Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

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 – y1) = m(x – x1

(y – y1) = ((y2 ̶ y1)/(x2 ̶ x1)) (x – x1)

(x2 – x1) (y – y1) = (y2 – y1)(x – x1)

(x2 – x1)y – (x2 – x1)y1 = (y2 – y1)x – (y2 – y1)x1

– (y2 – y1)x + (x2 – x1)y = (x2 – x1)y1 – (y2 – y1)x1

– (y2 – y1)x + (x2 – x1)y = x2y1 – x1y1 – x1y2 + x1y1

 – (y2 – y1)x + (x2 – x1)y = x2y1 – x1y2

 (y2 – y1)x – (x2 – x1)y = x1y2 – x2y1

 -1(y1 – y2)x – (-1)(x1 – x2)y = (-1)(x2y1 – x1y2)

(y1 – y2)x – (x1 – x2)y = x2y1 – x1y2

This form (y1 – y2)x – (x1 – x2)y = x2y1 – x1y2 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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