Two-point Form of a Line

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 y-intercept.

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., \(\frac{Difference of y-coordinates}{difference of x-coordinates in the same order}\)

Solved example on two-point 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.

10th Grade Math

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