Straight line in Two-point Form

We will learn how to find the equation of a straight line in two-point form or the equation of the straight line through two given points.

The equation of a line passing through two points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) is y - y\(_{1}\) = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)(x - x1)

Let the two given points be (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)).

We have to find the equation of the straight line joining the above two points.

Let the given points be A (x\(_{1}\), y\(_{1}\)), B (x\(_{2}\), y\(_{2}\)) and P (x, y) be any point on the straight line joining the points A and B.

Now, the slope of the line AB is \(\frac{y_{1} - y_{2}}{x_{1} - x_{2}}\)

And the slope of the line AP is \(\frac{y - y_{1}}{x - x_{1}}\)

But the three points A, B and P are collinear.

Therefore, slope of the line AP = slope of the line AB

⇒ \(\frac{y - y_{1}}{x - x_{1}}\) = \(\frac{y_{1} - y_{2}}{x_{1} - x_{2}}\)

⇒ y - y\(_{1}\) = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) (x - x\(_{1}\))

The above equation is satisfied by the co-ordinates of any point P lying on the line AB and hence, represents the equation of the straight line AB.


Solved examples to find the equation of a straight line in two-point form:

1. Find the equation of the straight line passing through the points (2, 3) and (6, - 5).

Solution:

The equation of the straight line passing through the points (2, 3) and (6, - 5) is

 \(\frac{ y - 3}{ x + 2}\) =  \(\frac{3 + 5}{2 - 6}\),[Using the form,  \(\frac{y - y_{1}}{x - x_{1}}\) = \(\frac{y_{1} - y_{2}}{x_{1} - x_{2}}\)]

⇒ \(\frac{ y - 3}{ x + 2}\) = \(\frac{ 8}{ -4}\)

⇒ \(\frac{ y - 3}{ x + 2}\) = -2

⇒ y - 3 = -2x - 4

⇒ 2x + y + 1 = 0, which is the required equation


2. Find the equation of the straight line joining the points (- 3, 4) and (5, - 2).

Solution:

Here the given two points are (x\(_{1}\), y\(_{1}\)) = (- 3, 4) and (x\(_{2}\), y\(_{2}\)) = (5, - 2).

The equation of a line passing through two points (x\(_{1}\), y\(_{1}\)) and (x\(_{2}\), y\(_{2}\)) is y - y\(_{1}\) = [\(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\)](x - x\(_{1}\)).

So the equation of the straight line in two point form is

y - y\(_{1}\) = \(\frac{y_{2} - y_{1}}{x_{2} - x_{1}}\) (x - x\(_{1}\))

⇒ y - 4 = \(\frac{-2 - 4}{5 - (-3)}\)[x - (-3)]

⇒ y - 4 = \(\frac{ -6}{ 8}\)(x + 3)

⇒ y - 4 = \(\frac{ -3}{ 4}\)(x + 3)

⇒ 4(y - 4) = -3(x + 3)

⇒ 4y - 16 = -3x - 9

⇒ 3x + 4y - 7 = 0, which is the required equation.






11 and 12 Grade Math

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