation of the rod is x - 4y + 34 = 0.
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