## Vedic Version of a Line From Two Points

Posted by mark schwartz on September 25, 2016

In *Vedic Mathematics* (revised edition, 1992) a very interesting algorithm is presented. It allows one to find the equation of a line in standard form by visually examining the values of the two points, doing a little mental calculation, and writing down the equation! One need not use the slope-intercept or the point-slope formula.

Given two points (a,b) and (c,d), the vedic version (pg. 343) is: x(b-d) – y(a-c) = bc – ad

A slight notation change gives the standard form (ax+by =c), thus (b-d)x – (a-c)y= bc – ad

For example, using the vedic version with (9,7) and (5,2) the equation is:

(7 – 2)x – (9 – 5)y = 7⦁5 – 9⦁2, giving 5x – 4y = 17.

I was curious about this because it looked familiar; basically, the difference in the y-values is the x-coefficient and the difference in the x-values is the y-coefficient. The constant is the ‘inner’ minus the ‘outer’, if you are familiar with FOIL. As I played with this, I realized that the vedic algorithm could be derived from combining the slope-intercept and the point-slope formulae. Starting with the point-slope formula, one gets:

(y – y_{1}) = m(x – x_{1}

(y – y_{1}) = ((y_{2 ̶ }y_{1})/(x_{2 } ̶ x_{1})) (x – x_{1})

(x_{2 }– x_{1}) (y – y_{1}) = (y_{2 }– y_{1})(x – x_{1})

(x_{2 }– x_{1})y – (x_{2 }– x_{1})y_{1} = (y_{2 }– y_{1})x – (y_{2 }– y_{1})x_{1}

– (y_{2 }– y_{1})x + (x_{2 }– x_{1})y = (x_{2 }– x_{1})y_{1} – (y_{2 }– y_{1})x_{1}

– (y_{2 }– y_{1})x + (x_{2 }– x_{1})y = x_{2}y_{1 }– x_{1}y_{1 }– x_{1}y_{2 }+ x_{1}y_{1}

_{ }– (y_{2 }– y_{1})x + (x_{2 }– x_{1})y = x_{2}y_{1 }– x_{1}y_{2}

_{ }(y_{2 }– y_{1})x – (x_{2 }– x_{1})y = x_{1}y_{2} – x_{2}y_{1}

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

(y_{1 }– y_{2})x – (x_{1 }– x_{2})y = x_{2}y_{1} – x_{1}y_{2}

This form (y_{1 }– y_{2})x – (x_{1 }– x_{2})y = x_{2}y_{1} – x_{1}y_{2} is the vedic form (b-d)x – (a-c)y = bc – ad.

Furthermore, this vedic form allows one to generate the equation of the line if given the slope and a point, or a point with a line perpendicular or parallel to a given line because a second point can be found from the given point and the slope.

Using the same example as above, if presented the point (9,7) and the slope 5/4, the second point is (9 + 4, 7 + 5), or (13,12), as well as (9 – 4, 7 – 5), or (5,2). Consider using this vedic version.

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