We will discuss here about the method of finding the equation of a straight line in the two point form.
To find the equation of a straight line in the two point form,
Let AB be a line passing through two points A (x\(_{1}\), y\(_{1}\)) and B (x\(_{2}\), y\(_{2}\)).
Let the equation of the line be y = mx + c ................... (i), where m is the slope of the line and c is the yintercept.
As (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) are points on the line AB, (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) satisfy (i).
Therefore, y\(_{1}\) = mx\(_{1}\) + c ................................ (ii)
and y\(_{2}\) = mx\(_{2}\) + c ................................ (iii)
Subtracting (iii) from (ii),
y\(_{1}\)  y\(_{2}\) = m(x\(_{1}\)  x\(_{2}\))
⟹ m = \(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\) ................................ (iv)
Substituting m = \(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\) in (ii),
y\(_{1}\) = [\(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\)]x\(_{1}\) + c
⟹ c = y\(_{1}\)  \(\frac{x_{1}(y_{1}  y_{2})}{ x_{1}  x_{2}}\)
⟹ c = \(\frac{ y_{1}(x_{1}  x_{2})  x_{1}(y_{1}  y_{2})}{ x_{1}  x_{2}}\)
⟹ c = \(\frac{x_{1}y_{2}  x_{2}y_{1}}{ x_{1}  x_{2}}\)
Therefore, from (i),
y = [\(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\)]x + \(\frac{x_{1}y_{2}  x_{2}y_{1}}{ x_{1}  x_{2}}\)
Subtracting y\(_{1}\) from both sides of (v)
y  y\(_{1}\) = [\(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\)]x + \(\frac{x_{1}y_{2}  x_{2}y_{1}}{ x_{1}  x_{2}}\)
⟹ y  y\(_{1}\) = [\(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\)]x + \(\frac{x_{1}(y_{2}  y_{1})}{ x_{1}  x_{2}}\)
⟹ y  y\(_{1}\) = \(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\)(x + x\(_{1}\))
The equation of the straight line passing through (x1, y1) and (x2, y2) is y  y\(_{1}\) = \(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\)(x + x\(_{1}\))
Note: From (iv), the slope of the line joining the points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) is \(\frac{y_{1}  y_{2}}{x_{1}  x_{2}}\) i.e., \(\frac{Difference of ycoordinates}{difference of xcoordinates in the same order}\)
Solved example on twopoint form of a line:
The equation of the line passing through the points (1, 1) and (3, 2) is
y  1 = \(\frac{1  2}{1  (3)}\)(x  1)
⟹ y – 1 = \(\frac{1}{4}\)(x – 1)
Also, y – 2 = \(\frac{2  1}{3  1}\)(x + 3)
⟹ y – 2 = \(\frac{1}{4}\)(x + 3)
However, the two equations are the same.
From Pointslope Form of a Line to HOME
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